Schmitt Triggers

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Nife Sima
Guest

Sun Feb 16, 2020 4:45 pm

On 10.02.2020 19:38, amdx wrote:
Quote:
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_at_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:

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
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:
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?

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.

amdx
Guest

Sun Feb 16, 2020 6:45 pm

On 2/16/2020 9:16 AM, Nife Sima wrote:
Quote:
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_at_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:

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
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:
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?

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

amdx
Guest

Tue Feb 18, 2020 3:45 pm

On 2/12/2020 1:44 PM, amdx wrote:
Quote:
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_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 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.
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.

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
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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