shorting input pins on an opamp

Thread Starter

anishkgt

Joined Mar 21, 2017
387
Would it be ok to short the input pins of the second comparator like this to get an inverted signal from the first. I am planning on to use MCP6L02T-E/SN as the comparator. The circuit is Low-Voltage-Cutoff except the cutoff is just indicated by an led from the output of the second comparator.
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Thread Starter

anishkgt

Joined Mar 21, 2017
387
Forgot about the voltage rating.
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Is it ok to have the input pins of the second opamp shorted and connected to the output of the first or is it ideal to have a resistor in series on both pins.
 

LesJones

Joined Jan 8, 2017
2,417
The output of an op amp depends on the DIFFERENCE in voltage between the inverting and non inverting inputs. This difference will be very small as the open loop gain is very high. They are almost always used with feedback which is normally negative. They can be used as a comparator with either no feedback or a small amount of positive feedback to give the comparator some hysteresis.

Les.
 

Thread Starter

anishkgt

Joined Mar 21, 2017
387
The output of an op amp depends on the DIFFERENCE in voltage between the inverting and non inverting inputs. This difference will be very small as the open loop gain is very high. They are almost always used with feedback which is normally negative. They can be used as a comparator with either no feedback or a small amount of positive feedback to give the comparator some hysteresis.

Les.
So i guess that would mean it would be a good practice to give a negative feedback then ? Giving a positive just inverted the output on the second comparator.
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LesJones

Joined Jan 8, 2017
2,417
Your new schematic is still wrong as R7 effectively connects both inputs together. Remove R6 and R7. Bias the plus input to Vcc/2 by adding two equal value resistors from the plus input. One to ground and one to Vcc. (The value of the resistors is not critical as long as they are both the same value. As someone said in a previous post if you swap over the inputs to U1 you would not need U2 to invert the signal. You would still need R2 connected between the plus input and the output. You would just swap over the wires that go to the junction of R1 and R9 and the one that goes to the junction of R4 and R10. R2 provides positive feedback to provide hysteresis on the switching point of U1.

Les.
 

OBW0549

Joined Mar 2, 2015
3,090
Is it ok to have the input pins of the second opamp shorted and connected to the output of the first or is it ideal to have a resistor in series on both pins.
You're not getting the important point, which I should think would be blatantly obvious: if you short the op amp inputs together, with or without resistors in series with them, THE OP AMP DOESN'T DO ANYTHING. It will just sit there with its output saturated to either V+ or V-.

Do yourself a favor: instead of blindly fumbling around aimlessly with op amps, read the two tutorials attached and learn something.
 

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BobTPH

Joined Jun 5, 2013
2,108
If you want a second comparator to output the opposite of the first one, connect the second comparator’s + to the first one’s - and vice versa.

Bob
 

Thread Starter

anishkgt

Joined Mar 21, 2017
387
Your new schematic is still wrong as R7 effectively connects both inputs together. Remove R6 and R7. Bias the plus input to Vcc/2 by adding two equal value resistors from the plus input. One to ground and one to Vcc. (The value of the resistors is not critical as long as they are both the same value. As someone said in a previous post if you swap over the inputs to U1 you would not need U2 to invert the signal. You would still need R2 connected between the plus input and the output. You would just swap over the wires that go to the junction of R1 and R9 and the one that goes to the junction of R4 and R10. R2 provides positive feedback to provide hysteresis on the switching point of U1.

Les.
Thank you all. I found a comparator that works within the voltage of 2-36v LM393ADR. Is there any advantage of using this compared to a lower voltage IC with a voltage divider ?

As someone said in a previous post if you swap over the inputs to U1 you would not need U2 to invert the signal.
U2 here is just to invert the signal from U1. As an indication when U1 has gone LOW or would it be ok to have an LED at the output of U1 in reverse biased to accomplish this ? That way i could eliminate U1 completely.
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LesJones

Joined Jan 8, 2017
2,417
As far as I can see from the data sheet the LT1018 should work without a problem. The inverter part of your schematic in post #15 is as I described and should work BUT the change to the scaling divider on the - input of U1 will cause a problem. This is because with X2 set to maximum (20 volts) the - input of U1 will be 10 volts which is above the Vcc level of 5 volts. U1 will not work with that condition.

Les.
 

LesJones

Joined Jan 8, 2017
2,417
A pull up resistor is required with the MIC6270YM5. If you remove the input scaling you would also have to increase Vcc as the inputs on comparators (Or op amps.) cannot go outside the supply rails.

Les.
 
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