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Schmitt Triggers

R

RobH

Guest
On 11/02/2020 21:19, amdx wrote:
On 2/11/2020 1:59 PM, RobH wrote:
On 11/02/2020 15:55, amdx wrote:
On 2/11/2020 9:17 AM, RobH wrote:
On 11/02/2020 14:04, amdx wrote:
On 2/10/2020 10:19 AM, amdx wrote:
On 2/10/2020 10:02 AM, Evgeniy wrote:
Hello, amdx!

On Mon, 10 Feb 2020 09:39:54 -0600
amdx <nojunk@knology.net> wrote:


Ok, I boobed by putting x2 incorrect resistors. Doh!!!
I used x2 47k resistors in error and have changed them to the 10k
resistors.

   Those are just a voltage divider, setting the voltage on Pin
2. I
suspect you could you could use a resistor value up to 500K and
have
little change in the voltage on Pin 2.
    I found the input resistance of a 741 is 2Mohm, so, I did the
calculations, if you use two 10k resistors and the voltage at
Pin 2 is
4.5v, changing them to 500k will only shift that voltage by 0.5v.
It would raise Pin 2 to 5v.
   (Side no, there may be different style 741s, with different
input
impedance, I don't know.)


It's not a good idea to set huge resistance on high-impedance imput:
AMP can receive radio noise on input. 10K is enough normal
resistance, I
think. With R1 and R2 47K it should work too, but I do not prefer
such
resistance: too high in high-impedance input.

All that is possible, 500k is not huge, It would actually be 250k
to ground, but, I was trying to teach him that, he is just setting
the voltage on pin 2 with those resistors. The ratio is important
not the absolute value.



  It looks like that there
was a problem with badly connected R2, because when leds are off
voltage
on pin 3 should be below ~6V (voltage on point A), but he had:
The voltage on pin 3 is 10.2 when Lo, but does fluctuate upto
10.5v
approx, and when Hi the voltage is 10.6v, and again does
fluctuate upto
10.9v.

  I think he is just letting light leak into his ldr giving him
that fluctuation.
                                      Mikek





------
With Best Regards,
Evgeniy Shtrenyov


    Rob, I still wonder,
  Do you want the the buzzer to buzz when light shines on the ldr?
  Your previous circuit was opposite.
                                Mikek


Mmm, strange that, but yes when the light shines on the ldr.

I was confused, back on track now.


  I have the led lighting up when light shines on the ldr,

Ok, when the ldr goes dark, Pin 6 of the 741 goes low.
Does it go below 1/3rd Vcc?

Remember you need the signal to go low to trigger the Pin 2 of the 555.

but I am not sure where to connect the buzzer to. Connecting it to
pin 3 on the 555 it buzzes with a low pitch and the led lights
dimly, then when I shine light on the ldr the buzzer really buzzes
and the led is in full brightness.

Did you remember to install the differentiator, The RC between Pin 6
of the 741 and Pin 2 of the 555?
   The 555 normally has a nice 0v or Vcc output.

If you add the RC and you still have a small voltage on the 555 Pin 3
when it should be 0V, try another 555.

                                Mikek

I've got a weird situation here??? Vcc is 8.19v

When the ldr is on the positive rail and the is is in the negative
rail, the voltage on pin 6 is as follows:
when the ldr is unshaded the voltage is 1.93v
when the ldr is shaded the voltage is 2.5v

When the ldr is in the negative rail and led in the positive rail
The voltage on pin 6 is as follows:
When the ldr is unshaded the volatge is 4.97v
When the ldr has light on it, the voltage is 7.46v

 What part of that are you uncomfortable with?

Are you sure you used the same resistor in both situations?

You changed your series resistor from your first circuit to second
circuit from 100k to 2.2k. That's a big change.
You/I/we need a little better understanding of the ldr you have.
In other words, what is the resistance in complete dark and what is the
resistance with the amount of light you expect to hit it in a normal
situation. I think the can do that with your DVM. Measure it in both
light and dark.
 Give me those numbers.

Now set up a circuit Vc--33k resistor--ldr--ground.
Now give me VC, and the the ldr/resistor voltage in the dark.
and
Now give me VC, and the the ldr/resistor voltage in the expected light.

I expect after doing the calculations I will find they are the same
resistances. Just making sure there is no current related characteristic.

                                   Mikek
Just to be sure we are talking about the same circuit, it is the schmitt
trigger circuit I am using and talking about.

What part of that are you uncomfortable with?
None really, I wasn't just sure if it was right that's all.
Are you sure you used the same resistor in both situations?
Yes I am completely sure.
You changed your series resistor from your first circuit to second
circuit from 100k to 2.2k. That's a big change.
No I didn't change the resistor, as it was a 100k potentiometer as per
the circuit you sent me.

I could not measure the resistance of the ldr correctly as it was either
off scale on the meter, or a negative resistance, depending on which way
round I had the positive and common leads on it.

The Vc of the resistor voltage in the dark is 4.8v
The Vc of the resistor voltage in the light is 6.8v


Apologies for any confusion I may have caused.
 

Guest
On Wednesday, 12 February 2020 10:29:32 UTC, RobH wrote:

I could not measure the resistance of the ldr correctly as it was either
off scale on the meter, or a negative resistance, depending on which way
round I had the positive and common leads on it.

The Vc of the resistor voltage in the dark is 4.8v
The Vc of the resistor voltage in the light is 6.8v


Apologies for any confusion I may have caused.
If you're measuring an LDR as negative resistance, something is seriously wrong with what you're doing.
 
R

RobH

Guest
On 12/02/2020 11:28, tabbypurr@gmail.com wrote:
On Wednesday, 12 February 2020 10:29:32 UTC, RobH wrote:

I could not measure the resistance of the ldr correctly as it was either
off scale on the meter, or a negative resistance, depending on which way
round I had the positive and common leads on it.

The Vc of the resistor voltage in the dark is 4.8v
The Vc of the resistor voltage in the light is 6.8v


Apologies for any confusion I may have caused.

If you're measuring an LDR as negative resistance, something is seriously wrong with what you're doing.
Fair enough, and I may just give up with it then.
 
