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Given this integrator circuit:

Circuit

And with Vin being a pulse wave which changes alternatively from 0V to 5V with a frequence of 100KHz, I need to find out the value of Vout after a large number of cycles.

I would like to get a hint on this problem, my intention is to calculate the Vout after each charge/discharge cycle using the formula V(t) = Vo*(1-e^(t/(R*C))) (charge) and V(t) = Vo*(e^(t/(R*C))) (discharge) for a large number of cycles using a computer program which does the calculations but I don't know how to take into account the remaining Voltage from the previous cycles in the charging formula.

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  • \$\begingroup\$ What do you mean by 'find the value of Vout after... for its value to stabilize'? The output will converge to a periodic waveform. Are you asking about the value of Vout sampled at the same period? \$\endgroup\$
    – copper.hat
    Sep 18, 2014 at 15:33

2 Answers 2

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That's not an integrator- it's a low pass filter. Just as well, as a true integrator has an arbitrary constant in the output so you would not be able to predict the steady-state voltage (in a real analog integrator it would tend to drift towards one rail or the other due to offset even if the input averaged zero).

Suggest you assume a starting output voltage V0 at the beginning of a cycle and show that it's in steady state (returns to the same voltage for the next cycle) rather than trying to model the transient start up condition.

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  • \$\begingroup\$ Yes - it is no real integrator. However, there is no analog circuit that can act as a real (ideal) integrator (unless it is a block within an overall sysytem with feedback). But each first-order lowpass can be used as an integrating device (with negligible errors) if the signal frequency is sufficiently above the lowpass pole which is app. at 160 Hz. Hence, the given circuit can be used for integrating a frequency of 100kHz. \$\endgroup\$
    – LvW
    Sep 18, 2014 at 7:43
  • \$\begingroup\$ Another way of saying this is that it works as an accurate integrator if there is negligible output voltage, since the resistor no longer acts as a current source when there is a voltage across it that differs from the input voltage. [This] (upload.wikimedia.org/wikipedia/commons/thumb/b/bd/…) is the classic inverting integrator circuit. \$\endgroup\$ Sep 18, 2014 at 11:48
  • \$\begingroup\$ May I add that the inverting integrator as shown in the wikipedia link resembles (for a non-ideal opamp) also a first order lowpass only? More than that, it cannot be used as a "stand-alone" unit because of lacking dc feedback (problem of offset voltage integration). Hence, a high-valued resistor is required in parallel to the feedback C. \$\endgroup\$
    – LvW
    Sep 18, 2014 at 12:59
  • \$\begingroup\$ @LvW, You are being a bit picky. A current source into a capacitor is an integrator. (Of course all components are non-ideal at some level.) \$\endgroup\$ Sep 18, 2014 at 13:23
  • \$\begingroup\$ G.Herold - thank you. Nevertheless, something wrong with my comment? Which current source are you speaking of? \$\endgroup\$
    – LvW
    Sep 18, 2014 at 16:51
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You can compute the solution explicitly.

Let \$v\$ be the output voltage and \$u\$ be the input. The ODE describing \$v\$ is \$\dot{v} = - {1 \over RC} v + {1 \over RC}u\$, with initial condition \$v(0)\$. The solution is \$v(t) = v_0 e^{- { 1 \over RC} t} + {1 \over RC}\int_0^t e^{- { 1 \over RC} (t-\tau)} u(\tau) d\tau\$. This works for any input \$u\$.

Suppose the input \$u\$ is \$T\$-periodic, and consider the circuit at times \$0,T,2T,...\$. Let \$v_n = v(nT) \$.

We have \$ v_{n+1} = e^{- { 1 \over RC} T} v_n + {1 \over RC} \int_{nT}^{(n+1)T} e^{ -{ 1 \over RC} ((n+1)T -\tau)} u(\tau) d\tau \$, and a change of variable shows that \$ \int_{nT}^{(n+1)T} e^{ -{ 1 \over RC} ((n+1)T -\tau)} u(\tau) d\tau \$ is independent of \$n\$, so if we let \$\sigma = {1 \over RC} \int_{0}^{T} e^{ - { 1 \over RC} (T -\tau)} u(\tau) d\tau \$, we get the equation \$ v_{n+1} = e^{- { 1 \over RC} T} v_n + \sigma \$, which converges to the steady state \$ \hat{v} = { \sigma \over 1 - e^{- { 1 \over RC} T} } \$.

In particular, if we start at the steady state, that is, \$ v_0 = \hat{v} \$, then the output will be periodic with period \$T\$ (that is, no transients).

From here on I am assuming that \$v\$ represents this steady state solution.

Note that \$v\$ is periodic, not constant, so I'm not sure what you meant by 'after ... value to stabilize'.

If you want to compute the average (DC) output of the steady state solution you can compute \$ {1 \over T} \int_0^T v(\tau) d \tau \$ (painful), or note that we can integrate the differential equation to get \$ \int_0^T \dot{v}(t) dt = v(T)-v(0) = 0\$, and so we see that the average values of the input \$u\$ is the same as the average value of the output \$v\$. This is no surprise since the DC voltage gain is 1.

So, if you want to compute the DC value of the output, it will just be the DC value of the input.

If you want to compute the steady state solution, pick the \$T\$-periodic input \$u\$, compute \$\sigma\$, then the relevant initial state \$\hat{v} \$, and then integrate using the solution above to get the periodic 'steady state' solution.

Note: The value of \$\hat{v} \$ depends on where \$u\$ 'starts'.

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  • \$\begingroup\$ Thanks for such an elaborated answer, unfortunately I haven't got enough knowledge about the subject to be able to understand some things you said, but according to what my teacher said, Vout converges to DC after a large number of cycles (that's what I meant when I said "for its value to stabilize"), 'cause given the high freq. of the Vin pulse (it is not a sinusoidal wave but a pulse) the capacitor doesn't have enough time to charge and discharge completely so this causes the Vout amplitude to decrease in every cycle, converging to a DC output after a time t which matches ∫0->t (Vin dt) \$\endgroup\$ Sep 18, 2014 at 22:04
  • \$\begingroup\$ Well, your teacher wasn't quite right. The output will also be periodic (and non-constant). However, the gain at the first harmonic is \$\approx 0.002 \$, so any ripple will be small. Note that in this example, the DC portion of the input and output signal will be the same (since the cap. is an open circuit at DC). \$\endgroup\$
    – copper.hat
    Sep 18, 2014 at 22:57
  • \$\begingroup\$ Sorry, I didn't mean you to switch accepted answer, I was just trying to help. Would you mind switching back please? \$\endgroup\$
    – copper.hat
    Sep 19, 2014 at 14:17
  • \$\begingroup\$ Much appreciated! \$\endgroup\$
    – copper.hat
    Sep 20, 2014 at 18:00

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