Schmitt Triggers

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.

With these measurements is the resistor a 33k ohm?

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.

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

Sorry forgot to put it in. >

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

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

I have a feeling it's not.

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

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

Just to make things simple watch the sales at Harbor Freight and pick up this driveway alert sensor.

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