A

amdx

Guest
On 2/12/2020 6:41 AM, RobH wrote:
On 12/02/2020 11:28, tabbypurr@gmail.com wrote:
On Wednesday, 12 February 2020 10:29:32 UTC, RobH  wrote:

I could not measure the resistance of the ldr correctly as it was either
off scale on the meter, or a negative resistance, depending on which way
round I had the positive and common leads on it.
The Vc of the resistor voltage in the dark is 4.8v
The Vc of the resistor voltage in the light is 6.8v
With these measurements is the resistor a 33k ohm?



Apologies for any confusion I may have  caused.

If you're measuring an LDR as negative resistance, something is
seriously wrong with what you're doing.


Fair enough, and I may just give up with it then.
No need for that.
First make sure the ldr is out of the circuit.
We want to measure the resistance of the ldr all by it's self.

When you try to measure yours (it may be different than the graph)
You can't use your fingers, the meter will measure you (especially in
the dark)
Make sure the ldr is dark (zero light can enter) and connect the meter
in the high resistance mode, (some meters need to me set for high
resistance) try measuring a 1M resistor to be sure. Connect to the ldr
using clip leads or whatever you have so you don't have your fingers
involved.
Record the dark resistance.
Then put your light on it and measure the light resistance.
record the resistance.

What are those numbers?

Here's a page with a graph showing how the resistance of an ldr changes
with the amount of light on it. It's about 1/4 page down.

The graph shows one that when dark was 1M ohm of resistance, and when
there is a LOT of light on it, it goes down to about 100 ohms.

Mikek
 
R

RobH

Guest
On 12/02/2020 13:33, amdx wrote:
On 2/12/2020 6:41 AM, RobH wrote:
On 12/02/2020 11:28, tabbypurr@gmail.com wrote:
On Wednesday, 12 February 2020 10:29:32 UTC, RobH  wrote:

I could not measure the resistance of the ldr correctly as it was
either
off scale on the meter, or a negative resistance, depending on which
way
round I had the positive and common leads on it.



The Vc of the resistor voltage in the dark is 4.8v
The Vc of the resistor voltage in the light is 6.8v


 With these measurements is the resistor a 33k ohm?





Apologies for any confusion I may have  caused.

If you're measuring an LDR as negative resistance, something is
seriously wrong with what you're doing.


Fair enough, and I may just give up with it then.

 No need for that.
 First make sure the ldr is out of the circuit.
We want to measure the resistance of the ldr all by it's self.

 When you try to measure yours (it may be different than the graph)
You can't use your fingers, the meter will measure you (especially in
the dark)
 Make sure the ldr is dark (zero light can enter) and connect the meter
in the high resistance mode, (some meters need to me set for high
resistance) try measuring a 1M resistor to be sure. Connect to the ldr
using clip leads or whatever you have so you don't have your fingers
involved.
 Record the dark resistance.
Then put your light on it and measure the light resistance.
 record the resistance.

What are those numbers?

Here's a page with a graph showing how the resistance of an ldr changes
with the amount of light on it. It's about 1/4 page down.

  The graph shows one that when dark was 1M ohm of resistance, and when
there is a LOT of light on it, it goes down to about 100 ohms.

                                         Mikek
Yes the measurements were taken with the 33k resistor in circuit

Ok I didn't realise you meant taking the ldr out of the circuit.

The resistance of the ldr in normal daylight is about 5k ohms, and in
darkness , about 32M ohms. If I moved the ldr into a black area, then
the meter went off or out of scale.
 
A

amdx

Guest
On 2/12/2020 7:33 AM, amdx wrote:
On 2/12/2020 6:41 AM, RobH wrote:
On 12/02/2020 11:28, tabbypurr@gmail.com wrote:
On Wednesday, 12 February 2020 10:29:32 UTC, RobH  wrote:

I could not measure the resistance of the ldr correctly as it was
either
off scale on the meter, or a negative resistance, depending on which
way
round I had the positive and common leads on it.



The Vc of the resistor voltage in the dark is 4.8v
The Vc of the resistor voltage in the light is 6.8v


 With these measurements is the resistor a 33k ohm?





Apologies for any confusion I may have  caused.

If you're measuring an LDR as negative resistance, something is
seriously wrong with what you're doing.


Fair enough, and I may just give up with it then.

 No need for that.
 First make sure the ldr is out of the circuit.
We want to measure the resistance of the ldr all by it's self.

 When you try to measure yours (it may be different than the graph)
You can't use your fingers, the meter will measure you (especially in
the dark)
 Make sure the ldr is dark (zero light can enter) and connect the meter
in the high resistance mode, (some meters need to me set for high
resistance) try measuring a 1M resistor to be sure. Connect to the ldr
using clip leads or whatever you have so you don't have your fingers
involved.
 Record the dark resistance.
Then put your light on it and measure the light resistance.
 record the resistance.

What are those numbers?

Here's a page with a graph showing how the resistance of an ldr changes
with the amount of light on it. It's about 1/4 page down.

  The graph shows one that when dark was 1M ohm of resistance, and when
there is a LOT of light on it, it goes down to about 100 ohms.

                                         Mikek

Yes the measurements were taken with the 33k resistor in circuit

Ok I didn't realize you meant taking the ldr out of the circuit.

The resistance of the ldr in normal daylight is about 5k ohms, and in
darkness , about 32M ohms. If I moved the ldr into a black area, then
the meter went off or out of scale.
So know that we know the resistance in the light is 5k, we know what
voltage different values of series resistor will cause.
This it the circuit, VCC--resistor--ldr--grd
If you use a 5K series resistor the Vcc will be split in half, (in
the dark). Let's assume 9v Vcc, so 4.5v. If you raise the resistor value
to 10k ohms, the voltage will drop to 3.3v, 15k and the voltage will
drop to 2.25k, 20k and the voltage will drop to 1.8v. if you use 100k
the voltage will drop to 0.43v.
When you go to dark the ldr resistance will increase and these
voltages i just mentioned will increase.
I suggest you build that ldr resistor circuit and measure the voltages
in light and dark. You should see about 8.9V and 0.5v. Remember the
light needs to be the same as when you measured the ldr with you meter.

