How can I analyse this circuit (analog PD controller)?

We had op-amps at school but I do not know how I can go about analysing this circuit because it has an additional node in the backwards branch. For me, I can not see why the capacitor is inside a closed loop, so I would not expect any voltage drop across it or current through it. Does somebody maybe know how you can go about analyzing this circuit? I want to know how I can calculate the output voltage (out_1-out_2), because the goal is to find the transfer function ((out_1-out_2)/(in_1/in_2)). Thanks a lot for a reply!

Edit: It says that the relationship between GND and out_2 is given by this Bode plot:

• what is the relationship between out_2 and gnd Sep 7 at 18:58
• About C1, remember we're talking about changing AC here, so there will be changing currents through C1 for every frequency except DC. What maths have you tried so far? Please click the edit link below the question to add in what you've tried. Sep 7 at 21:13
• The schematic is a bit unusual to me. Mainly because out_2 is not driven by anything. It is floating, which is perilous for any simulation, because the simulator will chose an operating point/initial condition for that, but it can be hijacked in a real circuit easily by unwanted coupling. Sep 8 at 8:47

Your circuit has no path to any fixed DC potential, such as ground. It looks like C1 and R4 provide such a path, but at DC C1 effectively disappears, and therefore there is no path to DC anywhere, and everything is floating.

To fix this, I'll assume that IN2 and OUT2, which are the same node, have a fixed potential, 0V, and the circuit becomes:

simulate this circuit – Schematic created using CircuitLab

We have shifted input and output signals down to ground, ensuring that the op-amp operates within its acceptable common-mode input voltage range. It may seem like this changes the behaviour of the circuit, but it does not, since all we have done is replace terms $$\(v_{OUT1} - v_{OUT2})\$$ and $$\(v_{IN1} - v_{IN2})\$$ with $$\v_{OUT1}-0=v_{OUT}\$$ and $$\v_{IN1}-0=v_{IN}\$$. The offset $$\v_{IN2}\$$ common to both could really be any potential within the op-amp's power supply range, but setting it to zero simplifies the arithmetic.

At DC capacitor C1 has infinite impedance, and the circuit can be redrawn and analysed as if C1 and R4 were not there:

simulate this circuit

Here the gain is calculated in the usual way. DC signals $$\V_{IN}\$$ and $$\V_{OUT}\$$ are related as follows:

\begin{aligned} V_{OUT} &= -\frac{R_2+R_3}{R_1}V_{IN} \\ \\ &= -4V_{IN} \end{aligned}

As input signal frequency rises, the impedance of capacitor C1 falls, and the combination of C1 and R4 may no longer be neglected. To keep analysis simple, remember the usual behaviour of an op-amp with negative feedback is to raise or lower its output by whatever amount is necessary to equalise the potential at its two inputs:

$$v_X = v_Y = 0$$

Note: We are also assuming that the feedback can never have a phase shift sufficient to become positive. That might be possible at very high frequencies, and with certain op-amps, but analysis at such frequencies is another topic.

Considering the op-amp to be a voltage source of $$\v_{OUT}\$$, which is always the exact potential needed to obtain the condition $$\v_X = 0\$$, the network consisting of R1, R2, R3, R4 and C1 may be redrawn, and analysed using the usual Kirchhoff's and Ohm's laws to establish the conditions present, given some arbitrary $$\v_{IN}\$$:

simulate this circuit

\begin{aligned} i_1 &= i_2 + i_3 \\ \\ i_1 &= \frac{v_{IN}-v_X}{R_1} = \frac{v_{IN}-0}{R_1} = \frac{v_{IN}}{R_1} \\ \\ i_1 &= \frac{v_X-v_P}{R_2} = -\frac{v_P}{R_2} \\ \\ i_2 &= \frac{v_P-0}{R_4 + Z_{C1}} = \frac{v_P}{R_4 + Z_{C1}} \\ \\ i_3 &= \frac{v_P-v_{OUT}}{R_3} \\ \\ \end{aligned}

I'll spare you the arithmetic, but combining those equations gets you a transfer function that looks something like:

$$H(s) = \frac{v_{OUT}}{v_{IN}} = -\frac{1}{R_1}\left(R_2+R_3+\frac{R_2R_3}{R_4+Z_{C1}}\right)$$

After plugging in $$\Z_{C1} = \frac{1}{sC_1}\$$, and tweaking into to a form that advertises poles and zeroes:

$$H(s) = -\left( \frac{sC_1(R_2R_4+R_3R_4+R_2R_3)+R_2+R_3}{sCR_1R_4+R_1} \right)$$

The numerator is zero when:

\begin{aligned} s &= -\frac{R_2+R_3}{C_1(R_2R_4+R_3R_4+R_2R_3)} \\ \\ &= -\frac{200+200}{1 \times 10^{-6}(200\times 300 + 200 \times 300 + 200 \times 200)} \\ \\ &= -2500 \end{aligned}

That corresponds to a zero at frequency $$\\omega_1 = 2500 rad/s\$$, or $$\f_1 = \frac{\omega_1}{2\pi} = \frac{2500}{2\pi} = 398Hz\$$. In other words, above this frequency gain increases at 20dB per decade.

