Please help me understand

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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

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 -

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?

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

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

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?

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

Hope This Helps! Rich

--- OK, the way I interpret that is if your customer wants the relay to energize when the input voltage is >4.9V you'll have him throw the switch one way, but if he wants the relay to energize when the input voltage is input voltage reaches the reference voltage. I built this and when I

--- No, you did a good job of describing where you want to go and what you want to use to get there.

The reason for the problem you're having is that in order for the output to switch properly, neither of the input voltages must go higher than 1.5V below the supply voltage.

That means that you must either raise the supply voltage if you can't change the level of your input signal, or you must change the level of your input signal so that the comparator can sense it properly.

I assume that you'd like to leave the supply alone, so what you must do, then, is decrease the level of your input signal.

In order to do that you need to use a voltage divider, which is nothing more than a couple of resistors in series with your input signal and grounded on one end: (View in Courier)

VINE1 | [R1] | +---->E2 | [R2] | GND>---+---->0V

what it's used for is to attenuate the voltage at the top of the divider, E1, and get a lower-voltage version of it at E2.

The ratio of the resistances determines how much the signal will be attenuated and, for example, two equal valued resistors will cause the voltage at E2 to be half the voltage at E1. The relationships between resistances and voltages is defined by:

E1 * R2 E2 = --------- R1 + R2

and by rearranging the equation any one of the resistances or voltage can be determined if the others are known.

Now, since we know that you're using a 5V supply and the LM393 won't work properly if either of the inputs to its comparators is higher than 3.5V, we'll need to do is drop that 4.9V down to 3.5V (or better yet, down to 3.25V just to err on the side of caution.

OK, what's next?

Well, looking at the divider again, with the voltages added, we have:

VIN4.9V | [R1] | +---->3.25V | [R2] | GND>---+---->0V

But we need to know the values of the resistors, so since the last one will come out automatically when we plug the value of the other one into the equation, what do we do to find the value of the first one? We guess! Well, actually, we don't, because we need to be concerned about how much bias current the comparator needs and how that current will affect the voltage dropped across the resistor and also what kind of a load the divider will present to the signal source. The LM393 needs 250nA max for bias current so if we guess that 10k will be OK, the voltage dropped across dropped across that

E = IR = 250E-9A * 1E4R = 2.5E-03V = 2.5 millivolts

and the error it will introduce will be:

Which isn't bad.

So now our voltage divider looks like this:

VIN4.9V | [R1] | +---->3.25V | [10k] | GND>---+---->0V

and to solve for R1 we rearrange:

E1 * R2 E2 = --------- R1 + R2

to:

R2 (E1 - E2) 1E4R (4.9V - 3.25V) R1 = ------------- = ---------------------- = 5076 ohms E2 3.25V

The closest standard 1% values are 4990 and 5110 ohms, and I chose

Anyway, the divider now looks like this:

VIN4.9V | [5110] | +---->3.25V | [10k] | GND>---+---->0V

Checking it out:

E1 * R2 4.9V * 1E4R E2 = --------- = ---------------- ~ 3.243 volts R1 + R2 5.11E3R + 1E4R

Pretty close.

So now, when you put 4.9V into the voltage divider you'll get an output from it which will keep the LM393 happy and which will make the comparator switch where you want it to. :-)

All you need now is a reference for "higher than 4.9V", "lower than

Since you're Google-Impaired and can't access the binary newsgroups I've emailed you a schematic of a circuit which should do what you want and I've also posted it to alt.binaries.schematics.electronic for anyone else who might be interested.

-- JF