Electronics > Projects, Designs, and Technical Stuff
Capacitance multiplier (#1116): Cool, but how does it work?
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GreyGnome:
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).

T3sl4co1l:
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
iMo:
The transistor is not fully opened. The Vce should be a couple of Volts, ie.2-3V, thus the Q1 is not saturated.
GreyGnome:
Thanks for the replies. I appreciate that you didn't make me feel too dopey. :-)
T3sl4co1l:
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
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