Cool thanks for your replies so far... please bear with my replies. I call my electronics background to date self-taught, no... badly self-taught
Also had an idea today at lunch toady, shorting the output over two 50W resistors in parallel (because 50W resistors are $3 each and 80W and above are uber-$$$). And if my dodgy napkin math is right, I'd need a resistance of 8.2Ω, but as I'm running two in parallel I'd need two 18Ωs or one 18Ω and one 15Ω (8.182Ω). Unless that logic is flawed.
A common way to control/limit current is to pass the load current through a low-value (e.g. 0.1Ω) sense resistor, compare the resulting voltage across the resistor against a reference voltage, then switch off or 'turn down' the supply according to the comparator output. Varying the reference voltage enables the current limit to be varied. Providing the resistor value and reference voltage are accurately known you don't need a dummy load to set the limit (though it's useful for test purposes).
Excuse me if I read that incorrectly, the circuit I'm using already has a current adjustment in the design. Unless I'm missing something?
Hi,
Welcome to the forum first of all. I think you're in the right place.
I understand your problem pretty well. You have a power supply and you want to use it to run something, but first you need to set the current level so you dont accidentally blow out the device being tested. It's pretty hard to do with a dummy load because the dummy load ends up being as big as the power supply. So you want an alternate method.
I know what you mean.
The simplest answer is if you dont need super accurate current setting (usually you dont) you can add a calibration plate to the power supply right behind the current set knob. It would be a round circular pattern like a clock, but instead of hours and minutes, you print amps and milliamps. So when you go to set the current level you dont need a load, you just dial it to the right value as you read off the plate.
Another method if you are building your own supply, is to use a dual ganged pot for the current control. The front pot portion is used for the normal current setting, and the back portion is used for viewing the setting of the pot on a digital voltmeter built right into the power supply. The calibration is such that the volt meter reads out in amps rather than volts. So when you turn the pot the volt meter gets say 1.000v and because of the mechanical linkage between pot sections and your predetermined calibration, that means 1.000 amps. This would be easy to implement provided you can get a dual gang pot.
I did consider that first option and it would be fine if I was using a standard 270° pot. But I'm using a multi-turn pot and that sends the scale idea out the window. Another option I'd like to have is a two pot setup, one for course adjustments and one for fine tuning; but that's another post me thinks.
The second option might be 'do-able' as I already employ a similiar setup on one of my homebrew radio control transmitters, in an attempt to show servo throw/rate.
I assume you mean "current limit" when you say current adjustment.
If so then you can just short the output with a push-button switch to adjust the current limit. But, of course, the power supply needs to be able to handle a short-circuit without failure for this technique (which a good power supply design will do). I have used laboratory supplies that use this technique.
Sorry, that just gives me the hebejebes. I did consider shorting the output with piece of aluminum rod but thought there must be a cleaver way.
The circuit says "Current Adjust Range" which I thought was a current limit (based on my first PSU which came in a kit form MANY years ago). Does Current Adjust Range mean something different to Current Limit? And... now that I've looked at the circuit it has a third variable I didn't notice before called "Full Adjustment". Now I'm scratching my head, perhaps I should just built it and find out what it means in my normal fashion; the hard/hot/smokey/"Well that didn't work" way.
You have to decide if your making a CC/CV power supply or a CV supply with current limit. If the latter you have to decide the type Foldback or Current limit.In a foldback supply if a 10 A supply was to deliver 10A, and the output was shorted, the voltage would nearly drop to zero and the current would be 10A.
Bear with me again to make sure I'm reading this right; by CC you mean Constant Current? And CV = Constant Voltage?
Stu