Bench PSU : output switch and protection - Page 1
Forum section: electronics / projects

author: Atom
author_rank: Regular Contributor
timestamp: May 24, 2018, 05:22:52 am
content: So first of all, hi to everyone.I started to build my diy power supply a long time ago and I've been going through a lot of designs , finally I settled on the HP 3620, I discarded all the things that weren't needed (prereg) for a basic test, now it's on the breadboard and it's working really well. Now I have a few questions since I'm a newbie.1. How should I be switching the output on and off ? P channel ? N channel mos (I have a 12 rail on top of the 0-30vdc main rail).. I would prefer to not use relays (because they can weld?)2. There are others countermeasures that I can take to not damage the PSU and the circuit attached to it?

author: Paul Price
author_rank: Super Contributor
timestamp: May 24, 2018, 03:53:45 pm
content: This power supply design has a 470uF cap directly across the output terminals. If the output voltage is set to some value, then this capacitor will try to discharge itself directly to the circuit it is suddenly connected to(if you turn on power to the circuit you are powring by using an output switch), even though this supply has constant current modes of operation. You don't need an output switch, you need to ramp up the output slowly and watch current draw as you bring the supply up to the desired output voltage level you want. This will best protect the P/S and the circuit it is connected to.

author: Atom
author_rank: Regular Contributor
timestamp: May 24, 2018, 07:03:54 pm
content: thanks a lot so your suggestion would be to not use an output switch and ramp up the voltage by watching the current increase ?by the way can someone help me to understand the current feedback loop? i understand the basics but that seems out of my leaguei was thinking to replace the shunt resistor with a 0.1 ohm or similar to decrease the voltage drop (and the power dissipated) because i wanted this psu to put out max 3 amps@1ohm/ 3 amps power disspiated 3^2 *1 = 9W@0.1ohm 3amps power dissipated 3^2 *0.1 = 0.9Wwell that's a difference but if i can't understand how the current feedback works i can't tweak it

author: Paul Price
author_rank: Super Contributor
timestamp: May 24, 2018, 10:53:28 pm
content: The current limiting circuit is fairly simple. As output current increases the voltage drop across R55 also increases, this resistor is the current sense shunt resistor. The op-amp below is biased by the current limit pot to not limit current until the voltage drop across the sense resistor is greater than the set point voltage, at which point the output of the op-amp shunts the output pass transistor base drive current away from the pass transistor as the power supply shifts from constant voltage to constant current.It is possible to lower the value of the sense resistor but you would have to change the u19a op-amp to a faster op-amp because it would have to have a higher gain to amplify the smaller voltages developed across a smaller valued sense resistor. The replacement op-amp must be able amplify the smaller sense voltage and yet slew fast enough to control current or else your power supply could change into a high-power oscillator at some constant-current or current limiting setting. The voltage set by the constant current adj. pot that sets the current limit by bias voltage would also have to be scaled-down(R68 to R68x10) to match the lower over current sense voltage thresholds. Something like the ADA4625 op-amp, for instance may be overkill, or maybe just the ticket, at USD 8 bucks a throw. The integrating capacitor across the op-amp also will likely also need to be lowered in value.