Power consumption of transformer based bench DC supplies, like the korad 3005 - Page 1
Forum section: electronics / beginners

author: exnihilodub
author_rank: Contributor
timestamp: February 22, 2025, 12:33:03 pm
content: Hi! Just jump to the TLDR section for the actual question, I felt the need to give background info on my situation and reasons.Details:Hi. I recently started building myself a puny lab, got myself a nice scope recently, and now in the phase of deciding which bench power supply to buy.Since I started this hobby as a guitar pedal builder, I currently still power my circuits with a 9VDC transformer/linear regulator based guitar pedal power supply made by a friend's friend. I kinda grew out of it, and want an adjustable power supply for my projects.I put every device in my room/lab behind a 3kVA UPS since I live in between two very remote villages, and the electricity here is finicky. We get 3-phase voltage here. Undervoltages as well as overvoltages due to bad weather and frail neutral connection (i guess) are common. I've put my UPS to the phase that "drops down" in case of unbalance, just to be on the safer side.Question / TLDR;So, details aside, I'm behind a UPS, and the power I can draw is limited. Right now, my power draw is low, never exceeds 25% unless i heat my soldering iron, hot air station, or 3d printer's hot end & bed.My question is, since power supplies like korad 3005D are transformer based, will they ALWAYS pull the power that's worth of 30V/5A from the back end, and give me what I need on the front terminals? Or does it roughly draw the power 1:1 with what I ask for on the front terminal?

author: Solder_Junkie
author_rank: Frequent Contributor
timestamp: February 22, 2025, 12:58:34 pm
content: What you are asking is what is the efficiency of a linear power supply.Depending on how much power you really need from a bench power supply, you could use a modest 1 or 2 Amp power supply. I built one many years ago and have found it invaluable. Mine is 0-20V, 0-2A. The current limit is adjustable down to just a few mA, which is very useful for test/repair as it minimises any possible damage.There are many such power supplies, a surplus one will be fine. Schools, laboratories, etc. use bench power supplies and they are often seen on the second hand market. One such bench supply is this TTi one:https://www.aimtti.com/go/er/index.php?p=elr-specSJ

author: Furna
author_rank: Frequent Contributor
timestamp: February 22, 2025, 12:59:33 pm
content: Why do you think it should pull full power even when idling?My KA3005P pulls around 12.5W in idle; pretty inefficient IMHO but it is cheap enough to justify the inefficency.

author: golden_labels
author_rank: Super Contributor
timestamp: February 22, 2025, 06:13:19 pm
content: In the simplest case it’s a fixed DC source, followed by a huge energy-wasting element, followed by your device. The “huge energy-wasting element” is the part controlled by the PSU to achieve given output voltage and current.Given the supply voltage is fixed (e.g. 35 V)(1)and everything else is in series, the total power depends entirely on current actually drawn at the moment. If you draw full 5 A, then it’s 35 V · 5 A = 175 W. If it’s 1 A, you draw 35 W. And if you draw 10 mA, it’s 350 mW. At idle it should draw nothing, but of course there is all the control circuitry. And in a linear power supply, which is designed to work through wasting energy, nobody cares about a few or a dozen watts heating air.Now, there is one more catch. Pretty much any relatively modern linear power supply has a multi-tap transformer. They will use the lowest voltage tap possible. The click you hear while voltage changes are relays chosing a different transformer tap. If that’s the case, the actual power used will be reduced, because the source voltage gets reduced.Summing it up: It’s neither the full 30V/5A nor 1:1. It’s something inbetween, mostly determined by the current you draw. If you draw nothing, it’s going to be just the power control circuits consume.(1)Note that a 30 V power supply needs slightly higher source voltage.

author: Furna
author_rank: Frequent Contributor
timestamp: February 22, 2025, 06:21:33 pm
content: golden_labels post deservers a "thank" since it really answers to the OP question and also (basically) explain linear PSU.I just realize neither of previous posts (mine included) gave an answer to the OP.

author: bdunham7
author_rank: Super Contributor
timestamp: February 22, 2025, 07:13:45 pm
content: Any discussion of this should include power factor. Your linear PSU will draw somewhat more power, perhaps 20-50%, than it puts out but it will draw much more current than a simple power ratio implies. The RMS-to-average ratio will be something like 2:1 at higher loads and even worse at low loads. You'd need to carefully measure this to know precisely what the numbers are, but the main issue for you is that your UPS will have both a VA (Volts * Amps) and power (Watts) maximum rating. With non PFC (Power Factor Corrected) loads like your PSU, the former is almost always the limiting factor. Also, the source impedance of your UPS is probably much higher than regular mains, so the distortion effects of the harmonics from the PSU rectifiers will have a greater effect on the voltage that all of your other devices see.

author: BeBuLamar
author_rank: Super Contributor
timestamp: February 22, 2025, 07:31:13 pm
content: So can you just bypass the UPS? You don't really need the UPS for the power supply.

author: exnihilodub
author_rank: Contributor
timestamp: February 22, 2025, 10:04:36 pm
content: thanks! this is the info what I was looking for

author: exnihilodub
author_rank: Contributor
timestamp: February 22, 2025, 10:17:14 pm
content: I was thinking that because i thought the current over the primary was not affected by the load on secondaryI thought the power draw from the source would be as if there's only 100m in the path. doing a quick test on falstad to see how transformers work was enough to see my error