I found a schematic which had this bit of circuitry between the +9 V (which apparently the author assumed to be > 9 V, more like 10-11 V) barrel connector and the +9 V supply rail.

I tried to simulate what's happening in LTspice, but I don't think I modelled it right, as even the forward voltage drop of the polarity protection diode is not visible in the output. My current guess is that the base-emitter voltage is driven to saturation (somehow) so that it's just a ~1 V linear dropdown regulator?

Enter image description here

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    \$\begingroup\$ Looks like a capacitance multiplier for ripple removal \$\endgroup\$ Nov 14, 2020 at 13:50
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    \$\begingroup\$ youtube.com/watch?v=wopmEyZKnYo \$\endgroup\$
    – G36
    Nov 14, 2020 at 13:51
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    \$\begingroup\$ "... even the forward voltage drop of the polarity protection diode is not visible in the output." No current --> no voltage drop. Add a load resistor. \$\endgroup\$
    – Transistor
    Nov 14, 2020 at 13:57

6 Answers 6


The circuit topology is that of a capacitance multiplier designed to remove ripple from a 'dirty' regulated power supply. The name stems from the fact that the value of the capacitance seen from the output terminals is a multiple (not exactly beta times) of the capacitance seen from emitter and base.
In its more 'general form' it allows for a re-scaling of the output voltage via a voltage divider, like this:

Cap multiplier

Here I added a 0.2 Vpp 50 Hz ripple to the 9V input voltage and I get 6.3 V output with just 0.5 mVpp ripple superposed (this is an old simulation I did years ago). Here is the result of the simulation from the moment the sinusoidal signal is applied:

simulation from start up

As you can see from the top graph the output voltage is basically a flat line at 6.3V.
In your circuit the diode add a polarity inversion protection, so you will see a lower voltage, according to the current supplied.

The single resistor version
(edited after reading other answers and comments)
To address the difference between the multiplier with R1 and R2 and your design, note that the capacitor sees a resistance across its terminal in your design as well. It's the resistance seen from the base of the transistor. That is enough to keep the transistor out of saturation (as long as the ripple is small, as it usually is).

Here is a simulation with a 6.3V input with ripple of 0.2 Vpp @ 50 Hz and a load of 220 ohm; I also added a Schottky diode as polarity inversion protection and to bring the voltage at almost exactly 5V.

Cap multiplier with implicit R2

Note that this design is highly dependent on the value of RL since the setpoint depends on the implicit voltage divider former by R1 and (basically) beta times RL.

As noticed in another answer, this circuit requires a bit of time to reach its steady state value, so that along with ripple suppression it adds a soft-start functionality. This might be an intended function (or even the intended function) of your circuit. The simulation with the DC supply turned on at t=0 shows that:


But even with a single resistance, a capacitance multiplier still works well as a ripple suppressor circuit if the load is low enough (more on this later) and the ripple is reasonably small. Here is the output of the same simulation a few seconds after startup:

ripple suppression

Ripple is basically negligible and completely symmetric. Also, the transistor does not even near saturation, with Vce always above 1V.

Vce and Ib

Why "capacitance multiplier"?
(edited after reading other answers and comments)
From the point of view of the load, the capacitance appears multiplied by a factor K. Of course the capacitor does not get physically bigger, nor does its energy storage capacity change. What this circuit does is boosting the current in the load so that the capacitor voltage that is replicated in the output (minus Vbe) does not need to change as much as if the charge had to be supplied entirely by the cap.
To show that, under suitable conditions, the corner frequency is lowered with respect to a passive RC filter, here are two circuits side by side:

comparison passive active

(I used an ideal transistor in order not to bother with BJT parasitic capacitances.)
Now, loading is the key to understand the shift in corner frequency: the simple RC circuit is heavily loaded by RL, so much that the steady state output voltage is about one hundredth of the input voltage (that's about 40 dB attenuation). The time constant is (R1//RL)*C and the corner frequency of this circuit is about 160 Hz.
Note that the boosted circuit does not attenuate that much. The capacitor is loaded with beta + 1 times the value of RL, which for a beta of 100 gives an equivalent load of about 1 kohm; this brings the steady state voltage at half the value of Vin, i.e. an attenuation of 6 dB. The time constant is accordingly increased to (R1//(beta+1)RL)*C and so so the corner frequency drops to about 3 Hz. A factor of more than 50.

shift in corner frequency

In the above Bode plot, the passive circuit's response is plotted in blue, while the capacitance multiplier circuit's response is in brown. From the point of view of the load, it is as if the capacitance has been multiplied by more than fifty times. If we plot the response of a passive circuit with 50+ times the capacitance we get a bode plot that shows the attenuation of the R1-RL divider, but has the same corner frequency as the capacitance multiplier (it's shifted 40 dB below).