Here is a voltage divider calculator to show you how the differing
resistances will change the voltage. Put in your Vcc. Set R1 value as
your whatever your series resistor is, and R2 as the resistance of your
ldr, (in this case 5k ohms in the light.) Change values to see how it works.
I'm sorry if this all takes you the long route to making the thing
work, but I think you should understand how it works so you can optimize
to your lighting conditions.
Mikek
Mikek
 
R

RobH

Guest
On 12/02/2020 15:20, amdx wrote:
On 2/12/2020 7:33 AM, amdx wrote:
On 2/12/2020 6:41 AM, RobH wrote:
On 12/02/2020 11:28, tabbypurr@gmail.com wrote:
On Wednesday, 12 February 2020 10:29:32 UTC, RobH  wrote:

I could not measure the resistance of the ldr correctly as it was
either
off scale on the meter, or a negative resistance, depending on
which way
round I had the positive and common leads on it.



The Vc of the resistor voltage in the dark is 4.8v
The Vc of the resistor voltage in the light is 6.8v


  With these measurements is the resistor a 33k ohm?





Apologies for any confusion I may have  caused.

If you're measuring an LDR as negative resistance, something is
seriously wrong with what you're doing.


Fair enough, and I may just give up with it then.

  No need for that.
  First make sure the ldr is out of the circuit.
We want to measure the resistance of the ldr all by it's self.

  When you try to measure yours (it may be different than the graph)
You can't use your fingers, the meter will measure you (especially in
the dark)
  Make sure the ldr is dark (zero light can enter) and connect the meter
in the high resistance mode, (some meters need to me set for high
resistance) try measuring a 1M resistor to be sure. Connect to the ldr
using clip leads or whatever you have so you don't have your fingers
involved.
  Record the dark resistance.
Then put your light on it and measure the light resistance.
  record the resistance.

What are those numbers?

Here's a page with a graph showing how the resistance of an ldr
changes with the amount of light on it. It's about 1/4 page down.

   The graph shows one that when dark was 1M ohm of resistance, and
when there is a LOT of light on it, it goes down to about 100 ohms.

                                          Mikek

Yes the measurements were taken with the 33k resistor in circuit

Ok I didn't realize you meant taking the ldr out of the circuit.

The resistance of the ldr in normal daylight is about 5k ohms, and in
darkness , about 32M ohms. If I moved the ldr into a black area, then
the meter went off or out of scale.

So know that we know the resistance in the light is 5k, we know what
voltage different values of series resistor will cause.
This it the circuit, VCC--resistor--ldr--grd
  If you use a 5K series resistor the Vcc will be split in half, (in
the dark). Let's assume 9v Vcc, so 4.5v. If you raise the resistor value
to 10k ohms, the voltage will drop to 3.3v, 15k and the voltage will
drop to 2.25k, 20k and the voltage will drop to 1.8v. if you use 100k
the voltage will drop to 0.43v.
 When you go to dark the ldr resistance will increase and these
voltages i just mentioned will increase.
 I suggest you build that ldr resistor circuit and measure the voltages
in light and dark. You should see about 8.9V and 0.5v. Remember the
light needs to be the same as when you measured the ldr with you meter.

Here is a voltage divider calculator to show you how the differing
resistances will change the voltage. Put in your Vcc. Set R1 value as
your whatever your series resistor is, and R2 as the resistance of your
ldr, (in this case 5k ohms in the light.) Change values to see how it
works.
 I'm sorry if this all takes you the long route to making the thing
work, but I think you should understand how it works so you can optimize
to your lighting conditions.
                             Mikek
                             Mikek
There was no page which you referred to in the previous that I could see
any link to, and there is no link to a voalage divider page if that is
what you mean.

All I want to do now is add a buzzer which will work when the led comes on.

Thanks
 
A

amdx

Guest
On 2/12/2020 9:29 AM, RobH wrote:
On 12/02/2020 15:20, amdx wrote:
On 2/12/2020 7:33 AM, amdx wrote:
On 2/12/2020 6:41 AM, RobH wrote:
On 12/02/2020 11:28, tabbypurr@gmail.com wrote:
On Wednesday, 12 February 2020 10:29:32 UTC, RobH  wrote:

I could not measure the resistance of the ldr correctly as it was
either
off scale on the meter, or a negative resistance, depending on
which way
round I had the positive and common leads on it.



The Vc of the resistor voltage in the dark is 4.8v
The Vc of the resistor voltage in the light is 6.8v


  With these measurements is the resistor a 33k ohm?





Apologies for any confusion I may have  caused.

If you're measuring an LDR as negative resistance, something is
seriously wrong with what you're doing.


Fair enough, and I may just give up with it then.

  No need for that.
  First make sure the ldr is out of the circuit.
We want to measure the resistance of the ldr all by it's self.

  When you try to measure yours (it may be different than the graph)
You can't use your fingers, the meter will measure you (especially in
the dark)
  Make sure the ldr is dark (zero light can enter) and connect the meter
in the high resistance mode, (some meters need to me set for high
resistance) try measuring a 1M resistor to be sure. Connect to the
ldr using clip leads or whatever you have so you don't have your
fingers involved.
  Record the dark resistance.
Then put your light on it and measure the light resistance.
  record the resistance.

What are those numbers?

Here's a page with a graph showing how the resistance of an ldr
changes with the amount of light on it. It's about 1/4 page down.

   The graph shows one that when dark was 1M ohm of resistance, and
when there is a LOT of light on it, it goes down to about 100 ohms.

                                          Mikek

 >Yes the measurements were taken with the 33k resistor in circuit
 
 >Ok I didn't realize you meant taking the ldr out of the circuit.
 
 >The resistance of the ldr in normal daylight is about 5k ohms, and in
 >darkness , about 32M ohms. If I moved the ldr into a black area, then
 >the meter went off or out of scale.

