Capacitance multiplier (#1116): Cool, but how does it work? - Page 1
Forum section: electronics / projects

author: GreyGnome
author_rank: Newbie
timestamp: December 19, 2019, 10:01:38 pm
content: Hello,I'm not sure if I should ask a new question or resurrect an old thread/video, so I've opted for a new question.I watched #1116 the other day, about the "capacitance multiplier", and it is very cool. I get that it works, but I don't know how it works. To me, it seems that once the base is turned on by the resistor, the transistor is conducting. If the transistor is conducting then the voltage on the emitter is just the voltage on the collector minus 0.6 for the base-emitter junction.Also, Dave says that a MOSFET would incur a larger voltage drop... but I would expect it to be smaller, since (assuming a use a hexfet or something similar), there's no junction voltage to overcome. Rather, it would have a portion of an ohm of resistance, and therefore less of a voltage drop.What am I missing? Thanks. (I've read about emitter followers a bit, but they generally talk about them as amplifiers... I still don't intuitively understand this circuit).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 19, 2019, 10:14:04 pm
content: So the base draws no current? That'd be fantastic.MOSFETs have a G-S voltage in the same way BJTs have B-E voltage. It varies with type, and is generally larger (reflecting the lower gain in comparison). Some are negative (depletion mode), which comes in handy sometimes.Tim

author: iMo
author_rank: Super Contributor
timestamp: December 19, 2019, 10:46:48 pm
content: The transistor is not fully opened. The Vce should be a couple of Volts, ie.2-3V, thus the Q1 is not saturated.

author: GreyGnome
author_rank: Newbie
timestamp: December 20, 2019, 12:35:58 am
content: Thanks for the replies. I appreciate that you didn't make me feel too dopey. :-)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 20, 2019, 12:39:35 am
content: Note by the way, even though a BJT is constructed in layers, and the collector is stuck on top of the base layer, and the base is +0.6V or so to the emitter -- the collector can still saturate to near 0V. This is because there's a built-in potential in the B-C junction, that acts to suck electrons out of there. You can't measure this potential externally, of course (that would be getting voltage for free!), but it explains why this is so.Hence, you can have a capacitor multiplier, using a BJT, with the base voltage nearly the input voltage, and the emitter voltage just ~0.6V less, and the transistor isn't running out of operating range or anything, it's not banging its head into saturation.Tim

author: MagicSmoker
author_rank: Super Contributor
timestamp: December 20, 2019, 12:20:41 pm
content: I think an alternate explanation for the capacitance multiplier is more instructive. The RC network feeding the base of the BJT is a low pass filter with a very low cutoff frequency, but with a DC impedance too high to be useful as a power source (because of the high resistance value). The latter can be reduced by connecting the output of the RC filter to the base of a BJT emitter follower which has near unity voltage gain but high current gain, so it faithfully reproduces the voltage at its base while drastically reducing the loading on the RC networking supplying it (by the DC current gain, beta). In effect, the capacitor value is multiplied by Beta while the resistor value is reduced by Beta, resulting in no change in the cutoff frequency (R goes down by the same ratio that C goes up), but a drastic reduction in the DC output resistance (by Beta).

author: iMo
author_rank: Super Contributor
timestamp: December 20, 2019, 12:50:26 pm
content: If you replaced the capacitor with a zener, you would get a standard voltage regulator - working the same as the capacitive multiplier when talking the suppression of the ripple (provided the input voltage with the ripple is higher than zener voltage, aprox).By replacing the zener with the C the "reference voltage" is floating and averaged, the "effect" of averaging not depending on the input voltage (but RC). The Capacitance is "multiplied" by beta the same way as the stabilization effect of the zener is multiplied by beta of the transistor, imho.

author: tggzzz
author_rank: Super Contributor
timestamp: December 20, 2019, 03:18:54 pm
content: A standard way of analysing circuits is in terms of their input impedance Zin, output impedance Zout, and gain G, and the source impedance Zs.To a first approximation for an emitter follower in this circuit:voltage gain G = 1source impedance = R1//C1, i.e.R1/(1+sR1C1)where s = jωinput impedance = high relative to Zsoutput impedance = Zs/β , where β is the BJT current gainNow an equivalent output impedance Zs/β could also be (roughly) obtained by using R1/β and C1*β. The latter is the source of the term "capacitance multiplier"

author: bson
author_rank: Unknown
timestamp: December 21, 2019, 08:16:42 pm
content: i = C dv/dtThe current gain of the BJT increases i without any change to v. Hence dv/dt doesn't change, and C scales linearly with i.