Dual Complementary Rectifier PSU - Page 1
Forum section: electronics / general

author: virusx2
author_rank: Newbie
timestamp: April 25, 2024, 11:22:32 pm
content: Hi all,i have an 60VA transformer with 2x15V secondary, rated at 2A each. I want to use it on a dual symmetrical dc power supply with outputs, +15V,-15V.The circuit is with filtering capacitor and LM7815/LM7915 on each rail.Except of the dual supply loads such as op-amps and transducers, i have some single-ended loads from +15V to common 0V and from -15V to common 0V.Lets say i need 1.3A from +15V to common 0V, 50mA from -15V to common 0V, and 150mA symmetrical load on+15V/-15VI am a little confused. Can you help me calculate the the total AC RMS current needed from the transformer?I attached the circuit.Thank you.

author: David Hess
author_rank: Super Contributor
timestamp: April 26, 2024, 12:38:37 am
content: Quote from: virusx2 on April 25, 2024, 11:22:32 pmJust going by my intuition:1.30/2+0.05/2+0.15=0.65+0.025+0.075=0.75 AmpsHowever the power factor for a diode-capacitor filter is only about 0.65, so the actual RMS current will be 0.75/0.65=1.15 amps. This comes about because the capacitor input is only drawing current in pulses during the peaks of the line cycle.

author: bdunham7
author_rank: Super Contributor
timestamp: April 26, 2024, 01:01:02 am
content: It isn't a simple calculation and you would need to know the DC resistance of the transformer secondary, the size and ESR of the filter caps and other stuff to do it mathematically. Practically, for a FWB you'll see RMS-to-average ratios of 1.5 to 2. However, since you are drawing your power mainly from the +15V rail, that will be supplied alternatively from each half of the transformer so I think the results might be a bit different. Using larger filter caps actually makes the RMS-to-average ratio worse (higher) so if you are close to the transformer limits, don't make them any bigger than needed. In any case, I'm pretty sure your 60VA transformer is sufficent for the task you've specified.

author: virusx2
author_rank: Newbie
timestamp: April 26, 2024, 04:28:04 am
content: Thanks a lot!I don't want to calculate the currents strictly mathematically, but i want to go with some “design” equations to be as close enough as i can.Maybe there are different rations of DC-to-RMS for the single ended loads and the common mode +-15V?

author: David Hess
author_rank: Super Contributor
timestamp: April 26, 2024, 11:23:18 am
content: Quote from: virusx2 on April 26, 2024, 04:28:04 amThere are limits to an unbalanced load because the entire output of the transformer cannot be drawn from half of the secondary, however your load is not even close to this.This also leads to a difference in the rating between the full-wave centertap configuration with 2 diodes and the full-wave bridge configuration with 4 diodes because the former increases resistive losses by only using half of the secondary on each half cycle, and some transformers specify this difference in ratings. Your configuration is somewhere in between, however the increased inefficiency from the low power factor of your capacitor input rectifier overwhelms these considerations.

author: virusx2
author_rank: Newbie
timestamp: April 28, 2024, 12:11:25 pm
content: Ok, if someone could analyze this with some equations i would be glad

author: sleepy2000
author_rank: Newbie
timestamp: April 29, 2024, 08:30:12 pm
content: Another thing to be considered is the loss on the linear voltage regulators. If your winding output is 15V ac rms. Then the peak is 15*1.414=21.21V. So there is some quite some voltage drop and loss on your regulator besides you load.As mentioned in previous posts, the dc to rms ratio varies a lot for different circuit parameters. If you only need a design, the get a simulator and run your case should give you a quick and accurate result. If you really want some analytical solution, there is some way to go.