So know that we know the resistance in the light is 5k, we know what
voltage different values of series resistor will cause.
This it the circuit, VCC--resistor--ldr--grd
   If you use a 5K series resistor the Vcc will be split in half, (in
the dark). Let's assume 9v Vcc, so 4.5v. If you raise the resistor
value to 10k ohms, the voltage will drop to 3.3v, 15k and the voltage
will drop to 2.25k, 20k and the voltage will drop to 1.8v. if you use
100k the voltage will drop to 0.43v.
  When you go to dark the ldr resistance will increase and these
voltages i just mentioned will increase.
  I suggest you build that ldr resistor circuit and measure the
voltages in light and dark. You should see about 8.9V and 0.5v.
Remember the light needs to be the same as when you measured the ldr
with you meter.

Here is a voltage divider calculator to show you how the differing
resistances will change the voltage. Put in your Vcc. Set R1 value as
your whatever your series resistor is, and R2 as the resistance of
your ldr, (in this case 5k ohms in the light.) Change values to see
how it works.
  I'm sorry if this all takes you the long route to making the thing
work, but I think you should understand how it works so you can
optimize to your lighting conditions.
                              Mikek
                              Mikek


There was no page which you referred to in the previous that I could see
any link to, and there is no link to a voalage divider page if that is
what you mean.
Sorry forgot to put it in.
http://www.ohmslawcalculator.com/voltage-divider-calculator
All I want to do now is add a buzzer which will work when the led comes on.

Thanks
Does adding the 555 circuit work for you?

Remember at one time you had the 555 setup so that touching a grounded
wire to Pin 2 of the 555, it would make the buzzer buzz for 8 seconds
and then it would stop.
So you need your Resistor/ldr circuit feeding your 741 schmitt
trigger, feeding an RC circuit driving Pin 2 of the 555 and Pin 3 of the
555 driving your buzzer.

Mikek
 
R

RobH

Guest
On 12/02/2020 18:10, amdx wrote:
On 2/12/2020 9:29 AM, RobH wrote:
On 12/02/2020 15:20, amdx wrote:
On 2/12/2020 7:33 AM, amdx wrote:
On 2/12/2020 6:41 AM, RobH wrote:
On 12/02/2020 11:28, tabbypurr@gmail.com wrote:
On Wednesday, 12 February 2020 10:29:32 UTC, RobH  wrote:

I could not measure the resistance of the ldr correctly as it was
either
off scale on the meter, or a negative resistance, depending on
which way
round I had the positive and common leads on it.



The Vc of the resistor voltage in the dark is 4.8v
The Vc of the resistor voltage in the light is 6.8v


  With these measurements is the resistor a 33k ohm?





Apologies for any confusion I may have  caused.

If you're measuring an LDR as negative resistance, something is
seriously wrong with what you're doing.


Fair enough, and I may just give up with it then.

  No need for that.
  First make sure the ldr is out of the circuit.
We want to measure the resistance of the ldr all by it's self.

  When you try to measure yours (it may be different than the graph)
You can't use your fingers, the meter will measure you (especially
in the dark)
  Make sure the ldr is dark (zero light can enter) and connect the
meter
in the high resistance mode, (some meters need to me set for high
resistance) try measuring a 1M resistor to be sure. Connect to the
ldr using clip leads or whatever you have so you don't have your
fingers involved.
  Record the dark resistance.
Then put your light on it and measure the light resistance.
  record the resistance.

What are those numbers?

Here's a page with a graph showing how the resistance of an ldr
changes with the amount of light on it. It's about 1/4 page down.

   The graph shows one that when dark was 1M ohm of resistance, and
when there is a LOT of light on it, it goes down to about 100 ohms.

                                          Mikek

 >Yes the measurements were taken with the 33k resistor in circuit
 
 >Ok I didn't realize you meant taking the ldr out of the circuit.
 
 >The resistance of the ldr in normal daylight is about 5k ohms, and in
 >darkness , about 32M ohms. If I moved the ldr into a black area, then
 >the meter went off or out of scale.

So know that we know the resistance in the light is 5k, we know what
voltage different values of series resistor will cause.
This it the circuit, VCC--resistor--ldr--grd
   If you use a 5K series resistor the Vcc will be split in half, (in
the dark). Let's assume 9v Vcc, so 4.5v. If you raise the resistor
value to 10k ohms, the voltage will drop to 3.3v, 15k and the voltage
will drop to 2.25k, 20k and the voltage will drop to 1.8v. if you use
100k the voltage will drop to 0.43v.
  When you go to dark the ldr resistance will increase and these
voltages i just mentioned will increase.
  I suggest you build that ldr resistor circuit and measure the
voltages in light and dark. You should see about 8.9V and 0.5v.
Remember the light needs to be the same as when you measured the ldr
with you meter.

Here is a voltage divider calculator to show you how the differing
resistances will change the voltage. Put in your Vcc. Set R1 value as
your whatever your series resistor is, and R2 as the resistance of
your ldr, (in this case 5k ohms in the light.) Change values to see
how it works.
  I'm sorry if this all takes you the long route to making the thing
work, but I think you should understand how it works so you can
optimize to your lighting conditions.
                              Mikek
                              Mikek


There was no page which you referred to in the previous that I could
see any link to, and there is no link to a voalage divider page if
that is what you mean.


 Sorry forgot to put it in.
http://www.ohmslawcalculator.com/voltage-divider-calculator


All I want to do now is add a buzzer which will work when the led
comes on.

Thanks

 Does adding the 555 circuit work for you?

Remember at one time you had the 555 setup so that touching a grounded
wire to Pin 2 of the 555, it would make the buzzer buzz for 8 seconds
and then it would stop.
 So you need your Resistor/ldr circuit feeding your 741 schmitt
trigger, feeding an RC circuit driving Pin 2 of the 555 and Pin 3 of the
555 driving your buzzer.

                              Mikek
On that schmitt trigger circuit you sent me there is a 555 ic in the
circuit.

When I put the live side of the buzzer into pin 3 and the negative side
to ground it buzzes continually, choose whether led is lit or not.
What is an RC circuit.