Why is loading key? Well, the bigger RL is, the less current it requires, the less important is the current boost. The "multiplication" effect gets smaller and smaller as the load gets bigger and bigger. Therefore, for example, if we increase the load to RL = 100k, the cap won't even notice the difference in loading and in fact the Bode plot of the passive circuit and the boosted one will not differ at all, neither in the steady state voltage, nor in the the corner frequency. You also won't notice a difference in absolute ripple reduction between the circuits.

Passive and boosted responses with RL = 100 kohm
The frequency response of the passive and boosted RC filter for RL = 100 kohm are basically the same

But when the loading is significant, the circuit (even with one resistor) does act as if the capacitance is multiplied. This effect is akin to the bandwidth widening associated with negative feedback. The difference is that instead of introducing attenuation, the transistor circuit removes it.

You might want to read the 1985 article from the Designer's Notebook column of Radio Electronics:

"A simple solution to power supply's ripple: A Capacitance Multiplier"
Designer's notebook
Radio Electronics 3, 1985

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    \$\begingroup\$ A little off-topic, but are these really the colors you use in LTspice? \$\endgroup\$
    – pipe
    Nov 16, 2020 at 15:49
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    \$\begingroup\$ @pipe these are the colors I use everywhere, all my windows have a green background: txt, doc, keepnote, microcap, ltspice,... all programs that allow me to change the background colors. It helped my a lot with eyestrain. I couldn't tolerate that glaring white background so many applications try to force on their users. (So, when I grab a screenshot of a circuit in Keepnote it merges nicely with its background) \$\endgroup\$ Nov 16, 2020 at 15:52
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    \$\begingroup\$ Ok, I have to admit that's pretty cool, but still... \$\endgroup\$
    – pipe
    Nov 16, 2020 at 15:54
  • \$\begingroup\$ Could D1 be needed to block C1 discharging through the B-C junction during low ripples? There's no series resistance for that path, so a low-enough ripple could give negative Vce and turn that junction into a forward-biased diode. (The circuit in the question has no Cin filter cap on the unregulated side, only the small 0.1uf across the load. But without a diode, your Cin could also discharge along with C1.) \$\endgroup\$ Nov 16, 2020 at 20:01
  • \$\begingroup\$ @PeterCordes seems plausible. It all depends on what is powering it. The diode might be a cost-effective solution to avoid having a couple more components in the wall-wart that went with the product. I have seen wall wart made of a transformer and one diode. \$\endgroup\$ Nov 16, 2020 at 20:25

"Capacitance multiplier" is a striking but somehow confusing name for beginners who can take it literally. So it needs to be well clarified what exactly this circuit does.

At first glance, there is nothing special here - the voltage of an RC integrating circuit is buffered by an emitter follower. Instead of using a capacitor, the same could be done with a Zener diode... and even by a battery (like in the picture below)…

Battery backup voltage regulator

Fig. 1. Battery backup voltage regulator

… but we do not name them "Zener diode multiplier" and "battery multiplier"...

So, the question arises, "Why has it not been done that way?" Let's try to find the answer...

Both Zener diode and battery "produce" a constant reference voltage which is much lower than the input unfiltered voltage. Thus, a significant voltage drop Vce appears across the collector-emitter part of the transistor and significant power is dissipated. Also, the DC input and output voltage are constant which in some cases can be undesirable.

The clever trick here is that the reference voltage across the capacitor is self-adjusted to stay close to (to follow) the input voltage (Vc = V1 - Ib.R1); so, the voltage drop Vce < 1 V. As a result, the output voltage follows only the input DC voltage and not the ripples… and the circuit acts as a kind of a buffered RC low pass filter.

