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New Siglent SPD4000X series power supply
ArdWar:
--- Quote from: MildInductor on October 07, 2024, 02:58:19 am ---Is this overshoot likely to damage any general IC circuitry? Say if I was supplying 3.3V to a general purpose IC that was only rated to operate up to 3.3V. Would the spill over be enough to damage it generally speaking?
--- End quote ---
With regards to this specific exercise, no. It should not damage your circuit IF you use your PSU in constant voltage mode.
The aforementioned overshoot only happen (so far) when PSU are configured (and expected) to startup directly into constant current mode. That isn't exactly normal nor usual mode of operation.
Even then, you normally shouldn't set the CV setpoint that far high (say, 8.0V here) when you expect your I*R product to never get that high (say, 40mA*50Ω = 2.0V here). In this situation a sensible user should set the CV setpoint at something like 3.0V.
MildInductor:
I see, thank you both!
mhsprang:
--- Quote from: ArdWar on October 07, 2024, 09:38:11 am ---The aforementioned overshoot only happen (so far) when PSU are configured (and expected) to startup directly into constant current mode. That isn't exactly normal nor usual mode of operation.
--- End quote ---
Why isn't that a normal mode of operation? It is, after all, advertized as one of the modes a PSU can operate in. CV means "constant voltage" whereas CC means "constant current". What makes the meaning of constant different in both definitions?
--- Quote from: ArdWar on October 07, 2024, 09:38:11 am ---Even then, you normally shouldn't set the CV setpoint that far high (say, 8.0V here) when you expect your I*R product to never get that high (say, 40mA*50Ω = 2.0V here). In this situation a sensible user should set the CV setpoint at something like 3.0V.
--- End quote ---
But what if I don't know the expected maximum voltage? I'll give you a real life example: I once had to repair a small heater that looked like a wood stove. This heater contained a string of orange LEDS, to light the fake flames. This string was connected to mains using a big ceramic resistor that got flaming hot and eventually burned. To find out the proper value of this resitor, I needed to push 20 mA through the string of LED's without knowing the actual VFWD and the number of LED's in the potted piece of plastic. In that case, I could have taken a power supply, set to CC at 20 mA, connect the LED and hit the Output button in order to read the total VFWD of the entire string. Is that such an odd application of a PSU?
And with some of the PSU's in this thread, I could have blown all the LED's, hypothetically.
2N3055:
--- Quote from: mhsprang on October 07, 2024, 12:49:43 pm ---
--- Quote from: ArdWar on October 07, 2024, 09:38:11 am ---The aforementioned overshoot only happen (so far) when PSU are configured (and expected) to startup directly into constant current mode. That isn't exactly normal nor usual mode of operation.
--- End quote ---
Why isn't that a normal mode of operation? It is, after all, advertized as one of the modes a PSU can operate in. CV means "constant voltage" whereas CC means "constant current". What makes the meaning of constant different in both definitions?
--- Quote from: ArdWar on October 07, 2024, 09:38:11 am ---Even then, you normally shouldn't set the CV setpoint that far high (say, 8.0V here) when you expect your I*R product to never get that high (say, 40mA*50Ω = 2.0V here). In this situation a sensible user should set the CV setpoint at something like 3.0V.
--- End quote ---
But what if I don't know the expected maximum voltage? I'll give you a real life example: I once had to repair a small heater that looked like a wood stove. This heater contained a string of orange LEDS, to light the fake flames. This string was connected to mains using a big ceramic resistor that got flaming hot and eventually burned. To find out the proper value of this resitor, I needed to push 20 mA through the string of LED's without knowing the actual VFWD and the number of LED's in the potted piece of plastic. In that case, I could have taken a power supply, set to CC at 20 mA, connect the LED and hit the Output button in order to read the total VFWD of the entire string. Is that such an odd application of a PSU?
And with some of the PSU's in this thread, I could have blown all the LED's, hypothetically.
--- End quote ---
You couldn't because you probably would not have enough voltage to light them in a first place.
As to good practice, you would set voltage to some low value and set 20mA.
Connect DUT, enable output and slowly ramp up the voltage...
mawyatt:
--- Quote from: MildInductor on October 07, 2024, 02:58:19 am ---Is this overshoot likely to damage any general IC circuitry? Say if I was supplying 3.3V to a general purpose IC that was only rated to operate up to 3.3V. Would the spill over be enough to damage it generally speaking?
--- End quote ---
Overshoot Voltage can and will damage/destroy certain components, of coarse this depends on the component and amount of Over Voltage. CMOS failure is caused by gate oxide puncture, many of today's advanced chips have very thin and hence low breakdown oxides (some BJTs have thin base regions). 2 decades ago we were dealing with custom chips that ~3 volts would certainly destroy, they normally operated at ~1.2 volts!!
Another potential effect is when Current Limit is engaged such as the discussion mentioned here:
https://www.eevblog.com/forum/testgear/lab-power-supply-turn-on-and-off-characteristics/
Anyway, IMO ANY Overshoot Condition is a potential disaster in the waiting. A perfect Power Supply should not Overshoot during Turn ON or Turn OFF under any conditions. Actual supplies will deviate from perfect behavior is various ways, some better than others.
Best
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