There is a wire from pin 6 of the 741 to pin 2 of the 555 via a 0.01uf
capacitor and x2 100k resistors voltage divider.
With a source voltage of 8.06v the voltage at pin 2 is 7.19v, and at pin
3 of the 555 it is 7.40v
 
A

amdx

Guest
On 2/12/2020 12:48 PM, RobH wrote:
On 12/02/2020 18:10, amdx wrote:
On 2/12/2020 9:29 AM, RobH wrote:
On 12/02/2020 15:20, amdx wrote:
On 2/12/2020 7:33 AM, amdx wrote:
On 2/12/2020 6:41 AM, RobH wrote:
On 12/02/2020 11:28, tabbypurr@gmail.com wrote:
On Wednesday, 12 February 2020 10:29:32 UTC, RobH  wrote:

I could not measure the resistance of the ldr correctly as it
was either
off scale on the meter, or a negative resistance, depending on
which way
round I had the positive and common leads on it.



The Vc of the resistor voltage in the dark is 4.8v
The Vc of the resistor voltage in the light is 6.8v


  With these measurements is the resistor a 33k ohm?





Apologies for any confusion I may have  caused.

If you're measuring an LDR as negative resistance, something is
seriously wrong with what you're doing.


Fair enough, and I may just give up with it then.

  No need for that.
  First make sure the ldr is out of the circuit.
We want to measure the resistance of the ldr all by it's self.

  When you try to measure yours (it may be different than the graph)
You can't use your fingers, the meter will measure you (especially
in the dark)
  Make sure the ldr is dark (zero light can enter) and connect the
meter
in the high resistance mode, (some meters need to me set for high
resistance) try measuring a 1M resistor to be sure. Connect to the
ldr using clip leads or whatever you have so you don't have your
fingers involved.
  Record the dark resistance.
Then put your light on it and measure the light resistance.
  record the resistance.

What are those numbers?

Here's a page with a graph showing how the resistance of an ldr
changes with the amount of light on it. It's about 1/4 page down.

   The graph shows one that when dark was 1M ohm of resistance, and
when there is a LOT of light on it, it goes down to about 100 ohms.

                                          Mikek

 >Yes the measurements were taken with the 33k resistor in circuit
 
 >Ok I didn't realize you meant taking the ldr out of the circuit.
 
 >The resistance of the ldr in normal daylight is about 5k ohms, and in
 >darkness , about 32M ohms. If I moved the ldr into a black area, then
 >the meter went off or out of scale.

So know that we know the resistance in the light is 5k, we know what
voltage different values of series resistor will cause.
This it the circuit, VCC--resistor--ldr--grd
   If you use a 5K series resistor the Vcc will be split in half,
(in the dark). Let's assume 9v Vcc, so 4.5v. If you raise the
resistor value to 10k ohms, the voltage will drop to 3.3v, 15k and
the voltage will drop to 2.25k, 20k and the voltage will drop to
1.8v. if you use 100k the voltage will drop to 0.43v.
  When you go to dark the ldr resistance will increase and these
voltages i just mentioned will increase.
  I suggest you build that ldr resistor circuit and measure the
voltages in light and dark. You should see about 8.9V and 0.5v.
Remember the light needs to be the same as when you measured the ldr
with you meter.

Here is a voltage divider calculator to show you how the differing
resistances will change the voltage. Put in your Vcc. Set R1 value
as your whatever your series resistor is, and R2 as the resistance
of your ldr, (in this case 5k ohms in the light.) Change values to
see how it works.
  I'm sorry if this all takes you the long route to making the thing
work, but I think you should understand how it works so you can
optimize to your lighting conditions.
                              Mikek
                              Mikek


There was no page which you referred to in the previous that I could
see any link to, and there is no link to a voalage divider page if
that is what you mean.


  Sorry forgot to put it in.
 > http://www.ohmslawcalculator.com/voltage-divider-calculator


All I want to do now is add a buzzer which will work when the led
comes on.

Thanks

  Does adding the 555 circuit work for you?

Remember at one time you had the 555 setup so that touching a grounded
wire to Pin 2 of the 555, it would make the buzzer buzz for 8 seconds
and then it would stop.
  So you need your Resistor/ldr circuit feeding your 741 schmitt
trigger, feeding an RC circuit driving Pin 2 of the 555 and Pin 3 of
the 555 driving your buzzer.

                               Mikek





On that schmitt trigger circuit you sent me there is a 555 ic in the
circuit.

When I put the live side of the buzzer into pin 3 and the negative side
to ground it buzzes continually, choose whether led is lit or not.
What is an RC circuit.

Is pin 3 actually switching low to high when your circuit triggers it?

I have a feeling it's not.




There is a wire from pin 6 of the 741 to pin 2 of the 555 via a 0.01uf
capacitor and x2 100k resistors voltage divider.
With a source voltage of 8.06v the voltage at pin 2 is 7.19v, and at pin
3 of the 555 it is 7.40v
If you have that circuit wired correctly Pin 2 should set at 1/2 Vcc, or
4.0V. other wise you have something improperly wired.
I have emailed a schematic to you.
Mikek
 
N

Nife Sima

Guest
On 10.02.2020 19:38, amdx wrote:
On 2/10/2020 11:15 AM, RobH wrote:
On 10/02/2020 16:13, amdx wrote:
On 2/10/2020 9:51 AM, RobH wrote:
On 10/02/2020 15:39, amdx wrote:
On 2/10/2020 8:01 AM, RobH wrote:
On 10/02/2020 12:04, Evgeniy wrote:
Hello, RobH!

On Mon, 10 Feb 2020 09:26:22 +0000
RobH <rob@despammer.com> wrote:

On 10/02/2020 00:38, amdx wrote:
On 2/9/2020 5:11 PM, RobH wrote:
I have been looking  at schmitt trigger circuits and didn't
realise
how many variations there are.

Anyway I found this one:
https://reviseomatic.org/help/2-radio/Schmitt%20Trigger%20Light%20Sensor.php



The author here says that by adding a 22k resistor to the bottom
diagram turns into a schmitt trigger circuit as the top diagram.

  From other schmitt trigger circuits I have see on youtube
and other
sites, this one is so simple, but works. The switching from
off to on
is almost instant for the 2 leds.

   It's faster than the persistence of your eye!
But I have a question here, it says this:

Two LEDs are needed because, when the Op Amp pin 6 voltage is
low,
there may still be as much as two volts present.
This could be enough to light a single LED. With two LEDs,
four volts
are needed.
The Op Amp pin 6 voltage will be well below 4V so the LEDs
will be off.