So, the capacitance is not multiplied; it stays the same. Only the time behavior of the weak small capacitor is copied by a powerful variable voltage source. If we talk about a capacitor here, it is rather a "virtual capacitor" emulated by a voltage source. This technique is widely used to create various virtual elements; the gyrator (simulated inductor) is a typical example.

Finally, I want to share that I have fabricated the above explanations at the moment relying on my experience and common sense. I only saw this circuit somewhere in the 70's, I think under the name "electronic filter" or something like that... but then it did not impress me so much.

Now I have taken the trouble to watch two movies on YouTube (the Dave's movie suggested by C36 and the FesZ's movie) and I saw that I had basically guessed the idea.

EDIT 1: I would add another observation to @fraxinus's answer and the comments below it in favor of the single resistor R1 vs the voltage divider.

The ripple magnitude depends on the load current since the higher the current, the faster filter capacitor of the diode rectifier is discharged. Higher load current means a larger base current and higher voltage drop across the resistor R1. So the base voltage decreases and the voltage drop VCE (the voltage reserve) increases. As a result, the transistor will remain in active mode.

A kind of automatism is obtained - the reserve of voltage VCE decreases when the ripple increases, which keeps the transistor in active mode.

EDIT 2: @Horror Vacui said:

"There are capacitance multiplier circuits, which really does increase the effective capacitance on a given node through some form of feedback. Typically they measure the current through the capacitor and then draw a multiple of that current from the node where the capacitance is connected to."*

This text made me open an old thick book (Linear Applications Handbook, National Semiconductor, 1986) and go back to such аn exotic capacitance multiplier circuit. It was described there by Bob Widlar on page 67 in his article IC Op Amp Beats FETs on Input Current (AN 29). You can find this circuit here (bottom figure at page 17):

Capacitive amplifier

Fig. 1. Op-amp capacitive amplifier

Let's try to see the idea behind this circuit solution. Here is a possible explanation:

An RC integrating circuit (high resistance R1 and low capacitance C1 in series) is driven by an external voltage source through a resistor (e.g., an external pull-up resistor connected between the right end of R1 and the positive supply rail); so the voltage across C1 slowly rises. It is copied by an op-amp follower (LM 108) and returned through a low resistance R3 to the input of the integrating circuit. What is the point of this?

The RC integrating circuit serves only as a "shaping element" producing an input voltage for the op-amp follower thus making it behave through time like a capacitor. The voltage across this "virtual capacitor" is the same as the voltage across the true capacitor C1 but the current through it is R1/R3 times bigger. So, the virtual capacitor has R1/R3 times bigger capacitance… as though the C1 capacitance is multiplied; hence the name "capacitance multiplier".

It only remains to explain how the current through the virtual capacitor is increased R1/R3 times. I feel like I have seen this circuit trick... but it does not matter, we can figure it out ourselves now.

Look at R1 and R3. They are connected through one of their ends and the others are under the same voltage... so they are "virtually connected"... and as if the two resistors are connected in parallel. What is this network?

Yeah, it is a current divider where the current flowing through R3 is roughly R3/R1 times bigger than the current through R1. Only these currents go to different places here (the small current to C1, the big current to the op-amp output).

EDIT 3: At the top of the same page, you can see something intriguing - a very similar circuit solution of a negative capacitance multiplier:

Negative capacitance amplifier

Fig. 3. Negative capacitance amplifier (there is a small typo in the expression - R2 should be R1)

It is interesting that this circuit uses the same components... and the original capacitance C1 is multiplied the same R1/R3 times... but the result is a negative virtual capacitance. What does it mean?

The explanation is simple: The "positive capacitance multiplier" is a virtual capacitor "creating" a voltage drop (loss) across itself that is subtracted from the input current-creating voltage source while the "negative capacitance multiplier" is a virtual capacitor "creating" a voltage (gain) across itself that is added to the input voltage source. But how does the op-amp do this magic?

If this was a circuit with only a negative feedback introduced by R3 (the non-inverting input was grounded), then it would add zero voltage to the input circuit. There was a virtual ground at the inverting input (transimpedance amplifier). But here, an additional positive feedback is introduced by the R1-C1 feedback network. It causes the op-amp to lower its output voltage below ground until it reaches balance (equality between the op-amp input voltages). As a result, the circuit input voltage is reversed; hence the name of this "mystic" circuit consisting of the op-amp and the two resistors R1 and R3 - negative impedance converter with voltage inversion (VNIC).

