EEVblog Electronics Community Forum
Electronics => Projects, Designs, and Technical Stuff => Topic started by: mcinque on October 08, 2021, 09:38:42 pm
-
Hello all,
I recently had to use a last resort method to find a short in a pcb that had no obvious signs of overheating: low voltage with high amperage injection.
I used a programmable switching power supply Manson set to 1.5V up to 20A putting in series a 0.1R 25W resistor between the PSU and the circuit. PSU is capable of 30A.
In this way I was able to inject "bursts" of high currents, up to 20A at low voltage on various points of the shorted track for and forcing the "responsible" of the short to reveal himself, by using also freeze spray.
I'm not a power supply expert (well, I'm not an expert in many things, expecially in various branches of electronics) so I ask you: how stressful is a 0.1R load on a power supply?
It doesn't seem to me any different than a resistive load like a light bulb, but since I'm not an expert, I fear that I'm missing something in PSU technology.
I ask you to better understand technically what this extremely low load means for a power supply and if there are any differences if the power supply is switching or linear.
Thanks in advance for your replies!
-
It depends on the supply. Lab supplies are usually OK with a short and have no problem with low reistance. Supplies that don't like a short may not like the 0.1 Ohms either. For the 1.5 V 20 A rated supply 0.1 ohms are even less than the load for full power.
I would avoid using so much power just to find a short. This may do more damage: e.g. burn traces to a shortet capacitor. Thin traces may not even like 1 A. Just get more sensitive in finding the current flow.
-
It's rated for 20A and you're using it at 15, I don't see the problem here?
Tim
-
No stress if it's actually designed properly, but do you realize how hot this resistor will get? It's probably meant to go on a heatsink too.
In other words, be careful. Also, the wires you are using will have significant resistance compared to 0.1R, that is, you can't ignore their resistance in your calculations.
You'd be surprised how resistive cheap test leads can be.
-
For the 1.5 V 20 A rated supply 0.1 ohms are even less than the load for full power.
Thanks :-+
I would avoid using so much power just to find a short.
Indeed, mine was a last resort and I started from low currents to avoid fires :)
It's rated for 20A and you're using it at 15, I don't see the problem here?
Thanks :-+
No stress if it's actually designed properly
Well, this is hard to understand if it's properly designed for me, especially without schematics.
It's a Manson HCS-3602 (https://www.manson.com.hk/product/hcs-3602/) 1-32V, 0-30A. From my understanding and research, this PSUs models are neither too crappy nor HP/Agilent/AimTTi tanks.
but do you realize how hot this resistor will get? It's probably meant to go on a heatsink too.
Absolutely, but it's a power resistor already contained in aluminium and I use it for quick bursts only, it doesn't have time to get hot.
Also, the wires you are using will have significant resistance compared to 0.1R, that is, you can't ignore their resistance in your calculations.
You'd be surprised how resistive cheap test leads can be.
Indeed! Luckily for this kind of "power" use I have a couple of 2,5 mm2 lab grade banana wires that has very low resistance .
Thanks for your replies!
-
It's a CC-CV lab supply. By definition, you can't harm it by connecting any resistance to it, from 0 to infinite.
A very large inductive load could harm it, as could connecting a large external voltage, such as mains, to its output.
This being said, there are some so called bogus or scam lab supplies on the market, they can get damaged by normal lab supply use or just without any reason whatsoever. But AFAIK this brand isn't one of them.
So I'd remove the series resistor as an unnecessary element.
Your fault finding process is fine, I do that a lot, I wouldn't call it last resort, even.
If the supply blows up, return it and claim for money back.
-
A very large inductive load could harm it
Indeed, I keep a handy Schottky power diode just to avoid that when connecting motors or inductive loads.
as could connecting a large external voltage, such as mains, to its output.
Of course, that is a tough mistake for a 2 positive quadrant power supply to withstand. But are there really 2 positive quadrant, low-voltage power supplies that can withstand such tough tests?
This being said, there are some so called bogus or scam lab supplies on the market, they can get damaged by normal lab supply use or just without any reason whatsoever.
Oh yeah, I see there is a lot of crap on the internet especially on ebay/aliexpress. It's easy to see that many small "makers" online shops here purchase things from them and then resell those scams. I only buy from known distributors (RS, Distrelec, Farnell, Mouser) choosing only known brands and depending on my budget (unfortunately I can't afford REAL lab-grade instruments like Keysight/Agilent/HP, Rohde & Schwarz, Keithleyetc. :D)
So I'd remove the series resistor as an unnecessary element.
I saw that's needed to have both C.C. and C.V. because without the 0.1R, shorting terminals brings to C.C. (let's say I program it to 1.5V 4A, without introducing a minimal load, it goes to 0.2V 4A) :-//
Your fault finding process is fine, I do that a lot, I wouldn't call it last resort, even.
:-+
If the supply blows up, return it and claim for money back.
It's out of warranty now and I think it's hard to prove it blow out due to poor design anyway.
Even the big distributors I purchase from (RS, Distrelec, Farnell) I do not think they are very willing to accept a blown power supply and recognize it as defective.
Thanks for your reply!
-
I don't understand the purpose of added resistance when the supply is CC-CV. In such fault finding strategy, you want some constant current into the circuit, maybe with some max CV voltage to avoid turning semiconductor parts on if the short suddenly goes away (if that matters). Like 20A 0.5V, then minimum possible resistance, i.e., good probes. Everything resistance does is to add level of uncertainty to the voltage delivered because now if the short goes away the voltage jumps to the higher value you had to set to let enough current flow through that resistance in the first place.
-
that's correct, is useless, I used only because it allows me in someway to raise the injecting voltage.
Setting the PSU 1.5V 20A and shorting the probes on a thick trace, I get:
without series resistor: 0.2V 20A
with 0.1R series resistor: 1.5V 20A
But in the end, it doesn't really matter and I can use it without the resistor and having the same result on the pcb.
-
There is the advantage that, as you're probing, you don't incur the inrush current of whatever capacitance is on the supply's output. Which always seems to be substantial, though it doesn't need to be... (given that it's a switching supply, there will be some unavoidable inrush due to output filtering though).
Tim
-
That's true, I tend to cope with the inrush by starting at lower CV setting then raising it.
-
An alternate method for finding a heavy current draw in a circuit is to disconnect the load to the point where the current draw is low. Then reconnect it in stages until you hit the place where it does go out of control. This can often be accomplished by cutting traces - at least on a two layer PCB.
-
Another alternative, is using a 5 1/2 digit or better DVM, using needle probes as you get closer to the short you can see the resistance decreasing.
-
Another alternative, is using a 5 1/2 digit or better DVM, using needle probes as you get closer to the short you can see the resistance decreasing.
Expensive meter and still the contact resistance is an uncertain variable.
Much better to just run some high enough current into the short (like 1-10A), then you can use any cheap multimeter in mV range to measure exactly the same - voltage difference between any arbitrary points shows current flowing there. The advantage is, contact resistance of the needle probes is out of equation because voltage is measured with very high impedance.