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In the Art of Electronics (Horowitz & Hill) the following circuit is given as a solution to reduce ripple current caused be variations in input voltage (pg. \$ 69 \$):

Reducing Ripple in the Zener Regulator

"An alternative method uses a low-pass filter in the zener bias circuit (Fig. 2.13). \$R\$ is chosen to provide sufficient zener current. Then \$C\$ is chosen large enough so that \$RC >> \dfrac{1}{f_{ripple}}\$. (In a variation of this circuit, the upper resistor is replaced by a diode.)"

I don't understand the need for the upper resistor R. In order to minimize variations in input voltage, couldn't we just connect the capacitor directly from Vin to ground? Why do we need to make an LPF with R instead? Since it is mentioned that we could use a diode instead, is there really a point?

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  • \$\begingroup\$ Remember that \$V_{in}\$ is a low impedance source, without R the voltage won't be smoothed that much. Why it can be replaced by a diode is a mistery to me. \$\endgroup\$ Aug 3, 2014 at 14:26
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    \$\begingroup\$ Ok now I've got it: without the upper R there would be no smoothing, while putting a diode there would result in a sort of peak/envelope detector instead of a lp filter. The lower R is there to bias the zener, and that's it. \$\endgroup\$ Aug 3, 2014 at 14:29
  • \$\begingroup\$ Thank you! I'm still somewhat confused about the use of the diode but I will try to figure it out based on what you said. I do have one other question though. The text says to choose Rc so that the voltage drop across it is less than the drop across the lower R for the highest normal load current. Could you perhaps shed some light on that? \$\endgroup\$
    – Ammar
    Aug 3, 2014 at 14:46
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    \$\begingroup\$ About the diode see this. About \$R_C\$ that's to keep the collector voltage well above the base voltage thus working in the transistor active region. \$\endgroup\$ Aug 3, 2014 at 14:50
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    \$\begingroup\$ Call the top R R1, the bottom R2. If the source is zero impedance and the Zener impedance is much less than the resistor(s), the Thevenin equivalent seen at the capacitor is R1||R2, so the optimal (minimizing the capacitor size for a given smoothing) is R1==R2 = R. \$\endgroup\$ Aug 3, 2014 at 17:18

2 Answers 2

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note: This answer also addresses some issues mentioned in the comments of the question, have a look there too.

I'll redraw the schematic for you:

schematic

simulate this circuit – Schematic created using CircuitLab

Note that I've added \$R_S\$ ('s' as in source). Let's analyze the circuit starting from the output.

Q1 is in common collector configuration, meaning that its voltage gain is near unity while current gain is much higer, given that it's in active region. The voltage across \$R_{load}\$ is approximately the base voltage diminished by a \$V_{BE}\approx0.7\text{V}\$.

Assuming that DZ1 is working properly the base voltage is set by it. The rightmost R must provide enough current for the diode and for the BJT though.

The diode and transistor bias current is drawn from the LP filter output. The LP filter is a classic RC first order filter, note that its output sees a load approximately equal to R (DZ1 is a short circuit for small signals).

The LP filter is powered by the PSU, that also feeds the bjt collector through \$R_C\$.

Now to your questions:

Why is the leftmost R needed?
That's because \$R_s\$ is very small. Using only a capacitor would result in an LP filter but its corner frequency would be too high. A source output resistance is not something you want to rely on anyway, that's probably not well characterized.

Can we use a diode instead of the leftmost R? If yes, why?
You can use a diode instead of R effectively building a peak detector.

What is better? Diode or resistor?
Honestly, I am not sure. I guess the answer lies mainly in the PSU spec: keep in mind that a diode there would not limit the current going into the capacitor, that might be a problem. A diode would be slower to follow descending peaks of the PSU but that should not actually be an issue.

Why voltage drop across \$R_c\$ should be less than voltage drop across rightmost R?
That's to keep the transistor in active region. If the collector voltage goes too low (high collector currents -> high drop across \$R_C\$) the transistor may saturate and stop working properly. I think the circuit can work without \$R_C\$.

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An RC lowpass always consists of an R-C series connection (C grounded) - and the output is BETWEEN both parts. Without this resistor you have no frequency-dependent voltage divider (because the internal source resistance is assumed to be zero).

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  • \$\begingroup\$ Thank you for replying. Well that makes sense but isn't there always a source resistance in real life? Even if it is really small, it would create an LPF. However, would that be poor designing? EDIT: I see that the cutoff frequency would be too high then. \$\endgroup\$
    – Ammar
    Aug 3, 2014 at 14:40
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    \$\begingroup\$ Exactly - of course, thetre is always a source resistance. However, who knows the value? Thus, it is always better to DESIGN the cut-off frequency rather than to hope that it will work. \$\endgroup\$
    – LvW
    Aug 3, 2014 at 16:05

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