# Circuit paradoxes – Or are they?

-December 03, 2014

I am sure that at some point all of us have come across puzzling circuit situations that, at first sight, seem to be plain absurd.  I want to share some I’ve run into during my student days (usually very late at night).

A frequency-independent capacitive impedance?

It is a well-known fact that a capacitance in the feedback path of an inverting amplifier, reflected to the input, appears magnified by the Miller effect.  So, the impedance Zi seen looking into the inverting input node of the circuit of Figure 1 should be capacitive, and therefore decrease with frequency at the rate of –1 dec/dec.  Yet, the accompanying Bode plot reveals a frequency-independent input impedance of 16 Ω.  What’s going on here?  Has the Miller effect gone on strike?  And where do those 16 Ohms come from?

Figure 1 Frequency plot of Zi.  Has the Miller effect gone on strike?  (Or hasn’t it?)

A strange differential amplifier – to say the least

So long as the open-loop gain a of the op amp of Figure 2a is infinite, the circuit gives V(O) = V1 – V2, so tying the inputs together to make V2 = V1 as in Figure 2b gives V(O) = 0, indicating an infinite common-mode rejection ratio (CMMR = ∞).  But what if a ≠ ∞?  Well, it turns out that the circuit of Figure 2b gives V(O) = 0 for any value of a within the range ∞ > a > 0.  Can you use physical insight to explain why?

Figure 2
A difference amplifier capable of infinite CMRR with only a finite open-loop gain a?

Actually, there’s more to it, because this circuit keeps giving V(O) = 0 even as a turns negative, in which case feedback becomes positive.  This is illustrated in Figure 3 for the case of an op amp with a dc gain of a0 = –1 V/V.  To verify the stability of this circuit, assume the op amp has a 1-MHz pole frequency, and subject the circuit to a small current disturbance, after which V(O) returns to zero.  Can you explain why this circuit is stable even though feedback is positive?

However, if you make a0 even more negative, the circuit becomes unstable.  This too is illustrated in Figure 3 for a0 = –3 V/V, in which case the disturbance results in a diverging response.  Why is it so?  What is the borderline value of a0 between converging and diverging responses?

Figure 3
A stable circuit with positive feedback?

Can two wrongs make a right?

It is well known that op amp differentiators are to be avoided because of their notorious tendency to produce intolerable ringing and gain peaking [1].  It is also known that one should avoid using voltage comparators in negative-feedback operation because they are meant for open-loop operation and as such they lack the frequency-compensation requisites for stable negative-feedback operation [1].  Yet, the circuit of Figure 4 uses precisely a voltage comparator to provide fairly accurate and stable differentiation, as demonstrated by the accompanying waveforms.  How come?  What happened to the saying that two wrongs don’t make a right?  Where is the frequency-compensation network stabilizing the comparator?

Figure 4
A differentiator implemented with a voltage comparator.

Getting straight lines out of curves?

The circuit of Figure 5 is nonlinear because it includes diodes.  However, if we focus our attention to operation within the range –4 V < vI < +4 V, we see that all diodes are on, in which case they approximate short-circuit behavior.  (I have specified an unusually large saturation current for the SPICE diode model D, so the diode forward voltage drops for the currents of this circuit never exceed a couple hundred millivolts.)  Given that for –4 V < vI < +4 V all voltages are straight lines (see top traces), the resistance currents are also straight lines, by Ohm’s law.  Consequently, the diode currents, which (by KCL) seem to be combinations of resistance currents, ought to be straight lines as well.  Yet, the bottom traces reveal nonlinear diode currents!  What’s going on?  Has KCL gone on strike?  Or is this a SPICE artifact?  Or is it just a late-night hallucination?

Figure 5
Diode bridge circuit

Shouldn’t this circuit oscillate?

The circuit of Figure 6 simulates an amplifier with a dc gain of 80 dB, two pole-zero pairs, and an additional pole.  Moreover, it saturates at ±10 V.  Its Bode plots reveal two frequencies at which the output is delayed by 180° with respect to the input.  Using PSpice’s cursor facility, we find these frequencies to be about 27 kHz and 60 kHz.  Moreover, the gains at these frequencies are, respectively, V(O)/V(I) = –370 V/V and V(O)/V(I) = –48.3 V/V.

Figure 6
An open-loop gain with three poles and two zeros and ±10-V saturation voltages.

If we now apply unity-feedback around this amplifier as in Figure 7 (top), we expect any noise arising inside the feedback loop with frequency components of 27 kHz and 60 kHz to get magnified, respectively, by 370 V/V and 48.3 V/V each time it goes around the loop, thereby leading to two diverging responses.  Because of the saturation limits of ±10 V, we expect the circuit to achieve a steady-state situation with two modes of oscillation in the vicinity of 27 kHz and 60 kHz, respectively.

Figure 7
Configuring the amplifier of Figure 6 for unity-gain operation.  Frequency response (top), and unit-step response (bottom).

Well, looking at the frequency and transient responses of Figure 7, we see that we have a fairly stable circuit.  Can you justify this intuitively?  Pretend you are explaining this circuit to an enthusiastic humanities major – possibly your significant other; so, no Nyquist criteria, no Cauchy arguments, no esoteric mathematical tools – use just physical intuition (if you can).

References

1. Design with Operational Amplifiers and Analog Integrated Circuits, 4th Edition, Sergio Franco, San Francisco State University

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