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First before we start, let me thank you in advance for your help. I know that I don't totally understand how to make this work but I must make it work.

Let me start from the beginning. I'm using an LM393 because it has two comparators. I'll use one for the low a reference and one for a high reference. The output will be depend on which reference voltage you want to deal with. The input voltages will be in the area of 4.4 to 4.9 volts. One reference will be in the lower range and the second in the upper range. If the 'user' wants to trigger the relay(driven by the output) at the lower they will place the switch in that position and if they want the upper they will flip the switch. In this way I can have one input to the chip and depending on which way the switch is set that side of the chip will do the work and trigger the relay.

I've seen on the web diagrams which cause an LED to burn when the input voltage reaches the reference voltage. I built this and when I try to make it work the LED will light but it does not when the input reaches the reference but far below that. For example when I set the input voltage to 2.3 volts and the reference voltage to 4.7 the LED will not light. As I bring the input voltage up the LED lights when it reaches something like 2.9 volts.

The LM393 uses a 5vdc to power it and the 'ground' is ground. There are mult-turn pots to set the reference voltage and I use a seperate

5vdc supply and a multi-turn pot to set the input voltage. I have no resisters in the circuit other than the pots.

I need to figure out why the LED lights long before the input voltage reaches the reference voltage and what I can do to my circuit so the output will 'output' only when the two voltages are the same or when the input voltage is higher than the reference voltage.

I guess I could have asked this question much simplier by simply begging for someone to offer a diagram with component values of a circuit that actually works and I could pattern mine from that.

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The 393 only has an input common mode range of 0 to (V+ - 1.5v). Neither input should be higher than 3.5 for proper operation. Use some voltage dividers to get your signals and references inside the common mode range first.

Steve

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Is there a comparator that will permit me to simply wire it up as in one that will permit input and reference ranges in the range of 4.6 -

4.9?

If not please clarify 'voltage divider'. I think this is a component or circuit which would reduce the incoming voltage. Are we talking about a simply resistor here? If so then all I need to do is figure out what the voltage drop must be and use that resister but I'm hoping there is a more accurate way of doing this. Do you have any suggestions as to how I can 'build' a voltage divider?

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sci.electronics.basic may be the appropriate group for these questions.

As long as your device doesn't pull appreciable current on the "Vout", a voltage divider is just two resistors:

Vin o-------. | .-. | | | |R1 '-' | Vout = Vin * R2 / (R1 + R2) o----------o | .-. | | | |R2 '-' | | === GND (created by AACircuit v1.28.6 beta 04/19/05

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Tim Wescott
Wescott Design Services```
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Is there a comparator that will permit me to simply wire it up as in

How much accuracy do you really need, and over what temperature range?

Accuracy depends on the resistor choices, and the source/load resistances, but you can make ~1% without much expense. Sounds like you are just tweaking this until it works, so just put a pot into one of the arms of the divider and you can adjust the divide ratio to any setting that you like. Just recognize you sacrifice characteristics like temperature stability. But from what you describe, it probably doesn't matter to you.

Alternately, you could raise the power supply voltage until all your signals fall inside the CM range. But I assume this is a bigger problem than reducing the signal levels. The 393 output is open collector, so you could still get 5V logic outputs if you keep the pull-up resistor on the 5V supply and raise the V+ voltage to some convenient value above ~6.5V

Steve

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Thanks,

If I build the circuit like I did before put supply 12vdc to the chip would it allow a reference voltage of 4.4 to 4.9? Please help me understand the pullup resistor on the 5v supply. I'm not exactly sure that I understand this. I guess it's a resistor which is at the supply but where does it connect and what does it do?

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I forgot to mention that accuracy up to +/-3 percent is not a problem.

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