Calculation for resistor and capacitor in parallel...

[Externet]

AC------------------------------||---------------------------load-------------------------AC

An AC load works with a limiting current from a series capacitor.
Brief interruptions of the AC supply can impose doubled voltage to the load as charge stored in the capacitor may sum the supply.
In order to prevent such, a resistor in parallel to the capacitor allows its discharge before the AC is restored.

------> What is the safest convenient discharge time to choose ?

If half an AC cycle of 60 Hz is chosen as discharge time; the discharge time would be 0.008 seconds.
If the capacitor is 0.5 uF
t=RC
0.008 = R 0.0000005
R= 16000 Ohms maximum

If an arbitrary interruption duration of 10 Hz is chosen, the discharge time would be 0.17 seconds.
For he same 0.5 uF capacitor:
t=RC
0.17=R 0.0000005
R=340000 Ohms maximum

Do I have it right?

 

[MrAl]

Hi there,

In one time constant the voltage decreases to 0.368 of the max, so for a 120vac line that would mean the voltage would decrease to about 63 volts after a time of one time constant. If the time constant is 0.008333 (one half cycle of the 60Hz line) then after one half cycle the voltage would decrease to 63 volts if the circuit was switched off at the peak voltage.
If the time constant was half of that (around 4.167ma) then the voltage would decrease to 0.135 of the max or about 23 volts. Similarly, if the time constant was one third of that the voltage would be around 8.5 volts.

But the other thing to consider is the normal running current and the power. For example, a 16k resistor at 120vac would pass around 7.5ma on it's own so even without the capacitor there would be current already. And also the power would be around 0.9 watts so you'd have to use at least a 2 watt resistor.
To get it to discharge even more after one half cycle going to 8k would mean 1.8 watts in the resistor, and it would already pass about 15ma even without the capacitor (into a low voltage load).

So the considerations include the normal current being passed by the resistor as well as the power in the resistor for sizing the resistor and also efficiency.

 

[Externet]

Thanks, Al.
The use of the 'bleeding' resistor has the double edge sword of conducting too much current and disipating power which goes against the purpose of using a limiting capacitor in the circuit; and for the safer choice of a fast discharge.

----> Should the paralleled resistor be a supply path of not more than 1% of the load current, leaving 99% trough the capacitor ? What is the rationale ?

The documents I have read about seem empirical, usually suggesting 0.5 MOhm as a typical norm. That puts the discharge time to about 12 cycles at 60 Hz. Which may appear to be 'normal' to expect on a brief AC interruption. But shorter interruptions than 12 cycles would not be safely handled.

I do not see the "4.167ma" relation in your post above. Could you clarify ? Is it actually meant 4.167 ms (0.004167 seconds) instead ?
I will take my calculations as good, and the discharge time to stay arbitrary per the designer... unless your better suggestion.

 

[Ratchit]

Externet,

I hope I have my head screwed on correctly about this. First of all, let's talk about no interruptions and no bypass resistor. The capacitor in series with the resistance will change the current and voltage phase. If the reactance and the resistance are equal, then the current will lead the voltage by 45°. If the reactance is much larger than the resistance, then the phase difference will approach 90°. Also, if the reactance and resistance are equal, then the voltage will be 0.707 of the source, not 0.500 like it would be if the reactance were replaced by a resistor of equal ohms. Let's assume you can live with the above conditions.

You threw out some values of capacitance (0.5uf), hertz (60), voltage (120 volts rms), and interruption time (12 cycles), so let's use those. You can calculate new values if they are different. The worse thing that can happen is that the interruption occurs when the cap is fully energized and starts up again around 12 cycles later before the cap is fully deenergized. In 3 time constants, the cap will deenergize down to 5% of it max value. Its max value is 1.414 of 120 = 170 volts. 5% of 170 is 8.5 volts. One third of 12 cycles is 4 cycles, so the time constant is 4/60 = .067 seconds. Therefore the determination equation is (Rp||Rl)*0.5E-6 = 0.067 secs , where Rp is the resistor in parallel with C and Rl is the load resistor. At worse, the recovery voltage will be 170+8.5 volts. After you figure out Rp, you can determine if you can live with its power dissipation. Maybe the load resistor will be enough to deenergize the capacitor sufficiently.

Ratch

 

[MrAl]

Thanks, Al.
The use of the 'bleeding' resistor has the double edge sword of conducting too much current and disipating power which goes against the purpose of using a limiting capacitor in the circuit; and for the safer choice of a fast discharge.

----> Should the paralleled resistor be a supply path of not more than 1% of the load current, leaving 99% trough the capacitor ? What is the rationale ?

The documents I have read about seem empirical, usually suggesting 0.5 MOhm as a typical norm. That puts the discharge time to about 12 cycles at 60 Hz. Which may appear to be 'normal' to expect on a brief AC interruption. But shorter interruptions than 12 cycles would not be safely handled.

I do not see the "4.167ma" relation in your post above. Could you clarify ? Is it actually meant 4.167 ms (0.004167 seconds) instead ?
I will take my calculations as good, and the discharge time to stay arbitrary per the designer... unless your better suggestion.

Hi,

Yes i mean 4.167ms not 4.167ma which was a typo.

Since you are only worried about the surge itself and not human safety, the way to handle this would be to clamp the output to some fixed level with something like a zener or two. For the DC circuit where this is usually used, that just means a single zener, or you could oversize the DC filter capacitor such that it can absorb one or two (or so) such surges without allowing the voltage to rise too high. The burden of the design is then shifted to the *series* resistor rather than the parallel one, and that only sees the surge for a short time anyway, and there must be some series resistor there to limit the max inrush current anyway.

We could look at a few practical circuits to see how to size the filter capacitor. It's not hard to do really. With max surge calculate max current and then how much more the filter capacitor charges in the given time, and that determines if it is go or no go depending on what your following circuit can tolerate.

I made an LED line operated light once where there was no parallel resistor at all just a series resistor, and when it was unplugged near the peak of the line (or whatever phase it was) if you put your finger across the two AC prongs that would complete the circuit with the still charged capacitor and light up the little white LED for a few seconds until the cap completely discharged