# High Current Pulse Opamp 2.5% Duty Cycle

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.

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?

• 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. Commented Jul 15, 2016 at 10:06
• @Claudio I have a series diode on between PS and capacitor so that there is no leakage going back to the PS.
– jasp
Commented Jul 15, 2016 at 10:08
• 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 Commented Jul 15, 2016 at 10:10
• But the current will limit the current for the capacitor to charge it during this pulses. I want to continuously charge the capacitor
– jasp
Commented Jul 15, 2016 at 10:11
• 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 Commented Jul 15, 2016 at 10:12