Dissipating capacitor stored energy when device turns off

Thread Starter

mike__MecE

Joined Feb 20, 2022
69
Hello, I have attached a simplified schematic of a larger and more complex model for the purposes of this question. The circuit receives V+ (48V) externally and a high side switch (BTS6163D) is used to turn the device power on and off. The 48V is inputted into a number of LM317HVs. Most of these LM317HV are used in the current limiting configuration and charge large capacitors on the output/load side (1000uF or more). One LM317HV is used to supply 5V to a microcontroller and OLED screen. I added protection diodes so that the large capacitor on the output of the LM37HV does not discharge into the output of the regulator when the V+ (48V) input drops to 0V (BTS6163D turned off). My issue is that the energy in the capacitors is entering the LM317HV powering the microcontroller and OLED. There is so much energy that the OLED is staying on for sometimes up to 30 seconds after the V+ (48V) power is switched off by the BTS6163D. (Qt:9 1000uF caps at 48V is more than 9 Joules; if the OLED and micro use 25-100mA, that's approx. 9-36 seconds to discharge). So my request is ideas to discharge the caps and stop that energy from powering the LDO for the OLED/micro. Thanks
 

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wayneh

Joined Sep 9, 2010
17,471
It's not obvious to me from your schematic how the LM317 powering the 5V supply is getting its power and how the capacitor might be a part of that. I think you over-simplified it?

Oh, maybe I get it now. The big, charged capacitor is essentially taking the place of the 48V supply and supplying power to the LM317 until it discharges?

Maybe you just need another switch on the power supply line to the 5V LM317, controlled by the same signal as your main power switch. This would disconnect it from the 48V bus when you turn that off.

Or, you could put your protection diode in series with the charging current, so that the cap cannot discharge into the regulator. Then you wouldn't need to shunt the cap to the Vin side of the regulator.
 
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ElectricSpidey

Joined Dec 2, 2017
2,631
I like wayneh's idea of placing the protection diode in series with the charging current, also a typical solution for this kind of condition is "steering" diodes before the 317s.

Please note that either solution won't discharge the caps, only prevent them from powering the micro and OLED.
 
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Thread Starter

mike__MecE

Joined Feb 20, 2022
69
Or, you could put your protection diode in series with the charging current, so that the cap cannot discharge into the regulator. Then you wouldn't need to shunt the cap to the Vin side of the regulator.
OK, so diode facing down at point B so that current can only flow into the cap from the regulator. I like that. That is what you meant right?
 

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Thread Starter

mike__MecE

Joined Feb 20, 2022
69
I like wayneh's idea of placing the protection diode in series with the charging current, also a typical solution for this kind of condition is "steering" diodes before the 317s.

Please note that either solution won't discharge the caps, only prevent them from powering the micro and OLED.
The steering solution would place a diode (pointing to right of page) at A, correct? so that current cant flow to the other LDOs.
 

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Thread Starter

mike__MecE

Joined Feb 20, 2022
69
Why do you need to discharge the capacitors?
I don't know. My original goal was just to avoid current flowing backwards through the LM317. Others suggested changing the diode placement so that current can only flow into the anode from the LDO. But considering it further, is it good practice to discharge a 1000uF-2000uF cap for safety reasons? I would prefer them to discharge when the power is turned off I think. I suppose I could just add a 100k resistor between the cap anode and ground - that would discharge the caps over a few minutes.
 

ElectricSpidey

Joined Dec 2, 2017
2,631
Both are correct.

If you choose to place the diode before the 317 I would also keep the protection diode across the 317.

Yes, you can place a large value resistor across the cap to provide a discharge path, also known as a "bleeder" resistor.
 
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