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I read about this phase shifter in "the Art of Electronics" and decided to run an LTspice simulation to verify if my understanding of the circuit is in line with reality.

enter image description here

According to LTspice the circuit is just below unity gain over 10Hz-100kHz band.

enter image description here

The dashed line shows phase. The continuous line shows amplitude, but notice the scale: Looking at µV's variation from a 1V(p) input.

My feeling however tells me R5 and C2 form a simple RC low pass filter and although fed by a symmetric input voltage tapped from the transistor's collector and emitter, I would still expect the output to drop at 6dB/oct from about 200Hz.

AC analysis:

  • As C2 current increases with frequency, the voltage across R5 increases.
  • Similarly the voltage across C2 decreases.
  • Amplitude fed into the R5/C2 at collector and emitter is flat with frequency.

What am I overlooking here, why is the output gain for higher frequencies still about unity? I am looking for a qualitative answer.

Version 4
SHEET 1 892 680
WIRE 288 16 112 16
WIRE 832 16 288 16
WIRE 112 48 112 16
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WIRE 288 160 288 144
WIRE -16 208 -208 208
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WIRE 224 208 112 208
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WIRE 528 208 512 208
WIRE 608 208 608 80
WIRE 608 208 592 208
WIRE 640 208 608 208
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WIRE 832 208 832 128
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WIRE -208 240 -208 208
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WIRE 288 288 288 272
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SYMBOL res 96 32 R0
SYMATTR InstName R1
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SYMBOL res 96 272 R0
SYMATTR InstName R2
SYMATTR Value 22k
SYMBOL res 272 32 R0
SYMATTR InstName R3
SYMATTR Value 2k203742
SYMBOL res 272 272 R0
SYMATTR InstName R4
SYMATTR Value 2k2
SYMBOL res 384 160 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 0 56 VBottom 2
SYMATTR InstName R5
SYMATTR Value 8k
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SYMBOL voltage -208 224 R0
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SYMATTR Value SINE(0 1 20000)
SYMATTR Value2 AC 1 0
SYMBOL cap 528 224 R270
WINDOW 0 32 32 VTop 2
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SYMATTR InstName C3
SYMATTR Value 10�
SYMBOL res 624 224 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 0 56 VBottom 2
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SYMATTR Value 10meg
SYMBOL cap 400 288 R270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName C2
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TEXT -242 410 Left 2 !.tran 0 0.2 0.1
TEXT -248 448 Left 2 !;ac dec 50 10 100000
\$\endgroup\$
9
  • \$\begingroup\$ You should post the plot, but in lieu of that, since you've already generated it, how about posting the circuit list so we don't have to generate it from your schematic? \$\endgroup\$
    – EM Fields
    Nov 29 '15 at 11:24
  • \$\begingroup\$ @EMFields I didn't include the plot because ... well it is flat. Hang on, I'll upload it. \$\endgroup\$
    – jippie
    Nov 29 '15 at 11:27
  • \$\begingroup\$ Does the voltage divider involving R6=10M have an effect? \$\endgroup\$
    – Chu
    Nov 29 '15 at 11:32
  • \$\begingroup\$ @Chu No I chose the values high enough to have very little effect. Notice that the output of the phase shifter is in the order of 10k, and the C3/R6-load order magnitude 10M. A factor 1000+ difference. \$\endgroup\$
    – jippie
    Nov 29 '15 at 11:36
  • \$\begingroup\$ OK, thanks... I'll get into it later. Right now it's about 0545 and my body is begging me to to put it to bed for a while. \$\endgroup\$
    – EM Fields
    Nov 29 '15 at 11:46
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My feeling however tells me R5 and C2 form a simple RC low pass filter and although fed by a symmetric input voltage tapped from the transistor's collector and emitter, I would still expect the output to drop at 6dB/oct from about 200Hz

At low frequencies R5 is dominant and the output is 180 deg phase shifted and unity gain (near enough) and at high frequencies C2 is dominant and produces a zero phase shift because the voltage feeding C2 is from the emitter.

There is no significant low pass filtering to consider here.

