Op-amp circuit help

Jun 29, 2006 5 Replies

Hello all, I've been trying without success to design a circuit that has 2 inputs and one output, such that the output is:- Vout = Vin1 + (Vin1-Vin2):-



Vin1----| >---- Vout Vin2----|



I'm using a single 5V supply and both input voltages come from op-amp voltage-followers, (originally derived from PWM signals). I need the output to swing from 0V to as close as possible to the 5V supply.



The circuit is intended as part of a closed-loop speed control system for a DC motor.



Can anyone help with ideas, links etc.?



... Humphrey


R1 R2 R1=R2= 10k-ish. Vin2---/\\/\\---+---/\\/\\---+ | _ | +--|- \\ | | >---+---> Vout = 2*Vin1 - Vin2 Vin1-------------|+_/

Look for an opamp described as rail-rail input/output.

Tony Williams.

Too easy. Looks good to me. Thanks, Tony. Do you think I'd strike any problems if I used 1M resistors instead of 10K, so that I could do away with the voltage-followers and take the inputs directly from the PWM filter capacitors? (Vin1 is the requested motor speed, a 20kHz PWM signal, and Vin2 is the motor speed feedback signal, PWM with a frequency ranging from 100Hz at 60RPM to 13.33kHz at 8000RPM, derived from a

100 pulse-per-revolution, 2", optical encoder disk Vin1 is smoothed with a 1uF cap and Vin2 is smoothed by a 10uF cap)

... Humphrey

Yes you could do. You could even probably reduce the circuit to just one filter. But if this is a first go then retaining the followers allows you to look at Vin1 and Vin2 separately.... useful for diagnostics.

Tony Williams.

Good one. Thanks again for the help, Tony.

Hi Humphrey, I think with your idea it will be very difficult to control the motor speed, because of long time constants. If your motor is a DC motor, you could actually use the fact that speed is proportional to applied voltage, if you correct for armature resistance. So just a current sensor in form of a resistor is needed instead of this encoder.

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These articles will give you some ideas I hope.

ciao Ban Apricale, Italy

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