# Ripple current in a linear power supply transformer

Posted on Feb 6, 2014

BTW, I guess I should add another input cap for high frequency noise though that`s hardly relevant to this question (and the schematic is just a very rough test circuit anyhow). The goals are 0 - 12 V at up to 2 amps (1. 5 would probably be good enough, though). The voltage source is 230 \$V_{rms}\$ since that`s what it`ll run on, and the transfo

rmer is set to simulate ~15 V RMS, so about 21 V peak. Here, the red voltage is the input to the voltage regulator, and the green/blue is current through two of the rectifier diodes. Note how the voltage is lowered a lot (from 15 Vrms - 2 diode drops) due to the series resistance combined with the 5. 5 A current peaks. This graph is at maximum output current (12 V / 6 \$Omega\$ load) = 1. 87 - 1. 99 A due to the output ripple; the input voltage is too low for it to regulate properly due to the drop on the secondary. What sort of series resistance would the transformer`s secondary have I`m looking at a 2x 10-15 V multi-tap transformer, with 2. 2 A per secondary rating (66 VA in total). The data sheet lists a few details, but not series resistance. Assuming a 1 \$Omega\$ series resistance on the secondary (as in the simulation above) and 0. 11 \$Omega\$ ESR on the smoothing electrolytics (some ballpark figures I found when searching), I end up with something like the above. With 0. 5 \$Omega\$ on the secondary, the output is great at 12 V and less (the target), but of course the 5+ amp spikes remain on the input side. Am I in the right ballpark with 0. 5 \$Omega\$ on the secondary, or is twice that closer to the truth I do realize that it differs between transformers, of course, but I can`t really find any figures and I have nothing to measure myself. but in this simulation, one works and the other doesn`t. Are the current spikes of...

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