Opamp with diode and capacitor

[Qacer]

Hey all,

I have this circuit. This is actually part of another bigger circuit with more components and branches, but since I can't understand this part, I feel stupid. The thing that is throwing me off is the diode and the VCC/2 bias. Although, I think that if an input signal is below VCC/2, the diode starts conducting.

Any help?

 

[BrownTown]

integrates the amount of the singal over VCC/2 maybe?

EDIT: i think the diode is there to provide a feedback path, but at the same time block current from just skipping the integrator all together. In the practicle integrator circuits I see on the web a resistor with a large value is used instead of the diode, however a diode may be a more ideal solution for the same problem.

 

[Qacer]

I tried doing this on paper using the equivalent circuit for opamps, but I got stuck since the equivalent circuit uses a dependent source. I like doing superposition, but I haven't quite figured out how to use the principle using dependent source. Apparently, it is possible to do that even though most text books say it's not.

I wanted to see the equations for this circuit on paper so I could explain to myself how it worked.

 

[BrownTown]

I dunno about all that, just look at it, the resistor will provide a current into the cap at [Vin-(Vcc/2)]/R, the voltage across a capacitor is the integral of the current into it. I know its an integrator jsut looking at it, the only question is the exact purpose of the diode. If you posted the whole circuit maybe it would help provide some additional context.

 

[Born2bwire]

Originally posted by: BrownTown
I dunno about all that, just look at it, the resistor will provide a current into the cap at [Vin-(Vcc/2)]/R, the voltage across a capacitor is the integral of the current into it. I know its an integrator jsut looking at it, the only question is the exact purpose of the diode. If you posted the whole circuit maybe it would help provide some additional context.

My guess is that it stops the integration after it reaches a certain level. If the diode is off, then it just acts as an integrating opamp circuit. But if the diode starts to conduct, then current is going to bypass the integrating feedback loop and stop building up charge on the capacitor. So if the capacitor builds up enough charge such that the voltage drop across it is equal to the bias voltage of the diode, then I would think that the diode would turn on and stop the integrator.

Normally, you have some sort of switch in parallel to the integrator's capacitor. This allows you to start and stop the integration, it seems to me that the diode allows for an automatic switch as opposed to the black box manual switch that is often shown in integration circuits.

This would only work given a negative direction of the current in the feedback loop. And that would be dependent upon your input signal. So another idea off the top of my head is that it allows you to only integrate the positive input with respect to VCC/2. If it is negative, then the feedback current bypasses the cap. If it is positive, the cap is charged. Really though, the OP is just better off using Spice to simulate these nonlinear circuit elements. Otherwise you have to consider the various diode situations and find which ones have a solution.

 

[Mark R]

Looks to me like an integrator, but one that can be reset by applying a 'negative' input.

So with input > Vcc/2 it will act as an inverting integrator, and produce a downgoing ramp.
Apply a voltage of < VCC/2 and the integrator will reset to your input voltage (with an offset = to Vf of the diode) albeit after some sort of a time lag.

Best option is to simulate it and see exactly what it does.

 

[Erik in sac]

To me, it looks as if the diode is simply for polarity protection @ DC. The Vcc is a reference voltage. That op-amp has a differential/comparator input.

 

[itachi]

the purpose of the diode, from what i can understand, is to reset the integrator, as mark stated.

when the diode is "off", the circuit acts as an integrator.. when it's "on", the output will be proportional to the log of the input.. the output will then be less than the input, causing the diode to turn "off" again.

 

[Casawi]

There could many reosons why the diode is there depending on what you wanna do. More info will maybe tell us more. That diode could be an LED for all I know.

 

[Casawi]

I looked into it more ... Depending on the value of the Cap on there if it is small enough to complete the circuit after the diode is "open".