Is a series resistor necessary when charging a capacitor?

AI Thread Summary
Connecting a 220 VDC source directly to a 250V 2200uF capacitor is generally acceptable for fast charging, though using a series resistor can help control current and smooth the DC output. A diode between the inverter and capacitor is unnecessary, as the inverter's rectifier already includes diodes that prevent reverse current flow. The current through the inductor will primarily be influenced by the inductance, coil resistance, and capacitor voltage. Residual voltage after discharge is expected due to diode drop, and while it can be minimized, it represents a negligible amount of energy compared to the total stored. Overall, a resistor is not essential for this application, but it could provide benefits in certain scenarios.
Slem
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I'm using 250V 2200uF capacitor to discharge high instantaneous current into an inductor as shown in the attached image.

I'm using a 12 Volt battery fed into an inverter that output 220 VAC which then is fed into a rectifier which outputs 220VDC.

My question; Can I connect the 220 VDC directly into the Capacitor terminals (activated by push button) without the use of a resistor in series? I just want fast charging.

And when i activate the discharge circuit would I decrease the current flowing through it because the capacitor is still connected to the charging circuit? (even though the switch is open)

I need as much current as possible to flow through the inductor . Would it help if i connect a diode between charging circuit and capacitor ? to block any minimal current wasted as heat flowing back. I'm not sure about what I just said I'm only guessing.
Thank you!
 

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In general, you do not need a resistor between the inverter and the capacitor, but there are good reasons for having one. If your inverter is just taking the output from the rectifier then you would probably want a resistor/capacitor or inductor/capacitor to act as a filter to give you a smoother DC output (which is immaterial here because it seems like you just want to charge the capacitor to get a current pulse). Another reason would be to control the current drawn from your source. You are stepping up the voltage by a factor of 18, which means that you need 18 times the current out of your source to get X amps out of your inverter. This may not be a problem if you are using a battery due to the internal resistance but something else to consider, perhaps.

No point in connecting a diode between the inverter and the capacitor. Without a return path you are not going to have current going back that way. Not to mention you already have diodes there from the inverter's rectifier stage.

The main thing that is going to control the current through the inductor is the inductance, the resistance of the coil, and the voltage across your capacitor. What's going to determine the time it takes to charge the capacitor is the power output of your source and the efficiency of your DC-DC converter.
 
Born2bwire said:
In general, you do not need a resistor between the inverter and the capacitor, but there are good reasons for having one. If your inverter is just taking the output from the rectifier then you would probably want a resistor/capacitor or inductor/capacitor to act as a filter to give you a smoother DC output

So my conclusion is that i don't really need a resistor for my application.
I just remembered another minor problem I had.. Whenever i discharged the capacitor I still had about 10V remaining in the cap . Is there a way I can make it discharge down to less voltage?
EDIT: I know perhaps by pressing the firing button longer (which isn't good for my application) I suppose my issue here is to decrease the time constant... But how ...
 
Slem said:
So my conclusion is that i don't really need a resistor for my application.
I just remembered another minor problem I had.. Whenever i discharged the capacitor I still had about 10V remaining in the cap . Is there a way I can make it discharge down to less voltage?
EDIT: I know perhaps by pressing the firing button longer (which isn't good for my application) I suppose my issue here is to decrease the time constant... But how ...


That's simply going to dependent on the time you allow to discharge, the inductance of the coil, and your resistance (most likely dominated by the coil). There is always going to be some residual voltage because of the diode drop. But you are talking about a miniscule amount of energy here. 10 V residual represents 0.2% of the energy that you originally stored up in the capacitor. Plus, that 10 V is only going to be able to drive a fraction of the current that you could at the pulse maximum.
 
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