Hmm. Not entirely happy with this new PSU.

It’s current limited, down to about 10mA or so (difficult to adjust for small changes) and up to 5A, so that’s fine. But it seems to take too long for the limiter to kick in. In the region of half or quarter of a second or so. This is much longer than I had expected.

Until I became careful about such things, I used to burn out the fuse in my DMM frequently. But have managed to avoid it for a few years now. Until today, while testing the current limiting on this PSU. It was set to 2.5v@15mA (I was experimenting with driving an LED directly). The LED glows very brightly for about half/quater second, then settles to 15mA. Replacing the LED with my DMM on 200mA range caused the fuse inside the DMM to blow.

I’m not convinced this PSU will be effective at protecting my Picaxe circuits during prototyping.

Nowhere in the specs does it refer to the reaction speed of the current limiter. Have I got a duff unit, will a replacement be the same? Am I expecting too much for £60? Since it’s unfit for purpose, should I ask for my money back? Any modding suggestions to reduce the delay to, say, a few mS? (I was already considering voiding the warrantee to replace the pots with multi-turn types)

 

It sounds to me like there are caps in the output.

For voltage supplies, this helps a great deal to reduce transients.

For current supplies, this is terrible.

Voltage regulated supplies must have extremely low output impedances.
Current regulated supplies must have extremely high output impedances.

You obviously have a voltage regulated supply that has current limiting.
It's doing it's job as advertised. Don't be miffed.

I would've quoted your reply and given more details, but you used custom fonts which make that a mess.

I don't bother going into such messes.

 

i wouldn't expect alot in a power supply for £60.

If you're looking for a decent power supply with good current limiting, look at getting a second hand TTi power supply i.e. PL330QMD. You won't be dissapointed.

 

Actually you can get some nice power supplies for that money, but you have to know what you're getting.

My personal preference is to make your own regulators. If you need a current source it will work the way you expect, and the power supply will provide the basic power needed.

Something else, if you know you're using it as a current source, short that sucker out when it isn't in use, and remove the short when you need the current.

 

Thanks all.

Sorry about the font. I compose in MS Word then paste onto forum. It’s just the default Word font.

 

Sorry about the font. I compose in MS Word then paste onto forum. It’s just the default Word font.

Thanks for that explanation. That may explain why lots of other people's posts also comes up with all the embedded font and size commands.

Composing in Word first is not a bad idea. Occasionally I'll lose a post if there is an internet "glitch".

I usually use Notepad as a "scratch pad" for copying/pasting text; it doesn't cause FONT and SIZE commands to be inserted into a post, and Notepad uses very few computing resources while running.

Back to your situation:
It does appear that your supply has capacitors in it's output. It is also likely that your supply is a switcher-type. If you remove the output capacitors, you will have quite a bit of ripple voltage, which would be undesirable.

Bill Marsden's solution is a very good one. Simply shorting the PSU's output before connecting the load will drain the charge on the output capacitors. It is not often that solutions are so simple and cost-effective.

Another idea is to construct a simple linear current limiter. LM317 positive voltage regulators can be used as a constant current source, for currents in the range of 10mA to 1.5A. Proper heat sinking is required for power dissipation over about 1W. Note that the LM317 will drop at least 3v across itself when used as a current regulator. If you require less than 10mA current, an LM317L can regulate properly down to 5mA, but is limited to 100mA maximum current.

Construction of the LM317 current regulator is very easy; it only requires 1 resistor and the regulator itself.
1) Connect the current source to the IN terminal of the regulator.
2) Connect a resistor between the OUT terminal and the ADJ terminal.
3) Connect the load to the ADJ terminal.

Calculation of the resistor:
R1 = 1.25/DesiredCurrent, where: 10mA <= DesiredCurrent <= 1.5A

Conversely:
OutputCurrent = 1.25/R1, where: 0.8333...Ohms <= R1 <= 125 Ohms

Measuring current:
As you have discovered (as I and many others have too), it is all too easy to blow the fuse on a meter while attempting to measure current directly.

Do yourself (and your meter!) a big favor, and get a precision non-inductive 1 Ohm resistor rated for at least 10W, preferably more. Put your resistor in series with the load, and measure the voltage drop across the resistor.
Since I=E/R, and R=1, your readings in mV will directly correspond to mA.

 

On the Maplin webpage there is the opportunity to ask questions. Why don't you ask them?

Here are a selection from their answers

Q- could two of these units be used in parallel to give 10amps? - Alastair

Answer- Yes this product does support that (output voltages would obviously need to be matched though).

Q- can this be used as a split rail supply - say, +/-20v - I notice the green "ground" terminal? - mike

Answer- Yes the product does support that.

Q- Do you stock spare parts for these units - Dave

Answer- unforutnately not

Q- Is this a linear or switched mode power supply? - steven ettles

Answer- This is a switched mode supply.

Q- what is the short circuit protection, a fuse or a circuit breaker? - tom

Answer- This is a fuse.

The claim about parallel operation for increased current in interesting as is the comment about spare parts.