limiting capacitor current on recharge

I'm designing a power surge circuit with hefty capacitors to help my power source with powering solenoids for short periods of time.
In the circuit is a ntc thermistor to limit the starting inrush current.

Now I want to make sure my problem solving is correct in practice:

After the capacitors discharge they will need to charge up again which will cause another high current surge, but since the circuit is already running the ntc thermistor would already be hot and have a low resistance.

So my solution:

I will increase my capacitors capacitance as to cause the needed discharge time for the needed current to be only say 2RC. Then when the capacitors will charge up again they will start from around 3RC and they wont draw as an insane current compared to Imax.

Will this solution work so long as I get my numbers right?

snuggyhuggy:
I will increase my capacitors capacitance as to cause the needed discharge time for the needed current to be only say 2RC. Then when the capacitors will charge up again they will start from around 3RC and they wont draw as an insane current compared to Imax.

I don't know what that means.

Why can't you just use a current limiting resistor?
If your charging current must stay below 5A and the fully-charged voltage is 12V, a 2.4 Ohm resistor will cover everything.

If you feel that slows the charging process down too much, look into constant-current circuits

Need to know the circuit - looks like your power supply might not be up to the job ?


(note: I forgot to add a switch )

I though of using a series resistor, but couldn't find one with proper wattage tolerance.
The need for the capacitor is less that the power supply cant handle it and more that I want to give a surge current for a quicker solenoid reaction (as the capacitor discharges once there is a voltage differential).

Plus I read that a sudden current spike in the circuit can cause the ac-dc power supply to recognize it as a short.

I know I can make a more efficient circuit with MOFSET, but my level is still low and I'd like to try to keep it simple and improve in the future

Why the voltage divider?

snuggyhuggy:
I though of using a series resistor, but couldn't find one with proper wattage tolerance.

Now that is hard to believe. But we can't help you finding one, since you don't reveal any numbers.
Hiw much current do the solenoids take for how long and how often?

couka:
Why the voltage divider?

Now that is hard to believe. But we can't help you finding one, since you don't reveal any numbers.
Hiw much current do the solenoids take for how long and how often?

The voltage divider is for the LED to safely make sure the capacitor has discharged completely on power off.

with ohms law the solenoids will take around 8A, but there is also the inductor factor of the solenoids which makes it so they aren't supposed to reach the point of such a current draw.

I have not gotten the parts yet so I have yet to test the timing in real world environment. Currently it seems I will have at the worst case scenario 3 solenoids powering at the same time for around 100ms

The capacitor has no value since I am undecided if I should use a capacitance twice what I need as to not enter a state of full discharge during runtime.

snuggyhuggy:
with ohms law the solenoids will take around 8A, but there is also the inductor factor of the solenoids which makes it so they aren't supposed to reach the point of such a current draw.

So, what do you actually mean by that muddled explanation?

If the power supply - presumably a switchmode regulated unit - is rated for the full expected current, you do not need to add anything.

If it is not rated adequately, you have a serious problem.

snuggyhuggy:
Plus I read that a sudden current spike in the circuit can cause the ac-dc power supply to recognise it as a short.

You read that, did you? :roll_eyes:

The power supply already has a capacitor on its output. On power-up that is fully discharged, if it "recognised" that as a short, it would never start up.

snuggyhuggy:
The voltage divider is for the LED to safely make sure the capacitor has discharged completely on power off.

Ok, these are 2 separate functions and should be implemented separately in the circuit:

snuggyhuggy:
with ohms law the solenoids will take around 8A, but there is also the inductor factor of the solenoids which makes it so they aren't supposed to reach the point of such a current draw.

This is DC, the inductance won't limit the current for any relevant time.

snuggyhuggy:
Currently it seems I will have at the worst case scenario 3 solenoids powering at the same time for around 100ms

Again, how often? Once a day? 5 times per second?

snuggyhuggy:
The capacitor has no value since I am undecided if I should use a capacitance twice what I need as to not enter a state of full discharge during runtime.

How do you know what capacitance you need?

Hi,
Very large VSDs have this problem when they are connected to their supply, the Bus capacitors are so large, the surge current sometimes exceeds the wiring and socket rating.

They limit the surge current by putting a current limit resistor in series with the supply rectifier and the cap.
A pair of relay contacts is wired across the resistor.
A circuit monitors the cap voltage and when the voltage is high enough, the contacts close and bypass the limit resistor.

I have seen circuits where the relay coil is across the capacitor and the coil characteristic controls at what cap voltage the relay activates.

Tom.. :slight_smile:

Paul__B:

with ohms law the solenoids will take around 8A, but there is also the inductor factor of the solenoids which makes it so they aren't supposed to reach the point of such a current draw.

So, what do you actually mean by that muddled explanation?

I'd take that to mean the solenoid pulse time is < L/R, although a real solenoid has a variable inductance and
various loss mechanisms.

couka:
This is DC, the inductance won't limit the current for any relevant time.

No, the OP specifically says short pulses, which is the typical situation in which capacitive boosting is used
to increase the response speed of a solenoid without cooking it. The inductance dominates in
such an application.

A good way to do this is charge the capacitor with a constant current limit, so that the high
voltage power supply is happy, and to reduce voltage droop on high duty-cycles (so long as the duty
cycle is below a threshold the cap will fully recharge to the same voltage before each pulse.)

How big are the capacitors to be charged? Voltage?

I've spent a lot of effort on this myself but I am working with supercapacitors up to 7F (one million times bigger than the 7uF in your diagram.)

For regular capacitors in the 10,000uF range it is not usually a problem. While the surge current is above the power supply rating it only flows for microseconds. If you've ever looked at a MOSFET datasheet you will see some crazy high peak current like 55A in a MOSFET that is useful for maybe 6A continuous.

The 55A rating has a time limit. Like 10us. So big capacitor charging currents can be tolerated for a short time. At these kinds of rise rates the inductance of straight power wires is significant and helps to cut the peaks.

When I get back to my computer I will attach some circuit ideas...