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I'm making a bench power supply, which will go from 0 to 17VDC and up to 30ADC at the output, which will have a voltage and current regulation (with potentiometers).

I am wondering, is it okay to use a push-button to short-circuit the Vcc and the GND pin, current going through the shunt resistor (enough for 30A or 50A)?

And the second part of this question here is, can I use (instead of 30A push button) a smaller push-button and few of MOSFET transistors, where I'd connect the current to the Emittor, some smaller voltage (I have stabilized +5 and +12VDC from the circuit) connect to the Base of MOSFET and the Colector to the shunt (where the other pin of shunt goes to GND)?

The thing (as I imagined) would work such as: I set the voltage, then I press the button. While the button is pressed, through the shunt goes max. current, as my circuit let's it through. While I'm measuring the current going through the shunt, I can regulate it and set it to desired value.

EDIT: Is okay to use shunt in series as this schematic applies?

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  • \$\begingroup\$ The MOSFET is a better idea than a push button - even though 17VDC won't do much to you it's still safer in case of some malfunction. Also the button can get burned after prolonged use on that much current. \$\endgroup\$
    – DELTA12
    Commented Apr 22, 2016 at 19:52
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    \$\begingroup\$ Most decent PSUs allow setting of the current limit with the output switched off. If this is a linear power supply then 17 V @ 30 A = 510 W (minimum) to be dissipated somewhere in the PSU. \$\endgroup\$
    – Transistor
    Commented Apr 22, 2016 at 19:58
  • \$\begingroup\$ @delta12: Thank you for your reply. Well, so you would recommend using this option? And which MOSFETs should I use, got any idea? You think, I could find any that could be triggered by 12 (or even, if possible) 5VDC? I can use even multiple, to reach desired current. \$\endgroup\$
    – Jakey
    Commented Apr 22, 2016 at 20:06
  • \$\begingroup\$ @transistor: Thank you for your reply, but..I'm not sure, what was your point. Yes, there is approximately 510W (or maybe few W more), but still..I'd use at least 30A or max. 50A shunt resistor. \$\endgroup\$
    – Jakey
    Commented Apr 22, 2016 at 20:07
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    \$\begingroup\$ Your proposal is incredibly brute force. It would be much safer and more elegant if you first understood how the current limiting circuit works and then come up with a way to use that to report a number. For example, if the limiter potentiometer changes a voltage that is fed to a current sense amp, then you can just monitor this voltage, do some computation then display a number based on this current limit control voltage. \$\endgroup\$
    – Vince Patron
    Commented Apr 23, 2016 at 0:11

1 Answer 1

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If I understood you correctly you are designing a bench supply that will work as a voltage source but the current is capped at some user defined value.

In that case you first set the voltage correctly. Then you have a very small resistor which is rated for high current (R1) (search for current sense resistor) and connect it to an operational amplifier (OA1). This ordinary analog signal (A) can now be connected to a comparator (OA2) where one input is connected to a potentiometer and the output of said comparator connected to the switching MOSFET. Providing those circuits with power is easy since you already have +5 and +12VDC.

schematic

simulate this circuit – Schematic created using CircuitLab

A few comments:

  • I could not be bothered to calculate anything but the values given seem quite reasonable (based on my mental model of generic OpAmps)
  • N-channel MOSFETs tend to be generally better
  • the amplification of the voltage over R1 is proportional to R3 (and a hard to measure amplifier spec); just calibrate your scale by measuring A with a know current. I feel like printing a scale (after calibration and some calculating) and gluing it next to the pot is a neat solution
  • R5 is there to provide some additional hysteresis to OA2, not sure what value and if needed at all
  • most amplifiers have low input current, you can choose a high potentiometer value
  • it is good practise to include a fixed resistance in series with the potentiometer so you do not short and burn it
  • You can display Signal A somehow, for example using an ADC and a display

I hope this was coherent enough; you are also likely to find good application notes when googling for current sense amplifier. It would be nice if people point out errors and such, I will likely have made some.

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  • \$\begingroup\$ Could the downvoter please explain themself? This is a good answer to my knowledge, else I would not have written it. \$\endgroup\$
    – caconyrn
    Commented Apr 23, 2016 at 15:59
  • \$\begingroup\$ Thank You for your comment, appriciate it! I think it would be great for many cases, but not mine, since I have already finished my power supply, only needs case. I like the idea, but still: Why cannot just use a relay or MOSFET in combination with resistor or shunt? Is any scheme from link below useful? The resistor R would be 1 ohm 50W (or a shunt). shrani.si/f/2v/4i/1tiQbPvF/stackex.png a) a 50A push-button b) using a 5V push button and a relay, which "short-circuits" through the resistor c) instead of relay is used a single 33A MOSFET d) instead of single, use of multiple MOSFETs. \$\endgroup\$
    – Jakey
    Commented Apr 25, 2016 at 0:08
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    \$\begingroup\$ I'm not the downvoter, but this is not a particularly good design, I know you only meant to be an example, but it is missing important things. Driving the gate of a MOSFET from an OpAmp is prone to oscillation for example... \$\endgroup\$
    – anrieff
    Commented Jan 30, 2018 at 21:29

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