There's also a pole, when the denominator is zero:

\begin{aligned} s &= -\frac{R_1}{C_1R_1R_4} \\ \\ &= -\frac{1}{C_1R_4} \\ \\ &= -\frac{1}{1 \times 10^{-6} \times 300} \\ \\ &= -3333 \end{aligned}

This represents a pole at frequency $$\\omega_2 = 3333 rad/s\$$, or $$\f_2 = \frac{\omega_2}{2\pi} = \frac{3333}{2\pi} = 530Hz\$$. It causes gain to decrease at a rate of 20db per decade at frequencies above 530Hz.

Because the zero we calculated above predicts a gain increase at this same rate, this pole's downward slope effectively cancels the upward trend from the zero, and we can expect a constant gain at frequencies well above 530Hz.

If we make the substitution $$\s = j\omega\$$, we can plug in values for frequency $$\\omega\$$ and find the gain at that frequency. We already calculated the gain at DC ($$\\omega=0\$$), but let's confirm it here:

\begin{aligned} H(s) &= -\left( \frac{sC_1(R_2R_4+R_3R_4+R_2R_3)+R_2+R_3}{sCR_1R_4+R_1} \right) \\ \\ H(j\omega ) &= -\left( \frac{j\omega C_1(R_2R_4+R_3R_4+R_2R_3)+R_2+R_3}{j\omega C_1R_1R_4+R_1} \right) \end{aligned}

As $$\\omega \to 0\$$, the imaginary parts diminish to nothing, the denominator tends to $$\R_1\$$, and the numerator tends towards $$\R_2+R_3\$$, giving us:

\begin{aligned} H(j\cdot 0) &= -\frac{R_2+R_3}{R_1} \\ \\ &= -\frac{200+200}{100} \\ \\ &= -4 \\ \\ \end{aligned}

That's what we saw before, but technically gain in the frequency domain is the ratio of amplitudes of output and input signals, which is the magnitude of this (potentially complex) result. So to be more correct:

\begin{aligned} \lvert H(j\cdot 0) \rvert &= \left\lvert -\frac{R_2+R_3}{R_1} \right\rvert \\ \\ &= 4 \\ \\ &= 20log(4) \text{ dB} \\ \\ &= 12.04 \text{ dB} \end{aligned}

Similarly, we can find the gain at very high frequency (assuming the op-amp has infinite bandwidth; in reality gain will drop off as the op-amp's limits are reached). As frequency approaches infinity, the transfer function's numerator and denominator are dominated by the imaginary terms, which become huge compared to the real parts. The function tends to:

\begin{aligned} H(j\cdot \infty) &= -\frac{C_1(R_2R_4+R_3R_4+R_2R_3)}{C_1R_1R_4} \\ \\ &= -\frac{1\times 10^{-6} \times (200 \times 300 + 200 \times 300 + 200 \times 200)}{1\times 10^{-6} \times 100 \times 300} \\ \\ G &= \left\lvert -\frac{16 \times 10^4}{3 \times 10^4} \right\rvert \\ \\ &= 5.333 \\ \\ &= 20 log(5.333) \text{ dB} \\ \\ &= 14.54 \text{ dB} \end{aligned}

To summarise, we expect a gain of 12.04dB at frequencies well below 398Hz, and 14.54dB at frequencies well above 530Hz. Here's a plot of the frequency response, created by simulating the circuit at the very top of this answer:

You can clearly see the op-amp's limited bandwidth causing gain to plummet after 100kHz, but prior to that, gains are as we calculated.

• Thank you for the detailed answer!
– n328
Sep 8 at 12:57
• +1 Nicely done! Sep 9 at 3:53
• Units must not be in italic.
– Carl
Sep 9 at 7:22
• @Carl Yes, they should not. It's tough remembering that every time, but I have corrected the dB unit format. Sep 9 at 7:29

It seems a weird thing ...
I don't have OP777 in my database, but LT1013 is a "similar" replacement.

The complete function (green) seems a little complicated.
But out_2 for what use?

The behavior is dependent, for example, on R5 resistor (stepped log, x10, from 1k to 1Meg) ...

Added the case where "out_2" is driven by "voltage" with internal resistor.

Stepping C ...

Try this topology ... (pot are used for varying "importance" of P or D effects)

• I will replace the Resistor by a power supply. It would be more "consistent". Keep on. Sep 8 at 9:02
• Hum ... a PD is "proportional-derivative" function. It is ... in the "frequency behavior" at 100 uHz -> 1 mHz ... Don't forget you have a 1 F capacitor ... Sep 8 at 9:33
• No. It works only for "1 decade" ... Bigger range would require another "topology". Sep 8 at 10:58
• Should read this dl.icdst.org/pdfs/files3/26329415489841fd7ba891bfc7e44c0f.pdf Sep 8 at 11:03
• ? Google "LECTURES ON PID CONTROLLERS" Sep 8 at 12:03