  • \$\begingroup\$ nice explanation! but can you please elaborate the line "As a result, the output voltage follows only the input DC voltage and not the ripples" \$\endgroup\$
    – partykid
    Apr 8, 2021 at 11:12
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    \$\begingroup\$ @partykid, Thanks for the response! I have good memories of your reflections on the role of the operational amplifier in the inverting integrator. Here I mean something obvious - that the RC circuit smoothes the input ripples and the output voltage does not react to them. Maybe I should remove this "As a result" in the beginning of the sentence... \$\endgroup\$ Apr 8, 2021 at 20:24

No, it is not a capacitance multiplier.

In order to be one, it needs a voltage divider to ensure enough Vce for the transistor.

Here, we don't have one and cannot reliably keep the transistor in active state.

What we have here is a soft power-on circuit. On power-on, the output voltage slowly (determined by C1R1 time) rises to near input voltage.

Yes, it does have some ripple smoothing effect, but it is asymmetric and not really capacitance-like.

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    \$\begingroup\$ The second resistance is the resistance seen from the base of the transistor (you need the load to see that happen). As a matter of fact, the 'transposed' load resistance interferes with the divider design in the circuit I have shown above. If your ripple is not excessive you can safely omit the second resistance and design the circuit so the BJT won't even near saturation. (Just did a simulation with RL = 220 ohm, R1=2.8k and C= 500 uF; with 6.3V input I had to add a Schottky diode to bring the voltage at 5V - Vce does not go under 0.9V). \$\endgroup\$ Nov 15, 2020 at 13:37
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    \$\begingroup\$ @fraxinus R1 provides base current. The emitter of Q1 then follows the base voltage less \$V_{be}\$. \$\endgroup\$ Nov 15, 2020 at 14:03
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    \$\begingroup\$ @MathKeepsMeBusy try to follow the base voltage with ~0.5V ripple at the input. It won't work. \$\endgroup\$
    – fraxinus
    Nov 15, 2020 at 14:09
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    \$\begingroup\$ @fraxinus see the edit in my answer. Yes, the ripple needs to be reasonable. 1Vpp of ripple in my eye is not reasonable. But if you simulate with 1Vpp sine ripple my second circuit, it still manages to kill it - somehow - and Vce does not go under 1V either. \$\endgroup\$ Nov 15, 2020 at 14:18
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    \$\begingroup\$ @fraxinus In practice, this kind of circuit works fine when you have a few millivolts of ripple, and you need to reduce that to microvolts. \$\endgroup\$
    – John Doty
    Nov 16, 2020 at 18:14

This circuit is a low-pass filter (R and C) followed by a voltage buffer. Some people call it a capacitance multiplier, but it does not multiply the capacitance.

There are three problems with the simple RC low-pass filter:

  1. The 3 dB cut-off will depend on the capacitance of the load. (\$C_{eff} = C_{filt} + C_{load}\$)
  2. The load current will cause a voltage drop. It's worse than a simple voltage drop, because it will change with the load current. This might lead to supply noise, and the time varying supply current of one circuit block will modulate the other blocks. This voltage drop is proportional to the resistance value, and hence it sets an upper limit on the resistance value, which supposed to be big for a low cut-off frequency.
  3. The small-signal resistance of the load (\$r=\delta{v_{out}}/{\delta{i_{load}}}\$) is parallel with the capacitor and shunts it at frequencies below \$1/(2\pi\cdot r \cdot C_{eff})\$. The attenuation below this frequency will be constant.

Once we add a voltage buffer after the RC filter, both of these problems are solved. The simplest voltage buffer is a source/emitter follower circuit or common drain/collector amplifier. If its voltage drop is not acceptable, than an opamp in unity gain configuration can be used. It has the very same function.

The opamp buffer is great, and it does not have any drawback function-wise. Why isn't it used here? Probably, because it might be considered a bit more expensive and it needs an additional supply and supply current.

