Elementary AC circuit analysis

I'm trying to calculate a voltage in a circuit, and having the problem that I can't quite remember simple circuit theory.

I have a circuit that boils down to something like this:

+-C-R1--+--R2--+ | | | AC R3 DC | | | | RL | | | | +-------+------+

where AC = an AC frequency source of frequency F, DC = a DC power supply and RL is a complex semiconductor load (where I know the current at a given value of DC, but in general it'll be nonlinear).

Reactance of capacitor C at F is approximately zero - it's simply a DC blocking capacitor. So everything is in phase, more or less.

Now I'm trying to work out the theoretical AC voltage across R3 due to the AC signal. That's no problem for the lefthand loop (I can assume the AC reactance of RL is zero). What I can't quite remember is what to do for the DC source. In DC analysis, voltage sources are treated as having infinite resistance. But what happens in AC analysis? I thought that they were treated as having zero AC resistance (low impedance, hence short circuit). But I'm not quite sure if I remember this right.

So if the impedance of the righthand loop as seen across the midpoints is given by (R3+RL) // (R2+R(DC)), what's a reasonable value to assume for R(DC) - zero or infinity?

What sort of reactance do DC voltage sources have in reality? I'd have thought fairly low, given all those smoothing caps floating around.

Thanks Theo

PS This isn't a homework question, this is the kind of thing you do in school and then never use again. So I've forgotten the vital detail.

Reply to
Theo Markettos
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A perfect dc supply will not change its voltage regardless of the current or change in current through it. So, by definition, its delta(v)/delta(i) (its impedance) would be zero.

A supply or battery's internal impedance depends on the supply itself. A brand new fully energized car battery has an internal impedance in the low milliohms (under 20mohms?). A 9V battery might have an impedance of over 1 ohm when it's fully energized, or several ohms when it's pooping out. A bench supply would certainly be below an ohm before it current limits.

For your analysis, it's probably safe to set the DC source's impedance to zero.

Bob

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Reply to
BobW

Thanks. That's what I thought I'd worked out from first principles, but my brain was having a day off :-)

Theo

Reply to
Theo Markettos

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Cool circuit, If I make RL an LED it looks like a simple modulation technique. (in which case I'd like to add a bit of inductance to R2. No sense wiggling the battery around at AC.)

">In DC analysis, voltage sources are treated as having infinite

Opps, DC or AC ideal voltage sources have zero impedance.

George Herold

Reply to
ggherold

Or you could replace R2 with a current source if the inductance needed proves too bulky.

Reply to
Jasen Betts

Well spotted, that's exactly what I'm doing! (but at higher frequencies)

That's an interesting idea... 'wiggling the battery around' was indeed what I was trying to avoid, especially with power supplies that are chock full of smoothing capacitors.

That was my brain fart. And the more I stared at it the more my head started going round in circles :)

Theo

Reply to
Theo Markettos

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"> Well spotted, that's exactly what I'm doing! (but at higher frequencies)"

Hmm, I didn't mention a frequency. I've only done this with a diode laser as load up to 200MHz (which is the highest 'easy' frequency I have available at the moment.) But I've got dreams of 3.X or 6.7 GHz. (the hyperfine resonance's of Rubidium.)

BTW a current source as DC driver is the way to go. (As mentioned by someone else.) The inductor in the DC path is still a good idea. (Though not mine, I stole it from Libbrecht and Hall, RSI, circa

1992)

George Herold

Reply to
ggherold

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