What is a typical internal impedance of voltage source?

[tca]

I remember doing this task in my electronics course (one semester, 20 years ago)... but I don't have the tools to do it again. So here is the question: What is the internal impedance of typical voltage source (capacitor, diodes bridge, transformer)? And of a 9V battery (it depends on the material that it is build)?

Ideally is zero, my educated guess says .1 to 10 Ohm?!

Thanks.

[R.G.]

I remember doing this task in my electronics course (one semester, 20 years ago)... but I don't have the tools to do it again. So here is the question: What is the internal impedance of typical voltage source (capacitor, diodes bridge, transformer)? And of a 9V battery (it depends on the material that it is build)?

Ideally is zero, my educated guess says .1 to 10 Ohm?!

Correct - ideally it is zero. However, they vary so much that I'm not sure there is a "typical" voltage source.

There is a wide variation, even in the same circuit, and it varies depending on what frequency you use to measure it. At DC, it would be calculable from the impedance of the AC-wall-socket, line cord, transformer primary and secondary wiring, wire resistances, and so on. But it gets tricky when you rectify and filter.

The diodes have a forward resistance that is dependent on how much current flows through them. At quite small currents, they're high resistance, and at larger currents they approach the resistance of the leads and semiconductor resistance.

The real kicker is the capacitor. The capacitor filter that's most common essentially IS the output impedance above a few times the AC power line frequency, so it declines with frequency until ESR and ESL take over. But the capacitor charges up each half cycle of AC power and then runs down as current is drawn from it between diode charging cycles. The amount that the cap voltage runs down is the ripple voltage, and there is a sag in the DC output by half the ripple voltage. This sag acts like a resistance, but isn't. It's an effect of the switched nature of power to the cap. It's calculable something like a switched-capacitor filter.

To measure the internal impedance, you measure the DC voltage at no load, then with some fixed DC load. The change in voltage divided by the change in current is the effective internal impedance.

[tca]

To measure the internal impedance, you measure the DC voltage at no load, then with some fixed DC load. The change in voltage divided by the change in current is the effective internal impedance.

That is what I recall/learned from a afternoon at the lab 20 years ago...

[R.G.]

Yep, that was the same.

[PRR]

We say "Voltage Source" when we "know" the voltage will NOT sag enough to matter.

All other cases, you gotta know what reality really is.

"Voltage Sources" come in ALL sizes.

If I lived close to the street which has the hydro dam, I could possibly assume my electric meter was a Pretty Perfect Voltage Source.

In fact I am far from the dam, though the power at the street is pretty solid. However I am a loooong way back from the street. My "240V Voltage Source" sags real bad. Pump kicks in, lights sag. 0.4 ohms impedance. Microwave drops the voltage 4V. Just eyeballing the 1,000 loop-feet of thin aluminum wire, I can account for about 0.38 ohms of that. The pole-transformer adds a little, so does 60Hz inductance, but really not enough to bother with.

A "good" value of internal impedance for a house feeder is under 0.05 ohms (2% drop on 240V 100A load).

I could simply use eight times more aluminum (about 3X more Copper); or move the house eight times closer to the street.

OTOH....

A cheep 9V battery may sag to half with 100 ohm load (something you CAN test with basic tools), yet can be assumed Voltage Source for most simple pedals.

In Hi-Fi interconnect, 1K is near-enough Voltage Source for most practical purpose.

In biasing tubes/FETs, a 100K source impedance is zero-enough.