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I am driving an IR LED (LED914 and LED907) in series at 4.5 A for 2.5% duty cycle, In the 28 V I connect a DC supply and set the current to 150 mA, the idea is to continuously charge the capacitor C17. Based on the calculation $$Irms= 4.5A(.025) = 112.5 mA$$ but I gave 150 mA from the DC power supply \$(\frac{28 V}{0.15 A})\$ just to make sure the capacitor will be replenish I tried up to 500mA just to get 30 pulses which is way away on the calculation.

**enter image description here** enter image description here

Channel 3 is the Voltage across the capacitor and channel2 is the voltage across the current sense (R28). As you can see the last 9 pulse voltage across the current sense is less than 0.450 mV, The IR LED \$Vf=4.6 V\$ (during this pulse),I supply 28 V/0.15 A on the capacitor C17 continuesly to charge the capacitor while discharging.

On bench the charge on the capacitor is getting drained. I checked the leakage on C17 and I am getting in \$µA (120 µA)\$, I am wondering why I cannot replenish the charge on the capacitor with the DC supply?

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    \$\begingroup\$ The current pulses are also being taken from your P.S. Maybe you should put a small resistor in series between the P.S. and the capacitor so the pulse is really taken from the capacitor and not from the P.S. As it is, maybe the P.S. is entering some kind of protection because of the high current pulses. \$\endgroup\$ – Claudio Avi Chami Jul 15 '16 at 10:06
  • \$\begingroup\$ @Claudio I have a series diode on between PS and capacitor so that there is no leakage going back to the PS. \$\endgroup\$ – jasp Jul 15 '16 at 10:08
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    \$\begingroup\$ A diode doesn't help, you need a resistor so during the high current pulses the P.S. won't feel high current at its output \$\endgroup\$ – Claudio Avi Chami Jul 15 '16 at 10:10
  • \$\begingroup\$ But the current will limit the current for the capacitor to charge it during this pulses. I want to continuously charge the capacitor \$\endgroup\$ – jasp Jul 15 '16 at 10:11
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    \$\begingroup\$ So? Make the calculation, during the pulse is the cap who provides the current, then you have a lot of time with no pulse for the cap to charge \$\endgroup\$ – Claudio Avi Chami Jul 15 '16 at 10:12
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Power supply current-limiting may not expect fast transient events at its output terminals - a good one should switch from "voltage output" mode to "current-limited output" very quickly (to protect the load), but many take some time to switch modes and stabilize. Some switch back to "voltage output" mode from "current-limit" mode much more slowly - this could be your problem. Power supply spec sheets often don't characterize transient response of mode-switching.
The solution that some have suggested adds a series buffering resistor between DC supply and the large reservoir capacitor. This resistor should be large enough that the average current drawn from the supply doesn't invoke current-limiting, in an attempt to avoid mode-switching transient issues. This solution does soften the DC supply regulation, so you need a bit more voltage going in. And this over-voltage becomes dependent on duty cycle (150us/6ms).
If you really trust your op-amp/MOSfet current regulator, invoking your DC supply's current limiter is redundant protection, and shouldn't be necessary. But be aware that some DC supplies just can't handle fast transient events - internal regulation can go crazy when driving pulse loads, especially where some inductance exists between supply and pulse circuit. (You might try twisting together the two supply wires between load & source).

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For a capacitor, I = C*dv/dt. Since I is 4.5 amps and C= 6800 uf, dv/dt is 662 volts per second. Therfore, since you turn on the FET for 150 uSec per cycle, the voltage should drop only about 100mV per cycle. You can perform the same calculation to determine that the voltage would increase 441mV per cycle with the half-amp power supply. Your scope says that you are discharging at a much higher rate. So the problem is not the charging, it is the discharging.

You might consider that you are greatly exceeding the current specification for this capacitor, which is 2 amps. I'm not sure what happens to the capacitance value when aluminum caps are discharged more rapidly, but you might try a cap with higher ripple current capacity like a film capacitor.

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  • \$\begingroup\$ The problem is the ESR of the capacitor which takes the charging to be longer. Also, the discharging rate is too fast compare to charging. For the 4.5A pulse which is beyond the ripple current of the capacitor, when you do the rms of 4.5A*Duty cycle it will be less than the ripple current specs. Correct me if I am wrong? \$\endgroup\$ – jasp Jul 18 '16 at 4:31

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