Simple capacitor charging circuit problem

[ZeroResistance]

I have the following circuit for charging a capacitor.
Basically to back to back 230Vac / 12V ac transformers so that I can check the output on a scope without earth problems.

Now at the output capacitor I am expecting 230V x 1.414 = 325V DC

Instead when I check with a multimeter across the capacitor I get around 25V DC. 

What am I doing wrong here?

[Benta]

What am I doing wrong here?

Your schematic looks OK. The only problem could arise if your cap is very small. What size/type/voltage is it?

[Rbastler]

Now at the output capacitor I am expecting 230V x 1.414 = 325V DC

For single wave rectification the output voltage is more like times 1.2
I would post a picture page from of the book were I read it, but with a lot of my stuff atm I don't have it here. It's some older german electronics book.

[AndyC_772]

Reverse leakage through your diode, perhaps. At negative peaks you have about 325V plus the capacitor voltage across it. If its reverse leakage means its effective resistance is comparable to your 220K resistor, then it won't be rectifying.

[Zero999]

Now at the output capacitor I am expecting 230V x 1.414 = 325V DC

For single wave rectification the output voltage is more like times 1.2
I would post a picture page from of the book were I read it, but with a lot of my stuff atm I don't have it here. It's some older german electronics book.

Where did you get 1.2 from?

The original poster was correct. The capacitor will charge up to the peak mains voltage which is equal to the RMS voltage multiplied by root(2) or roughly 1.414. There will be a small 1V to 2V drop for the rectifier, depending on the load current.

[alm]

I would second the question about the type of capacitor. If you unplug it, how fast does the voltage across the cap decrease? The cap might be leaky at that voltage. Try putting it across a stable DC voltage source, ideally of at least 25 V, and measure the current with a DMM (or put a 1kOhm resistor in series and measure the voltage across it). Also, what AC voltage do you measure at the secondary winding of the second transformer (i.e. before the diode)?

For single wave rectification the output voltage is more like times 1.2
I would post a picture page from of the book were I read it, but with a lot of my stuff atm I don't have it here. It's some older german electronics book.

These values are usually based on some expected ratio of capacitance to load current, and take into account the ripple that will remain at that current draw. This is pretty much a peak detect circuit, so I would expect it to get very close to the peak voltage, which would be about sqrt(2) times the nominal voltage if not heavily distorted.

[Rbastler]

Now at the output capacitor I am expecting 230V x 1.414 = 325V DC

For single wave rectification the output voltage is more like times 1.2
I would post a picture page from of the book were I read it, but with a lot of my stuff atm I don't have it here. It's some older german electronics book.

Where did you get 1.2 from?

The original poster was correct. The capacitor will charge up to the peak mains voltage which is equal to the RMS voltage multiplied by root(2) or roughly 1.414. There will be a small 1V to 2V drop for the rectifier, depending on the load current.

From and quite old book about electronics

[madires]

325V / 220kOhms = 1.5mA (peak current)

[alm]

What can you do with that figure unless you now the capacitance and leakage of the capacitor?

[ZeroResistance]

What am I doing wrong here?

Your schematic looks OK. The only problem could arise if your cap is very small. What size/type/voltage is it?

The cap is 2u2, 400V its seems to be either a polyester or polypropylene cap is is epoxy coated and has MPP marked on it.

[ZeroResistance]

Reverse leakage through your diode, perhaps. At negative peaks you have about 325V plus the capacitor voltage across it. If its reverse leakage means its effective resistance is comparable to your 220K resistor, then it won't be rectifying.

The cap discharges quite fast when I switch of the mains. I am checking this with a multimeter, so its discharging thru the multimeter resistance and the reverse leakage of the diode perhaps??

I changed the diode from 1N4007 to 1N5408 and now the voltage has moved up to 270VDC, I have checked the 1N4007 with a multimeter and nothing seems to be wrong with it.

[ZeroResistance]

Also, what AC voltage do you measure at the secondary winding of the second transformer (i.e. before the diode)?

I measure around 220Vac

[madires]

What can you do with that figure unless you now the capacitance and leakage of the capacitor?

It gives an idea about the order of magnitude we're talking about, and also where to look for explanations. I agree, that the knowledge about the capacitor used is important, but we can start without that from another angle of view. Experience teaches us that sometimes you don't have all the information you need, and that you have to work with what you have.

[AndyC_772]

I changed the diode from 1N4007 to 1N5408 and now the voltage has moved up to 270VDC

Excellent!

I have checked the 1N4007 with a multimeter and nothing seems to be wrong with it.

