0
\$\begingroup\$

I am trying to design a circuit which would interface AC signals from pick-up coils. Below is the circuit. enter image description here

R4, R3 are for DC bias, and R2 is for feeding that bias to inverting input of the OP amp.

R5 and R8 is for hysteresis regarding noise immunity.

V2 is the input signal.

V1 is power rail.

My question is that how should I determine the state of Vout(18V or 0V) when V2 is open?

I simulated it on CircuitLab. and below is the result V+ = 8.946 V V- = 8.955 V Vout = 1.783 mV

My concern is that V- is higher than V+ only by 9 mV, which I believe is leading Vout to near negative rail(0V).

Can I rely on current circuit for the output to be low always?

Or should I add some kind of pull-up or pull-down resistor for circuit stabilization?

I tried adding pull-up or pull-down resistors to V+ or V- but it prevents the signal from making Vout with 18V peak to Peak.

Below is the link to the circuit i made public

https://www.circuitlab.com/circuit/74bt8h/comparator/

\$\endgroup\$
2
\$\begingroup\$

To start, eliminate V2, C1, R2 and R5. Then the junction of R3 and R4 will have a Thevenin equivalent voltage of 9 volts, and a resistance of 5k. Adding R5 provides a hysteresis of $$ \Delta v = 18\times \frac{6k}{1.006 M} = .107\text{ volts} = 9 +/- .053\text{ volts} $$ this will also provide a hysteresis at the R2/R3 junction of $$ \Delta v = 18\times \frac{5k}{1.005 M} = .089\text{ volts}= 9 +/- .045\text{ volts} $$ With V2 still unconnected, the - input will be forced to the value of the R2/R3 junction. Since this is undetermined, the output voltage is undetermined.

Effectively, with V2 disconnected the comparator can be in either state, and there is no a priori way to determine which state that will be.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Is there any way I can force the state to determined state without losing its functionality? If not, will it stay in that state once it's determined even though I don't know what that state is? – \$\endgroup\$ – Steve Feb 3 '16 at 2:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.