How to calculate current limitation in an AC capacitive drop circuit

Thread Starter

arishy

Joined Apr 26, 2014
109
A very simple ac circuit:
1. ceramic cap 0f 0.22 uf 400v with 1 M ohm resistor( in parallel)
2. 8 Amp Bridge rectifier
3. 100 ohms 2 W current rush resistor

Source 240 v 50 Herz
z approximately 14Kohm
AC current around 18 mA

The voltage after the bridge is about 40 DC

If I put 8 A dc load, will it blow a fuse ( if installed)? This is my post question
My confusion is: what component ( if any ) from the above, will NOT allow me to do so
 

Marley

Joined Apr 4, 2016
475
I assume your circuit looks like this.
The current in the circuit is limited mainly by the impedance of the capacitor.
If with a 240V 50Hz supply you short together the DC output terminals the output current (in fact the current from the supply) is limited to about 16mA.
So you can't have a 8A DC load.
 

Attachments

BobTPH

Joined Jun 5, 2013
4,906
It is not reasonable to get 8A from that type of supply.

How did you choose the capacitor? If you had done the calculation for 8A it would be 500 times larger.

Bob
 

Ramussons

Joined May 3, 2013
1,264
A very simple ac circuit:
1. ceramic cap 0f 0.22 uf 400v with 1 M ohm resistor( in parallel)
2. 8 Amp Bridge rectifier
3. 100 ohms 2 W current rush resistor

Source 240 v 50 Herz
z approximately 14Kohm
AC current around 18 mA

The voltage after the bridge is about 40 DC

If I put 8 A dc load, will it blow a fuse ( if installed)? This is my post question
My confusion is: what component ( if any ) from the above, will NOT allow me to do so
A .22 uF @ 50 Hz will offer about 15 K Ohms. The 100 ohms resistor may as well be a short. It plays no role in current limiting.
The current will be approximately 240 / 15 mA.
 

Ian0

Joined Aug 7, 2020
5,132
The 100Ω resistor is essential. When the circuit is switched on there is a voltage step at the input, which results in a high current pulse through the capacitor. If the next part of the circuit is a zener - the zener blows up if there is no limiting resistor.
 

Kjeldgaard

Joined Apr 7, 2016
461
The circuit in #2 lacks a load to keep the voltage down at the output, typically a zener diode directly at the output.

Without this, the output will rise towards the peak voltage at the input.
 

Thread Starter

arishy

Joined Apr 26, 2014
109
This is a very good starting point to clear my confusion. The reason I picked the 0.22 uf is I want an out of 40 Volt DC.
It is now clearer to me; the limitation of the capacitive dropping concept. Your schematic is correct.
Bare with me on this issue. Can I redisign the circuit to give me High current?. If yes what is the value of the components AND how high can I go?
The parameters are 240v Source with 50 Herz. The bridge rectifier is "capable" of 8 AMP
 

DickCappels

Joined Aug 21, 2008
8,708
A .22 uF @ 50 Hz will offer about 15 K Ohms. The 100 ohms resistor may as well be a short. It plays no role in current limiting.
The current will be approximately 240 / 15 mA.
Agreed the 100 ohm resistor plays no role in the operation of the circuit, but the 100 ohm resistor is absolutely crucial. Imagine that before the circuit is connected to the mains that the capacitor is fully discharged, also imagine that the circuit connects to the mains during the peak of the mains waveform, which for 240 VAC is 339 volts peak. Without the resistor the capacitor will charge with the current limited by the impedance of the power line and the ESR of the capacitor. That kind of current spike is often able to kill components. I did the experiment and have the dead components to prove it.
 

Ian0

Joined Aug 7, 2020
5,132
I know this from experience, but if you have designed it without the current limit resistor, you can often make it sufficiently reliable by using a higher wattage zener diode.
 

Sensacell

Joined Jun 19, 2012
2,998
The problem with getting higher current out of a simple capacitive dropper is the inherent conflict between two operating conditions:

1) Inrush current, discharged capacitor, peak voltage about 325 Volts.

2) Normal operation - requires a Very Large capacitor with a low enough impedance to supply the current.

You see that this large capacitor will make your inrush current destructively enormous.
You cannot use a resistor to limit this inrush, because it limits the current, and will start to dissipate massive amounts of power.

High current capacitive dropper = not really workable in reality.
You would only use this design for a few mA at most.
 
Last edited:

Ian0

Joined Aug 7, 2020
5,132
The problem with getting higher current out of a simple capacitive dropper is the inherent conflict between two operating conditions:

1) Inrush current, discharged capacitor, peak voltage about 325 Volts.

2) Normal operation - requires a Very Large capacitor with a low enough impedance to supply the current.

You see that this large capacitor will make your inrush current destructively enormous.
You cannot use a resistor to limit this inrush, because it limits the current, and will start to dissipate massive amounts of power.

High current capacitive dropper = not really workable in reality.
That’s why high current droppers are usually inductive (ballasts for disaggregated lighting)
 

Thread Starter

arishy

Joined Apr 26, 2014
109
The current is proportional to the capacitance.
I ≈ V/(2πfC)
According to "my" calculations; to get 1 A you need a ceramic cap of wapping value 0f 14 uf !!!!!!!!!
I do not think such a "beast" exists.
It is obvious to me now; that this approach is used ONLY for small currents in the mAMP range.
 

Ramussons

Joined May 3, 2013
1,264
Agreed the 100 ohm resistor plays no role in the operation of the circuit, but the 100 ohm resistor is absolutely crucial. Imagine that before the circuit is connected to the mains that the capacitor is fully discharged, also imagine that the circuit connects to the mains during the peak of the mains waveform, which for 240 VAC is 339 volts peak. Without the resistor the capacitor will charge with the current limited by the impedance of the power line and the ESR of the capacitor. That kind of current spike is often able to kill components. I did the experiment and have the dead components to prove it.
The argument is logically valid. Extending the argument further means that any device connected at the bridge rectifier output will blow just because even that 100 ohm resistor will allow a 3 Amp surge through the load !!! Or am I missing something?
 

DickCappels

Joined Aug 21, 2008
8,708
No, you are not missing a thing. The surge limiting resistor gives you control over the maximum surge so you can take that into account.

Many rectifiers are rated for inrush currents that are much greater than their rated average current, but we also have to consider every other part in the circuit, including the fuse that @arishy should probably include if they ever actually build this circuit.
 

BobTPH

Joined Jun 5, 2013
4,906
Maybe I am blind today but I see no path for inrush current except through the load. Depending on what the load it, a limiter may ir may not be needed, right?

Bob
 

ronsimpson

Joined Oct 7, 2019
2,067
but the 100 ohm resistor
On the power line there may be high frequency noise and harmonics of the 50/60hz.
The " 0.22 uf 400v " capacitor, by definition, passes high frequencies to the bridge diodes and to the output. There could be enough energy in the harmonics to charge the load more than intended.
There will be spikes on the power line when large motors turn on and off. These spikes will want to pass through the cap.
The 100 ohm resistor and the output filter capacitor makes a low pass filter, to remove some of the high frequency energy.
 

BobTPH

Joined Jun 5, 2013
4,906
It occurred to me that you can get 8A or higher through your 0.22 uF capacitor. Just put an inductor of 46H in series with it so that it resonates at 50 Hz.

Bob
 
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