Your multimeter isn't applying hundreds of volts of reverse bias. If you have a high voltage supply, you could test the reverse leakage current under more representative conditions.

[ZeroResistance]

Excellent!

270V is better that 25V but I was expecting something around 320V any idea what could be the issue?
I am checking this with a multimeter probably a scope would give a better idea of what the signal looks like?

[MrAl]

Hello,

270vdc would imply an AC voltage of about 191 volts rms.

If you have a scope you can (carefully) check the waveform output of the transformer(s) as sometimes the waveform wont be perfect because of the properties of the core material and the excitation current.

If you suspect the diode leakage, try two in series.  With 230vac rms input the half wave diode will see a reverse voltage of about 650 volts.

It goes without saying these are dangerous voltages here.

[AndyC_772]

270V is better that 25V but I was expecting something around 320V any idea what could be the issue?
I am checking this with a multimeter probably a scope would give a better idea of what the signal looks like?

If the problem is diode leakage, then it may well be that your new diode is much better than the original one, but still not ideal.

Your transformer turns ratio may not be exactly what it says on the tin.

Why such a high value resistor? If you use a lower value, what happens to the capacitor voltage?

[ZeroResistance]

Your transformer turns ratio may not be exactly what it says on the tin.

Why such a high value resistor? If you use a lower value, what happens to the capacitor voltage?

I wanted to limit the peak power dissipation in the resistor. I tried a lower resistor with 1N4007 there was no difference. I think I tried 4 100k's in paralllel.

[alm]

The latter suggests a diode that is breaking down to me. If the cap was leaky, then I would expect increasing the current to increase the voltage across it.

Since the circuit is floating, you could do measurements with a scope. The signal from the second transformer could be interesting, to see how sinusoidal that 220 VAC is and what the amplitude is. Another interesting measurement would the voltage across the 220 kOhm resistor (ground clip between resistor and cap, probe tip between resistor and diode). This should be proportional to the current through the cap, and might show the diode breaking down. You would expect a positive voltage during for a few milliseconds near the maximum value of the AC signal, and possibly a negative voltage during the negative cycle if the diode is breaking down.

Obviously these are potentially dangerous (hands off) measurements and you should not be using more than one scope probe at a time.

[viperidae]

Are you sure it was a 1n4007 and not the 50v rated 1n4001?

Wouldn't that explain the 25v across the cap? Half way between +325v and -275v

[Damianos]

Its a simple circuit that is a good exercise, to show us that we must not overlook the parameters!

Simulating the circuit, to evade from the math calculations, we have a result that shows a maximum voltage of around 261V.

This is not strange, if we take in account that the capacitor, in each period of the input signal, has a short time to charge, from a high resistance source and a long time to discharge via the load plus the leakage current of the diode.

In the simulation we have to examine what happens many seconds after applying the input voltage, because the time constant of the circuit is very high relatively to the period of the signal.

[ZeroResistance]

Are you sure it was a 1n4007 and not the 50v rated 1n4001?

Wouldn't that explain the 25v across the cap? Half way between +325v and -275v

I just inspected the diode and its turns out to be 1N5819 schottky 
Don't know how it landed in the 1N4007 box.

The peak reverse voltage of this diode seems to be 40V, so there was probably no rectification happening, I would have probably got some distorted signal which the multimeter was showing as 25V on the DC range

I'm so sorry I set you'll on a wild goose chase 

Having said that I would still like to investigate why I'm getting only 270V instead of the agreed upon 325V

[StillTrying]

Having said that I would still like to investigate why I'm getting only 270V instead of the agreed upon 325V

The 10M ohm DVM accounts for most of it, and probably the diode capacitance for a bit more, plus maybe a flat topped mains waveform.
You could try a smaller resistance, ~56K, or 2 diodes in series.

[ZeroResistance]

Having said that I would still like to investigate why I'm getting only 270V instead of the agreed upon 325V

The 10M ohm DVM accounts for most of it, and probably the diode capacitance for a bit more, plus maybe a flat topped mains waveform.
You could try a smaller resistance, ~56K, or 2 diodes in series.

Currently I have used 4 100K in parallel (effective resistance 25K) and diode is 1N5408, will try with 2 of these diodes in series.

[MrAl]

Hi,

Where did i hear that before, oh yeah now i remember :-)

Hello,

270vdc would imply an AC voltage of about 191 volts rms.

If you have a scope you can (carefully) check the waveform output of the transformer(s) as sometimes the waveform wont be perfect because of the properties of the core material and the excitation current.

If you suspect the diode leakage, try two in series.  With 230vac rms input the half wave diode will see a reverse voltage of about 650 volts.

It goes without saying these are dangerous voltages here.