   What are the high and low voltages you get on Pin 6?

What actually happens is both leds come on at the same time
and the
voltage is 2.18v when on.

   Usually they will come on at very close to same time, there
will be a
slight difference of the voltage across them, but they will be
pretty
similar.
If you are using a common red LED they will have a voltage drop
of about
1.8v, so 3.6V for to with the rest dropped across your 560 ohm
resistor.
   The voltage drop is depending on the current through the
LED, but the
1.8V for a red LED is close.

I am using a 33k resistor instead of a 22k which I don't have,
if that
makes the difference.

   You tell us, record the voltage on Pin 3 and Pin 6 when Pin 6
switches, both high and low. Then put another 33k in parallel
with the
one you have and repeat the voltage records as above. Then you
can tell
me what difference changing the value of the feedback (your 33k)
resistor causes.
   I like this you are one upping me and removing even more parts.
You have been thinking.
                                              Mikek

The voltage on pin 6 , as the circuit is now, or per the
diagram/schematic, is 2.0v when Lo, and 4.0v when Hi

The voltage on pin 3 is 10.2 when Lo, but does fluctuate upto 10.5v
approx, and when Hi the voltage is 10.6v, and again does
fluctuate upto
10.9v.

Another 33k resistor was placed in parallel with the one across
pins 3
and 6. The Lo voltage was 2.0v and the Hi voltage was 3.2v on
pin 6.
On pin 3, the voltage when Lo was 10.2v and when Hi the voltage was
10.2v, and both leds were flickering rapidly.

Just want to say some words. Schematic is strange, and description
is strange too. Author said:
"Adding the 22K resistor adds some positive feedback and the on
and off
threshold voltages will now differ slightly. Adding this resistor
converts the circuit from a Comparator to a Schmitt Trigger."

That's not true, because threshold voltage will be the same: it
will be
half of power voltage, value at point A. Feedback does not change
threshold voltage, it just changes voltage/current through LDR, so
behaviour of schematic depends on Volt-Ampere characteristics of
LDR. If
characteristic is like classical resistor (resistance does not
depend
on voltage/current, just light; I have not work with LDR), it
will make
LDR resistance thresholds, which makes lighting thresholds.

The voltage on pin 3 is 10.2 when Lo, but does fluctuate upto 10.5v
approx, and when Hi the voltage is 10.6v, and again does
fluctuate upto
10.9v.

There is something wrong: with schematic on board or OP AMP or while
measuring. Check voltage at point A, directly on pin 2, may be R2 is
not connected properly. Voltage on pin 2 should be ~6V. Schematic
should
turn turn on/off leds when Voltage on point B is near voltage on
point
A. You may measure resistance of LDR when it is "in dark" and
when it
is "in light". Do not forget to remove it from schematic for
measuring.

In addition there is a mistake on the schematic of board: there
is no
22k resistor.


------
With Best Regards,
Evgeniy



Ok, I boobed by putting x2 incorrect resistors. Doh!!!
I used x2 47k resistors in error and have changed them to the 10k
resistors.

  Those are just a voltage divider, setting the voltage on Pin 2. I
suspect you could you could use a resistor value up to 500K and
have little change in the voltage on Pin 2.
   I found the input resistance of a 741 is 2Mohm, so, I did the
calculations, if you use two 10k resistors and the voltage at Pin 2
is 4.5v, changing them to 500k will only shift that voltage by 0.5v.
It would raise Pin 2 to 5v.
  (Side no, there may be different style 741s, with different input
impedance, I don't know.)



At pin 2 the voltage is 6.1v.
The voltage at pin 6 is 10.5v when both leds are on using x2 10k
resistors.Putting light onto the ldr turns the leds off and the
voltage at pin 6 is 1.98v
Where it should be x2 10k ones I used a 10k potentiometer, and
gives the same results

  Do you need to know how to shift the output level ( 6.1v to
1.98v) so you can put a buzzer on the output?
                        Mikek


Err, yes I would like to know that.

So would I! I might, from what you said below, it seems you need to
voltage shift and invert, I'm not sure that can be done with just one
transistor. ( but you know with what you have you could trigger a 555
and have clean 0 to 12v swing. (I haven't thought if it needs an
inversion.) But the 555 does do your level shift.

When the 12v is applied the leds are both on, and go off when light
is shone on to the ldr, so I'd like the buzzer to sound when the
leds go off.

  Let me restate that as a question. Do you want the the buzzer to
buzz when light shines on the ldr?
  Your previous circuit was opposite.

Thanks

Oh, you are now using 12V? That shifts the numbers for that voltage
divider up to 6v and without redoing my calculation, maybe to 6.7
with the 500k.
                               Mikek

I was only  using a 12v supply because that was what was shown on the
circuit on the website. Actually, I have now used a 9v battery supply
and the readings are now different.

On Pin 2 the voltage is 6.63v

If you are using two 10k resistors, that should split your supply
voltage in half, or 4.5V. Are you sure it's Pin 2 and are you sure you
have 2- 10k resistors. Also what it the exact part number and
manufacturer of the 741.

On Pin 6 the voltage is 7.36v.
As before, putting light onto the ldr to make the leds go ff, the
voltage is 1.9v


  Don't know why you got more output range with lower Vc.


Yes I would want the buzzer to buzz when light shines onto the ldr.

 I would just use the 555, but If if I took the time I think I could
make it work with two transistors. But I would need to experiment, not
design.
                                    Mikek




Thanks
I don't think there's need to overcomplicate this, there are buzzers
available that emit a solid tone when powered (with DC). Just go to your
local electronics shop and ask for one that emits a tone.

Cheers, Chris.
 
A

amdx

Guest
On 2/16/2020 9:16 AM, Nife Sima wrote:
On 10.02.2020 19:38, amdx wrote:
On 2/10/2020 11:15 AM, RobH wrote:
On 10/02/2020 16:13, amdx wrote:
On 2/10/2020 9:51 AM, RobH wrote:
On 10/02/2020 15:39, amdx wrote:
On 2/10/2020 8:01 AM, RobH wrote:
On 10/02/2020 12:04, Evgeniy wrote:
Hello, RobH!

