# Schmitt Triggers

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RobH
Guest

Tue Feb 11, 2020 4:45 pm

On 11/02/2020 15:32, RobH wrote:
Quote:
On 11/02/2020 15:17, 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_at_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:
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

th

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

Rob, I still wonder,
Do you want the the buzzer to buzz when light shines on the ldr?
Mikek

Mmm, strange that, but yes when the light shines on the ldr.
I have the led lighting up when light shines on the ldr, 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.

Oops, wrong circuit above, please ignore as that is the schmitt trigger
circuit you sent me.

Yes if it is possible.
This circuit using the 741 op amp has the led shining when light hits
the ldr, and this is the one where iI don't know which pins or where to
connect the buzzer. If it can't be done, then it can't be done, and if
it needs a 555 to make the buzzer work, then I may as well just use the
schmitt trigger circuit you sent me.

Thanks

RobH
Guest

Tue Feb 11, 2020 4:45 pm

On 11/02/2020 15:17, RobH wrote:
Quote:
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_at_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:
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

th

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

Rob, I still wonder,
Do you want the the buzzer to buzz when light shines on the ldr?
Mikek

Mmm, strange that, but yes when the light shines on the ldr.
I have the led lighting up when light shines on the ldr, 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.

Oops, wrong circuit above, please ignore as that is the schmitt trigger
circuit you sent me.

amdx
Guest

Tue Feb 11, 2020 5:45 pm

On 2/11/2020 9:17 AM, RobH wrote:
Quote:
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_at_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:
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?
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.

Quote:
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

RobH
Guest

Tue Feb 11, 2020 9:45 pm

On 11/02/2020 15:55, amdx wrote:
Quote:
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_at_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:
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?
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

amdx
Guest

Tue Feb 11, 2020 10:45 pm

On 2/11/2020 1:59 PM, RobH wrote:
Quote:
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_at_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:
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?
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?

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

RobH
Guest

Wed Feb 12, 2020 11:45 am

On 11/02/2020 21:19, amdx wrote:
Quote:
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_at_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:
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?
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?

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.

Quote:
What part of that are you uncomfortable with?
None really, I wasn't just sure if it was right that's all.

Quote:
Are you sure you used the same resistor in both situations?
Yes I am completely sure.

Quote:
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

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

Wed Feb 12, 2020 12:45 pm

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

Quote:
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

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.

RobH
Guest

Wed Feb 12, 2020 1:45 pm

On 12/02/2020 11:28, tabbypurr_at_gmail.com wrote:
Quote:
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

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.

amdx
Guest

Wed Feb 12, 2020 2:45 pm

On 2/12/2020 6:41 AM, RobH wrote:
Quote:
On 12/02/2020 11:28, tabbypurr_at_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

Quote:
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?

Quote:

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

RobH
Guest

Wed Feb 12, 2020 3:45 pm

On 12/02/2020 13:33, amdx wrote:
Quote:
On 2/12/2020 6:41 AM, RobH wrote:
On 12/02/2020 11:28, tabbypurr_at_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

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.

amdx
Guest

Wed Feb 12, 2020 4:45 pm

On 2/12/2020 7:33 AM, amdx wrote:
Quote:
On 2/12/2020 6:41 AM, RobH wrote:
On 12/02/2020 11:28, tabbypurr_at_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

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
Mikek
Mikek

RobH
Guest

Wed Feb 12, 2020 4:45 pm

On 12/02/2020 15:20, amdx wrote:
Quote:
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_at_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

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
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

amdx
Guest

Wed Feb 12, 2020 7:45 pm

On 2/12/2020 9:29 AM, RobH wrote:
Quote:
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_at_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

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
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.
Quote:
http://www.ohmslawcalculator.com/voltage-divider-calculator

Quote:
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.
trigger, feeding an RC circuit driving Pin 2 of the 555 and Pin 3 of the

Mikek

RobH
Guest

Wed Feb 12, 2020 8:45 pm

On 12/02/2020 18:10, amdx wrote:
Quote:
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_at_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

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
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.
trigger, feeding an RC circuit driving Pin 2 of the 555 and Pin 3 of the

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

amdx
Guest

Wed Feb 12, 2020 8:45 pm

On 2/12/2020 12:48 PM, RobH wrote:
Quote:
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_at_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

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
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.
trigger, feeding an RC circuit driving Pin 2 of the 555 and Pin 3 of

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.

Quote:
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

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