Simple Mains Rectifier Failed (Is it ESR?)

[Mechatrommer]

i slap this circuit into Mains and it miserably failed. i killed few 1N4007 and 1 big fat 6A10 tonight. fuse disintegrated into firework. tonight i did nothing but firing up many fireworks. any reason? the big fat 400V 33uF cap got heated. is it due to high ESR? if it is, how does this ESR mechanism shorting the circuit failing everything? i thought there should be a "pure capacitor" in series prohibiting Adc flowing? misterious!

ps: the big fat 6A10 diode hummed just for a moment before shorted and igniting the fuse.

[rbola35618]

Would putting a thermister help in reducing the initial surge current?

RB

[madires]

OMG :-) The inrush current for the cap isn't limited. The cap pulls whatever the power line delivers until the fuse or the diode blows. Is the hum of the diode 100 Hz? :-) There are some more issues, but PLEASE go for a transformer or two (back to back)!!!

[IanB]

I calculated the inrush current[1] and it is within limits so I don't know that this should cause the fireworks.

My guess is something like a simple wiring error or a faulty component.

Also note that 240 * sqrt(2) = 340 so any capacitors across the supply should be rated for at least 340 V.

The 500 k resistor is going to dissipate about 1/4 W, so it should be a 1/2 W or 1 W resistor.

[1] Calculated inrush current:

I = C * dV/dt

C = 33 uF
dV/dt = 340 / (0.25 * 1/50) = 68 kV/sec  (rise time from 0 to 340 V in a quarter cycle)

I = 33e-6 * 68e3 = 2.2 A

The 1N4007 has a peak non-repetitive surge current rating of 30 A, so this should not cause fireworks.

[saturation]

Hi mecha

On the idea this is for experimentation with all the suitable precautions, where the fuse is add a resistor in series with the fuse, say another 500kohm, to serve as a current limiter.  When you first power the circuit, the caps act like short circuits to the diode, and the full amps of your mains will be transiently applied to it.

Later replace this current limiting resistor with one that provides the necessary current, suitable rated for voltage and power dissipation.

[MartinX]

Plenty of gear rectifies the mains straight into the capacitor without any inrush current protection, like IanB says the diode should be fine and the fact that you heard humming for a while before the 6N10 failed indicates that you have a permanent fault condition, are you sure the cap is OK?

[madires]

[1] Calculated inrush current:

I = C * dV/dt

C = 33 uF
dV/dt = 340 / (0.25 * 1/50) = 68 kV/sec  (rise time from 0 to 340 V in a quarter cycle)

I = 33e-6 * 68e3 = 2.2 A

The 1N4007 has a peak non-repetitive surge current rating of 30 A, so this should not cause fireworks.

That would be the average current.

i(t) = j w C U0 e^jwt  (w = lower case omega).

But the current is not constant :-)

[IanB]

That would be the average current.

i(t) = j w C U0 e^jwt  (w = lower case omega).

But the current is not constant :-)

I'm confident about the calculation I performed for estimation purposes.

But OK, you have quoted a formula. Can you go ahead and use it to do a calculation and come up with your answer to the estimated peak inrush current?

[madires]

I'm confident about the calculation I performed for estimation purposes.

But OK, you have quoted a formula. Can you go ahead and use it to do a calculation and come up with your answer to the estimated peak inrush current?

Assuming that the cap is discharged and you connect the circuit to mains, the peak current is just limited by the resistance of wires, fuse, leads and so on. The worst case is if you connect the circuit while the mains voltage has the right polarity and is at its maximum. For that you get I = V / R = 240V * sq(2) / R = 339V / R. But that current flows just for a very short time and decreases fastly.

[sorin]

The peak current should be limited from dV/dt characteristic of the diode, the diode can't be open instantly

[IanB]

Assuming that the cap is discharged and you connect the circuit to mains, the peak current is just limited by the resistance of wires, fuse, leads and so on. The worst case is if you connect the circuit while the mains voltage has the right polarity and is at its maximum. For that you get I = V / R = 240V * sq(2) / R = 339V / R. But that current flows just for a very short time and decreases fastly.