On Mon, 10 Feb 2020 09:26:22 +0000
RobH <rob@despammer.com> wrote:

On 10/02/2020 00:38, amdx wrote:
On 2/9/2020 5:11 PM, RobH wrote:
I have been looking  at schmitt trigger circuits and didn't
realise
how many variations there are.

Anyway I found this one:
https://reviseomatic.org/help/2-radio/Schmitt%20Trigger%20Light%20Sensor.php



The author here says that by adding a 22k resistor to the bottom
diagram turns into a schmitt trigger circuit as the top diagram.

  From other schmitt trigger circuits I have see on youtube
and other
sites, this one is so simple, but works. The switching from
off to on
is almost instant for the 2 leds.

   It's faster than the persistence of your eye!
But I have a question here, it says this:

Two LEDs are needed because, when the Op Amp pin 6 voltage is
low,
there may still be as much as two volts present.
This could be enough to light a single LED. With two LEDs,
four volts
are needed.
The Op Amp pin 6 voltage will be well below 4V so the LEDs
will be off.

   What are the high and low voltages you get on Pin 6?

What actually happens is both leds come on at the same time
and the
voltage is 2.18v when on.

   Usually they will come on at very close to same time, there
will be a
slight difference of the voltage across them, but they will be
pretty
similar.
If you are using a common red LED they will have a voltage
drop of about
1.8v, so 3.6V for to with the rest dropped across your 560 ohm
resistor.
   The voltage drop is depending on the current through the
LED, but the
1.8V for a red LED is close.

I am using a 33k resistor instead of a 22k which I don't
have, if that
makes the difference.

   You tell us, record the voltage on Pin 3 and Pin 6 when Pin 6
switches, both high and low. Then put another 33k in parallel
with the
one you have and repeat the voltage records as above. Then you
can tell
me what difference changing the value of the feedback (your 33k)
resistor causes.
   I like this you are one upping me and removing even more
parts.
You have been thinking.
                                              Mikek

The voltage on pin 6 , as the circuit is now, or per the
diagram/schematic, is 2.0v when Lo, and 4.0v when Hi

The voltage on pin 3 is 10.2 when Lo, but does fluctuate upto
10.5v
approx, and when Hi the voltage is 10.6v, and again does
fluctuate upto
10.9v.

Another 33k resistor was placed in parallel with the one across
pins 3
and 6. The Lo voltage was 2.0v and the Hi voltage was 3.2v on
pin 6.
On pin 3, the voltage when Lo was 10.2v and when Hi the voltage
was
10.2v, and both leds were flickering rapidly.

Just want to say some words. Schematic is strange, and description
is strange too. Author said:
"Adding the 22K resistor adds some positive feedback and the on
and off
threshold voltages will now differ slightly. Adding this resistor
converts the circuit from a Comparator to a Schmitt Trigger."

That's not true, because threshold voltage will be the same: it
will be
half of power voltage, value at point A. Feedback does not change
threshold voltage, it just changes voltage/current through LDR, so
behaviour of schematic depends on Volt-Ampere characteristics of
LDR. If
characteristic is like classical resistor (resistance does not
depend
on voltage/current, just light; I have not work with LDR), it
will make
LDR resistance thresholds, which makes lighting thresholds.

The voltage on pin 3 is 10.2 when Lo, but does fluctuate upto
10.5v
approx, and when Hi the voltage is 10.6v, and again does
fluctuate upto
10.9v.

There is something wrong: with schematic on board or OP AMP or
while
measuring. Check voltage at point A, directly on pin 2, may be
R2 is
not connected properly. Voltage on pin 2 should be ~6V.
Schematic should
turn turn on/off leds when Voltage on point B is near voltage on
point
A. You may measure resistance of LDR when it is "in dark" and
when it
is "in light". Do not forget to remove it from schematic for
measuring.

In addition there is a mistake on the schematic of board: there
is no
22k resistor.


------
With Best Regards,
Evgeniy



Ok, I boobed by putting x2 incorrect resistors. Doh!!!
I used x2 47k resistors in error and have changed them to the 10k
resistors.

  Those are just a voltage divider, setting the voltage on Pin 2.
I suspect you could you could use a resistor value up to 500K and
have little change in the voltage on Pin 2.
   I found the input resistance of a 741 is 2Mohm, so, I did the
calculations, if you use two 10k resistors and the voltage at Pin
2 is 4.5v, changing them to 500k will only shift that voltage by
0.5v.
It would raise Pin 2 to 5v.
  (Side no, there may be different style 741s, with different
input impedance, I don't know.)



At pin 2 the voltage is 6.1v.
The voltage at pin 6 is 10.5v when both leds are on using x2 10k
resistors.Putting light onto the ldr turns the leds off and the
voltage at pin 6 is 1.98v
Where it should be x2 10k ones I used a 10k potentiometer, and
gives the same results

  Do you need to know how to shift the output level ( 6.1v to
1.98v) so you can put a buzzer on the output?
                        Mikek


Err, yes I would like to know that.

So would I! I might, from what you said below, it seems you need to
voltage shift and invert, I'm not sure that can be done with just
one transistor. ( but you know with what you have you could trigger
a 555 and have clean 0 to 12v swing. (I haven't thought if it needs
an inversion.) But the 555 does do your level shift.

When the 12v is applied the leds are both on, and go off when light
is shone on to the ldr, so I'd like the buzzer to sound when the
leds go off.

  Let me restate that as a question. Do you want the the buzzer to
buzz when light shines on the ldr?
  Your previous circuit was opposite.

Thanks

Oh, you are now using 12V? That shifts the numbers for that voltage
divider up to 6v and without redoing my calculation, maybe to 6.7
with the 500k.
                               Mikek

I was only  using a 12v supply because that was what was shown on the
circuit on the website. Actually, I have now used a 9v battery supply
and the readings are now different.

On Pin 2 the voltage is 6.63v

If you are using two 10k resistors, that should split your supply
voltage in half, or 4.5V. Are you sure it's Pin 2 and are you sure you
have 2- 10k resistors. Also what it the exact part number and
manufacturer of the 741.

On Pin 6 the voltage is 7.36v.
As before, putting light onto the ldr to make the leds go ff, the
voltage is 1.9v


   Don't know why you got more output range with lower Vc.