Consider that the voltage at the emitter is the same as the voltage at the base and call it Vin. The voltage at the collector is -Vin: -

schematic

simulate this circuit – Schematic created using CircuitLab

Therefore the voltage at the midpoint of R and C (Vout) is: -

\$V_{IN} - \dfrac{2V_{IN} \cdot X_C}{X_C+R} = V_{IN}(1-\dfrac{2X_C}{R+X_C})\$

Therefore Vout/Vin = \$\dfrac{R-X_C}{R+X_C}\$

This is all without using complex numbers but if you did the analysis of top and bottom of the equation above, the magnitudes are equal: -

enter image description here

The above picture is when R and Xc are vectorially added together. Now if Xc was negative (as per the numerator in the equation), Xc would, of course point upwards BUT, the length of -Xc and R added together is exactly the same magnitude.

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  • \$\begingroup\$ I indeed overlooked the fact that I can neglect C2 for low frequencies and R5 for high frequencies, thnx. That certainly answers the question for low and for high frequencies. Now what is left for me is trying to wrap my mind around the intermediate frequencies, say 100-1000Hz. \$\endgroup\$
    – jippie
    Nov 29 '15 at 11:54
  • \$\begingroup\$ jippie, see the transfer function in my post. It gives the answer (when both parts are effectiv). \$\endgroup\$
    – LvW
    Nov 29 '15 at 12:45
  • \$\begingroup\$ @RespawnedFluff jw is inbuilt into Xc i.e. Xc = \$\dfrac{1}{j\omega C}\$ \$\endgroup\$
    – Andy aka
    Nov 29 '15 at 17:56
  • \$\begingroup\$ I realized that, but then you also claim "without using complex numbers"... Your Vout/Vin is a complex number! If you calculate its magnitude... it's exactly 1, of course. \$\endgroup\$
    – Fizz
    Nov 29 '15 at 17:58
  • \$\begingroup\$ @RespawnedFluff it was without using complex numbers - at that point Xc could be just regarded as "Z". I don't understand your point. \$\endgroup\$
    – Andy aka
    Nov 29 '15 at 17:59
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Here is the LTspice illustration of the equivalent circuit that LvW analyzed (compared to the original transistor circuit):

enter image description here

Note that in the original circuit Vc and Ve do shift a little bit (in the same direction) at the transition, but this can be ignored as you see.

And if you want "proof by LTspice" of this latter superposition problem, you could do the following by stepping the AC voltage sources:

enter image description here

In these plots I've selected just the (Vc=1, Ve=0), (Vc=0, Ve=1) and finally (Ve=1, Vc=1) data.

The intuition behind solving this latter equivalent circuit this is simple: whatever magnitude you low-pass from the Vc source you add it to the 180-shifted magnitude that you high-pass from the Ve source. So their sum is constant in general. The formal proof of this is in LvW's answer. If you want to make LTspice "prove" this sum you need to put both partial circuits explicitly on the same schematic as shown below. (I don't know how to make LTspice add parametrized results for different parameter values.)

enter image description here

To pick a particular abscissa point to verify this numerically, say, corresponding to the -3dB magnitude (\$=\frac{1}{\sqrt{2}}\$, which is the same for both filters obviously) the phase of the low-pass-ed signal is -45 degrees there, and the phase of the high-passed one is -135 degrees. As expected this gives a magnitude of 1 and -90 degrees phase:

$$ \frac{1}{\sqrt{2}} \angle {-45}^{\circ} + \frac{1}{\sqrt{2}} \angle {-135}^{\circ} = \Big(\frac{1}{2} - \frac{1}{2}j \Big) + \Big(-\frac{1}{2} - \frac{1}{2}j \Big) = -j = 1\angle {-90}^{\circ} $$


Finally, here's a re-stated proof that doesn't resort to s-domain (since you don't really need it here), nor use original definitions of reactance. You have two voltage dividers, one for each of those sources

$$ V_{co} = V_{c} \frac {\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}} = V_{c} \frac {1}{1+j\omega RC} $$

$$ V_{eo} = V_{e} \frac {R}{R+\frac{1}{j\omega C}} = V_{e} \frac {j\omega RC}{1+j\omega RC} $$

Since \$V_c=-V_e\$, and by superposition:

$$ V_{sum} = V_{co} + V_{eo} = - V_{e} \frac {1}{1+j\omega RC} + V_{e} \frac {j\omega RC}{1+j\omega RC} = V_{e} \frac{-1+j\omega RC}{1+j\omega RC} $$