You're on the right lines, but I see you have picked a different formula to work with from the one you originally gave.

When we have a problem like this to solve we first should look at the configuration of the system and build a picture of what we expect it to do. Then we can come up with a suitable model that will describe the system responses we want to understand and do some calculations with that model.

If we do that with the circuit illustrated we see a diode charging up a capacitor from an AC supply. Current will only flow through the diode when the forward voltage across the diode is positive. Once the capacitor charges up to the peak mains voltage there will be no more positive voltage difference across the diode and no continuous current will flow. So any continuous current formulas will not apply and we need a transient analysis.

Next, we want to determine the maximum current that may flow. Since we are looking for a maximum we can make some conservative assumptions. One such assumption is to assume that all resistance in the circuit is zero. That will produce the biggest current.

If all resistances are zero, the capacitor being charged is the only circuit element we need to consider.

We need a transient or dynamic formula for the capacitor so we can calculate what kind of current will flow through it. The dynamic formula for a capacitor is the one I gave above:

I = C dV/dt

This says that the current in the capacitor is proportional to the size of the capacitor and proportional to the rate of change of voltage across its terminals.

The capacitor has a fixed size of 33 uF, so we just need to estimate the maximum rate of change of voltage.

If our mains supply is 240 V AC, 50 Hz then the peak voltage at the top of the waveform is 240 * sqrt(2) = 340 V. Looking at the shape of the waveform, the voltage can rise from zero to 340 V in one quarter of a cycle, which would be 1/200 second. This gives us our dV/dt value as 340 / (1/200) = 68000 V/s. This is not exact, but it is good enough for our purposes. If the result is too close to the edge we should go back and refine the calculation.

We can now plug the numbers into our formula:

I = C dV/dt = 33e-6 * 68000 = 2.2 A

This is the maximum surge current likely if we happened to connect the mains supply right at the zero crossing point and the capacitor was fully discharged.

Since 2.2 A is much less than the diode rating of 30 A we don't need to refine the calculations and we can sit comfortably. The diode should not blow up.

[madires]

The peak current should be limited from dV/dt characteristic of the diode, the diode can't be open instantly

Right! If we want to calculate the real peak current we would end up with a quite complex function. That's more of an academic problem than a quick simplified calculation :-) 

[madires]

I = C dV/dt

This says that the current in the capacitor is proportional to the size of the capacitor and proportional to the rate of change of voltage across its terminals.

Or you could say that the voltage at the caps terminals is proportional to the integral of the current over time. Ok?

For a cap the voltage follows the current. So there has to be a current first before the voltage rises.

The question is now, what happens at the very first moment after connecting the circuit?

[sorin]

is not possible that the 6A10 is destroyed the fuse should prevent that, unless you use 100A fuse

[madires]

is not possible that the 6A10 is destroyed the fuse should prevent that, unless you use 100A fuse

A fuse needs to heat up to break, e.g. it needs a specific amount of thermal energy generated by the current flowing through it's internal resistance. The high inrush current of the cap flows only for a very short period of time. That might not generate enough thermal energy to brake the fuse.

[sorin]

1A fuse have a coeficent 0.6=I²*t
so definitely you don't use a fuse
assuming that the cap i fault and is short, the diode don't blow

[IanB]

For a cap the voltage follows the current. So there has to be a current first before the voltage rises.

That is a misconception. The statement that current leads voltage or voltage lags current only applies when considering a steady state situation with a cyclic AC wave like a sine wave. It is an outcome of solving more fundamental equations, it is not a description of how capacitors work.

The question is now, what happens at the very first moment after connecting the circuit?

That is indeed the question, and it highlights something I missed in my first analysis. When evaluating the transient surge current we should consider the worst case scenario, and that occurs not when the switch is closed at the zero crossing point, but when the switch is closed at the peak voltage. In that case the full 340 V will be applied across the capacitor terminals and the instantaneous current will be limited only by the resistances in the circuit including the capacitor ESR. However, this peak current will only last a short time until the capacitor charges up and this is why the capacitor time constant comes into the picture.

Let's look at it that way.