If a bipolar transistor is used as an emitter follower (common collector amplifier) its base current still flows through the resistor. Also the loading effected of the circuit is still there, just mitigated by the current gain of the bipolar transistor (\$\beta\$). The voltage drop limit on that resistor is also only relaxed by \$\beta\$. A MOS device has ideally infinite \$\beta\$, but its current capability is lower than a similar bipolar transistor --- though there are excellent high current MOS devices nowadays ---, and it will have a higher gate-source capacitance than the base-emitter capacitance of a bipolar transistor with similar current rating. Due to the buffer effect, one can use \$\beta\$ times higher resistor for the same capacitor, which leads to a \$\beta\$ times lower cut-off frequency. This is equivalent of \$\beta\$ times higher capacitance value for the resistor, hence the name. Typically one uses the highest available capacitance there, so I guess this also contributed to the proliferation of this name.

Does it multiply the capacitance? No, the cut-off frequency is still determined by the RC product, but the loading effect has been removed, and there isn't any voltage drop trade-off on the series resistor. This is why I am personally against this misleading name.

There are capacitance multiplier circuits, which really does increase the effective capacitance on a given node through some form of feedback. Typically they measure the current through the capacitor and then draw a multiple of that current from the node where the capacitance is connected to. These circuits are working within the bandwidth of the feedback loop, but in many cases it is enough. They are also usually used in filter circuits.

I have not mentioned D1 before, because it's not critical in the answer. I've assumed that question is more about the transistor and RC. What does D1 do there? Either the input is an AC source and D1 is a one-way rectifier or the output could be higher than the input under some circumstances and D1 blocks any reverse current.

  • \$\begingroup\$ Valuable observations... I agree with them. In addition to the mentioned capacitance multiplier circuits, we can consider the well-known op-amp inverting integrator as another capacitance multiplier. The trick here is that the op-amp output voltage is added in series to the voltage drop across the capacitor thus compensating it. The result is something like a "bottomless capacitor" - the current continously charges the capacitor but, as though, it does not charge... and the voltage (virtual ground) does not change... \$\endgroup\$ Nov 15, 2020 at 21:49
  • \$\begingroup\$ about the name: it's just a name, I don't think anybody would expect the capacitor to become physically bigger (jk). But the effect of transforming Rload into beta+1 times Rload alters the time constant of the capacitor, let's say of a factor K. So, if in the R1CRL filter the time costant was tau = Req C, in the capacitance multiplier it will be tau' = K Req C. From the point of view of the load you can see it as tau' = Req (K C), so it's as if the capacitance it sees has been multiplied by K. \$\endgroup\$ Nov 16, 2020 at 3:55
  • \$\begingroup\$ As for the corner frequency: you can see by killing the attenuation, it shifts much in the same way the bandwidth is widened by negative feedback (actually in reverse, since with the transistor we remove the attenuation). As for the feedback trick we can consider it 'a trick' because it does not change the physics of the capacitors, it just changes the position of the 'plateau', but it can surely be seen as a change in corner frequency (it all depends on how much RL loads the capacitor - or the divider in the version with R1 & R2). So, it won't be beta times, but a more complicated expression. \$\endgroup\$ Nov 16, 2020 at 4:00
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    \$\begingroup\$ @Circuitfantasist exactly. It is also called Miller effect :). Many students are actually paralyzed by this name, though it is nothing else than forcing an input dependent voltage to the other end of the capacitor. This effect widely used to internally compensate opamps. \$\endgroup\$ Nov 16, 2020 at 10:00
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    \$\begingroup\$ @Circuitfantasist those works seems interesting. They does not fit in my coffee break :), but I do hope I can read through them soon. I will get back to you once I did that. \$\endgroup\$ Nov 17, 2020 at 11:00

I've seen this used as a "super capacitor", to provide slightly lower VDDs, but with much cleaner (much less 60 Hz and 60,000 Hz supply ripple). In particular, this was in a set of RF ICs that provided GSM basestation functions, where a deterministic spurious would prevent FCC (and European) certification.

After accepting the few hundred millivolts of VDD reduction, the clean VDD was a delight to use.

Where does 60,000Hz ripple appear? switch_regs.

  • \$\begingroup\$ 60 kHz supply ripple? \$\endgroup\$ Nov 16, 2020 at 14:49

Your circuit is an active low pass filter aka "capacitance multiplier". It's purpose is to reduce voltage ripple.


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