It makes more sense to express the transfer function relative to \$V_e\$ than to \$V_c\$ because the former is in-phase with the input signal (transistor's base).

$$ \frac{V_{sum}}{V_{e}} = \frac{-1+j\omega RC}{1+j\omega RC} $$

Now this is a complex number of the form \$ \frac{-z^*}{z}\$ (where * denotes the complex conjugate), so its magnitude is always 1; this is a trivial math fact since \$z=a+bi\$ and \$-z^*=-a+bi\$ have the same magnitude, namely \$|z|=a^2+b^2 = (-a)^2+b^2=|-z^*|\$.

The phase is a different matter, and it's not constant since it is basically the function

$$ f(x) = \arg\Big(\frac{-1+jx}{1+jx}\Big) $$

enter image description here

Clearly at \$x \to 0\$ it is the phase of \$-1\$ so \$\pm \pi\$, i.e. plus or minus 180 degrees. So it starts [at low frequency] in exact opposite phase of the emitter voltage, i.e. in phase with the collector voltage. As \$x \to \infty \$ the limit is zero, so it becomes in phase with the emitter voltage at high enough frequency.

To do get more insinght by hand, first note that \$\arg(z_1/z_2) = \arg(z_1)-\arg(z_2)\$ (See eq 10 in this handout and proof idea). So

$$f(x) = \arg(-1+ix) - \arg(1+ix) $$

For \$x>0\$, the right half of this subtraction this is just \$ \arctan x \$, but the left is \$ \pi + \arctan (-x) \$. Since \$\arctan (-x) = -\arctan(x)\$, you finally have

$$ f(x) = \pi - 2 \arctan (x) $$

This is the same function but only for \$x>0\$. I guess that's why WA didn't automatically simplify it like this.

enter image description here

To actually make it look like in the Bode plot, you need to do a semilog plot, which I don't know how to do in WA (I think that requires the paid version), but... gnuplot does this easily:

gnuplot> set xrange [1:1E5]       
gnuplot> set logscale x                      
gnuplot> plot (180 / pi) * (pi - 2 * atan(2 * pi * 8E-4 * x)) 

8E-4 is the RC constant in this case.

enter image description here

I somewhat unwisely set Vc to be 0 phase in the previous LTspice sims/plots when I should have done that with Ve. With that change the theory agrees with the simulation perfectly:

enter image description here

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It is a first-order allpass filter with the following properties:

  • Both output voltages (magnitudes) are equal (collector resp. emitter): Therefore, v,c=-v,e.

  • Using the superposition principle, the output voltage across the common load resistor R6 (sum of a lowpass and a highpass output) is

    v,out=v,e[sR5C2/(1+sR5C2)]-v,e[1/(1+sR5C2)]

  • Hence: v,out/v,e=(sR5C2-1)/(sR5C2+1).

    This first-order allpass function has a unity magnitude (equal magnitudes for numerator and denominator) for all frequencies - as long as the frequency-dependence of the transitor gain can be neglected.

Note: The influence of the very large load resistor R6 was neglected.

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5
  • \$\begingroup\$ I haven't used s-domain for over 20 years, so your second bullet is a bit too summarized for me to grasp. \$\endgroup\$
    – jippie
    Nov 29 '15 at 12:17
  • \$\begingroup\$ But it is nothing else than simple math (common denominator for two expressions). And, as the next step, the magnitude is found for s=jw. Thats all. \$\endgroup\$
    – LvW
    Nov 29 '15 at 12:48
  • \$\begingroup\$ @jippie: wolframalpha.com/input/… The part assuming R, C, w positive being relevant here. If you're still confused about that, ask M.SE cause it's just a simple complex math problem at this point. \$\endgroup\$
    – Fizz
    Nov 29 '15 at 13:53
  • \$\begingroup\$ @LvW I think the formula should be H(s) = \$\dfrac{sCR-1}{sCR+1}\$ \$\endgroup\$
    – Andy aka
    Nov 29 '15 at 17:28
  • \$\begingroup\$ @Andy, thank you - you are right. The magnitude remains the same, but the phase starts at -180 deg. I have corrected the formulas. \$\endgroup\$
    – LvW
    Nov 29 '15 at 17:59

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