Hi,

When V1 > V2(Vouptut neg. sat.), shouldnt there be an intial negative voltage at V2, plate 'X', as the capacitor hasnt had time to charge?

The timing waveforms are on the .pdf.

Thanks.

 

The op-amp has a very high i/p impedence. Try anylising the circuit by looking only at C1 and R3. What voltage would be present at this junction prior to pushing the button? Where does that potential come from?

 

All I see when looking at C1 and R3 is a differentiator circuit.

How does the high Z of the op-amp affect what is expedcted from
this differentiator?

steve

 

Hi,

Just to throw in a guess, I'd say the potential at V2 would be a race between any leakage through C1 and leakage out terminal V2. That makes it about as likely to be positive in sign as negative. The value of R3 and the type of capacitor will be significant, as will the spec's on the op amp.

 

All I see when looking at C1 and R3 is a differentiator circuit.

Exactly. What is the derivitive of a horizontal line? What is the voltage at the junction of C1 and R3 if the button has not been recently pressed? (Hint: what would it be if we pulled the IC out?)

 

This is what the associated text says:

...With S1 open, the +input is grounded via R3, ie V2 = 0V and the -input is positive owing to the small voltage at the junction of the voltage divider. Therefore V1 is greater than V2 and the op amp is neg. saturatedm ie Vo = -Vs. Also C1 is charged with plate X positive relative to plate Y(Vo). (More exactly, X is at 0V and Y at -Vs approx).


Im probably wrong, the op-amp is holding V2 low, trying to get a zero differential between its two inputs?

Steve

 

Hi,

The op amp will not take an active interest in its environment. The inputs go to a differential amplifier. It would be more correct to say that it would take an equivalent voltage of opposite sign at V2 to balance the op amp.

 

The op-amp isn't really doing anything to influence V2 - not directly anyway. V2 is an input.

 

Upon power-up of this circuit (assuming +Vs and -Vs come up at essentially the same time), V1 will be +2.5V and more positive than V2 which is grounded through R3. This will cause the the output (Vo) to go negative, which will drive V2 negative for a time based upon the time constant of R3/C1. However, since V1 remains positive relative to V2, the ouput will not change and remains negative.

Now, once everything is stable, pushing that switch just turns you into a trouble maker.