Yes I would want the buzzer to buzz when light shines onto the ldr.

  I would just use the 555, but If if I took the time I think I could
make it work with two transistors. But I would need to experiment, not
design.
                                     Mikek




Thanks

I don't think there's need to overcomplicate this, there are buzzers
available that emit a solid tone when powered (with DC). Just go to your
local electronics shop and ask for one that emits a tone.

Cheers, Chris
He already is using a buzzer, what he needs is a level shifter, I
advised that a555 will do that, but could also be done with two
transistors. But, not be me.
Mikek
 
A

amdx

Guest
On 2/12/2020 1:44 PM, amdx wrote:
On 2/12/2020 12:48 PM, RobH wrote:
On 12/02/2020 18:10, amdx wrote:
On 2/12/2020 9:29 AM, RobH wrote:
On 12/02/2020 15:20, amdx wrote:
On 2/12/2020 7:33 AM, amdx wrote:
On 2/12/2020 6:41 AM, RobH wrote:
On 12/02/2020 11:28, tabbypurr@gmail.com wrote:
On Wednesday, 12 February 2020 10:29:32 UTC, RobH  wrote:

I could not measure the resistance of the ldr correctly as it
was either
off scale on the meter, or a negative resistance, depending on
which way
round I had the positive and common leads on it.



The Vc of the resistor voltage in the dark is 4.8v
The Vc of the resistor voltage in the light is 6.8v


  With these measurements is the resistor a 33k ohm?





Apologies for any confusion I may have  caused.

If you're measuring an LDR as negative resistance, something is
seriously wrong with what you're doing.


Fair enough, and I may just give up with it then.

  No need for that.
  First make sure the ldr is out of the circuit.
We want to measure the resistance of the ldr all by it's self.

  When you try to measure yours (it may be different than the graph)
You can't use your fingers, the meter will measure you (especially
in the dark)
  Make sure the ldr is dark (zero light can enter) and connect the
meter
in the high resistance mode, (some meters need to me set for high
resistance) try measuring a 1M resistor to be sure. Connect to the
ldr using clip leads or whatever you have so you don't have your
fingers involved.
  Record the dark resistance.
Then put your light on it and measure the light resistance.
  record the resistance.

What are those numbers?

Here's a page with a graph showing how the resistance of an ldr
changes with the amount of light on it. It's about 1/4 page down.

   The graph shows one that when dark was 1M ohm of resistance,
and when there is a LOT of light on it, it goes down to about 100
ohms.

                                          Mikek

 >Yes the measurements were taken with the 33k resistor in circuit
 
 >Ok I didn't realize you meant taking the ldr out of the circuit.
 
 >The resistance of the ldr in normal daylight is about 5k ohms,
and in
 >darkness , about 32M ohms. If I moved the ldr into a black area,
then
 >the meter went off or out of scale.

So know that we know the resistance in the light is 5k, we know
what voltage different values of series resistor will cause.
This it the circuit, VCC--resistor--ldr--grd
   If you use a 5K series resistor the Vcc will be split in half,
(in the dark). Let's assume 9v Vcc, so 4.5v. If you raise the
resistor value to 10k ohms, the voltage will drop to 3.3v, 15k and
the voltage will drop to 2.25k, 20k and the voltage will drop to
1.8v. if you use 100k the voltage will drop to 0.43v.
  When you go to dark the ldr resistance will increase and these
voltages i just mentioned will increase.
  I suggest you build that ldr resistor circuit and measure the
voltages in light and dark. You should see about 8.9V and 0.5v.
Remember the light needs to be the same as when you measured the
ldr with you meter.

Here is a voltage divider calculator to show you how the differing
resistances will change the voltage. Put in your Vcc. Set R1 value
as your whatever your series resistor is, and R2 as the resistance
of your ldr, (in this case 5k ohms in the light.) Change values to
see how it works.
  I'm sorry if this all takes you the long route to making the
thing work, but I think you should understand how it works so you
can optimize to your lighting conditions.
                              Mikek
                              Mikek


There was no page which you referred to in the previous that I could
see any link to, and there is no link to a voalage divider page if
that is what you mean.


  Sorry forgot to put it in.
 > http://www.ohmslawcalculator.com/voltage-divider-calculator


All I want to do now is add a buzzer which will work when the led
comes on.

Thanks

  Does adding the 555 circuit work for you?

Remember at one time you had the 555 setup so that touching a
grounded wire to Pin 2 of the 555, it would make the buzzer buzz for
8 seconds and then it would stop.
  So you need your Resistor/ldr circuit feeding your 741 schmitt
trigger, feeding an RC circuit driving Pin 2 of the 555 and Pin 3 of
the 555 driving your buzzer.

                               Mikek





On that schmitt trigger circuit you sent me there is a 555 ic in the
circuit.

When I put the live side of the buzzer into pin 3 and the negative
side to ground it buzzes continually, choose whether led is lit or not.
What is an RC circuit.




 Is pin 3 actually switching low to high when your circuit triggers it?

 I have a feeling it's not.




There is a wire from pin 6 of the 741 to pin 2 of the 555 via a 0.01uf
capacitor and x2 100k resistors voltage divider.
With a source voltage of 8.06v the voltage at pin 2 is 7.19v, and at
pin 3 of the 555 it is 7.40v

If you have that circuit wired correctly Pin 2 should set at 1/2 Vcc, or
4.0V. other wise you have something improperly wired.
 I have emailed a schematic to you.
                                 Mikek
Just to make things simple watch the sales at Harbor Freight and pick
up this driveway alert sensor.
> https://www.harborfreight.com/wireless-driveway-alert-system-93068.html
Often on sale for $11, I see they now have several other models, so
the sales might not happen like the used to. Just use a 25& off coupon.
I have several, they work great, battery life is good, and very happy
with how well they work. I modified one for a special use at the
business we once had. It was a very open area with lots of traffic, so I
needed it to only trip in a very small area, only were my customers had
to walk. I limited the area it could see with a short piece of 1/2" pvc.
I then modified the receiver to trigger a walkie talkie so I could be a
great distance from the business and get a signal that I had a customer.

Mikek
 
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