Output impedance of voltage multiplier/inverter - Page 1
Forum section: electronics / projects

author: ezalys
author_rank: Frequent Contributor
timestamp: November 25, 2022, 07:27:54 pm
content: One thing I've had a hard time understanding is what the disadvantages of charge pumps and voltage multipliers are relative to boost/buck/forward converters. I'm currently thinking of a design that uses an AC/AC wall brick but I'm in need of positive and negative supplies. Attached is a simple circuit for obtaining positive and negative rails from a single low voltage AC supply. When is this the right supply to use? How do I understand that it delivers 19 V at 150 mA?I should say that I'm a fan of dirty, simple, reliable solutions. My first impulse is that I see 1000 uF and think 1/(2*pi*(60 Hz)*(1000 uF)) = 2.6 ohms and think that ah, you need big capacitors to get low output impedance. I have a design where I don't mind huge caps, but want to avoid custom magnetics.

author: magic
author_rank: Super Contributor
timestamp: November 25, 2022, 08:35:27 pm
content: This circuit may reverse bias C1 during startup under load or when the output is overloaded or shorted. Maybe bypassing C1 with a diode would be a good idea.Output impedance at high frequencies is that of a capacitor.Output impedance near line frequency is some hairy stuff, not sure I want to think about it. If you need it low, more capacitance certainly helps.Output impedance at DC (and maximum practical average load current) is perhaps the more interesting question. As long as output voltage is more than line voltage, I think it's not a huge error to pretend that charge transfer from C1 to C2 occurs in a short moment during line voltage peak. Before the transfer, we can pretend there is ~21V on C1 and X on C2. After the transfer, both capacitors become equalized to (21+X)/2 (ignoring diode loss) since they are equal in value. This gives the amount of charge delivered to C2 per mains cycle as (21-X)/2·1000µF, and hence the average load current is easy to calculate. You see that it depends on X, i.e. how much voltage ripple you are willing to tolerate on C2. "Average" output impedance at DC is then 0.75·X/Iaverageor something like that.

author: TimFox
author_rank: Super Contributor
timestamp: November 25, 2022, 09:48:30 pm
content: Output voltage: ignoring diode drop and secondary resistance, the top two diodes form a voltage doubler, the (no load) output of which is the peak-to-peak voltage of the AC drive: 8 V rms x 2.828 = 22.6 V. Less two diode drops is roughly 21 V. The other two diodes form another voltage doubler to get the identical voltage, negative with respect to common.In real life, the transformer secondary voltage may be higher than specified, since it is specified the maximum rated current.Also, the no-load output I mentioned above is the peak voltage of the ripply waveform. The load current determines the valley of the sawtooth (at line frequency, not twice line frequency). The mean value of the DC will be between those extremes.I use Duncan Amps freeware PSUD2 to estimate these circuits at line frequencies (50 or 60 Hz).To use it, you need the no-load secondary voltage and the effective resistance.It includes a voltage-doubler rectifier as a possible circuit; that would solve half of your circuit problem.https://www.duncanamps.com/psud2/The computation engine includes either a resistive load, a current load, or a stepped current.With the current load, it will give an error at the beginning of the simulation, which you can ignore as unphysical with the current load.Comparing no-load to current load will give you the output resistance.