If we assume the system resistance including ESR is 1 ohm, then 340 V will produce a current of 340 A. Now the charge on a capacitor is given by Q = CV, so at 340 V the capacitor will have absorbed a charge of 33e-6 x 340 = 0.011 coulombs. Since 340 A is 340 coulombs per second, the time to charge the capacitor will be at fastest 0.011/340 = 32 microseconds. Now this may or may not be more than the diode can handle. We might reasonably average this current out over 8.3 ms as this is how the datasheet quotes the maximum surge current. If we do that we get 340 A * 0.032 / 8.3 = 1.3 A.

So we still seem to be well inside margins and most likely don't have to do more sophisticated analysis. I'm giving above a simplified way of thinking through the problem, but for more detailed analysis there are the design references that others have quoted.

[Mechatrommer]

PRECAUTION: for beginners watching this circuit, if you dont have enough clue, dont do it at home, or do it under supervision of an adult or expert. we are dealing with 330V of DC potential of naked copper here. more than enough to get you killed under the right (wrong) condition. i did put transparency in between my eyes and this circuit in every powering up.

hmm this one seems controversial. let me put the complete picture of what i've experienced last night... actually i want to build a "Mains FG" using bjt opto "switcher" (lets call it "switcher" for simplicity i dont know the precise term, because it will be switched later on), but first, build from simple starting point and test the circuit first. all components are salvaged except resistors and pc817. here's the schematics...


and i firstly built the switcher circuit on the board (perf?) like the following... the rectification circuit is on another board (bottom most pic)


failed! few 1n4007 died on the other board (i was using full bridge rectifier, 2 died at a time, and later just using the half bridge to save the 1n4007). i was about to post a thread about the above schematic, but i found the reason early that the pcb is carbonizing and shorting, i just realized the truthfullness of mike's comment on the other thread. the carbon created a nice longer lasting neon light before BANG! so i redo on another board, dead bug style, should have alot of creepage now. i believed the C4106 and PC817 are already dead by this time since all leads are shorted. but for the sake of RnD i continued since my curiosity is that i cant find any low impedance dc path for the circuit to go BANG. here's the later soldered revision (same circuit like above)..


i put sticky flux on every components so i can visually see if they got heated (smoke), success! no BANG albeit the shorted bjt and opto. but only for 10-20 seconds, the resistors smoking up before another big BANG. all resistors and capacitors survived, persistently the diode and fuse got sacrificed. fyi, diode died without a BANG, it simply silently got shorted, the BANG is on the fuse (small diy copper wires since the original fuse is gone in the first BANG i dont have much spare for this testing so its DIYed).

and here's the subject matter. i removed the "switcher" leaving the rectification pcb alone. i cant find any carbonizing occured. the story and schematic is in the OP. in the bottom part of the picture below is the said capacitor and few dead bodies of diode. i've checked the capacitance and leads continuity of both capacitors involved, both indicating good result. i dont have esr meter.




few responds from this thread:

1)

Would putting a thermister help in reducing the initial surge current?

i know its not the proper circuit, but it still a mistery for me. the circuit did not failed immediately so i believed its not inrush issue. components yielding is more probable. but in the final test i was using the big fat 6A10 which also failing, reducing the possibility of components yield (maybe? i mean how can a big guy failed even it is old?).

2)

It sounds like you have something wired incorrectly

i've provided the picture for you to judge, i've checked every circuit's nodes to comply with the drawing. this is the first error i checked in every failure and this is also the reason i removed the switcher circuit to rule out incorrect wiring. i believe i wont make mistake for this very simple half bridge rectifier. but eventually it also went BANG, not immediately.

3)

The question is now, what happens at the very first moment after connecting the circuit?

temperature-wise i dont know. i was not touching the circuit while powering up (safety reason ) but as already said, the circuit did not fail immediately.

4)

is not possible that the 6A10 is destroyed the fuse should prevent that, unless you use 100A fuse

i was using DIY fuse later on (copper wire) i put the smaller i have, but i guess its still too big to fail first before the diode, its quite thin though about a hair sized wire.

the diode hum i mentioned was very faint its like vibration on the wooden bench, but if i'm not mistaken, thats the usual mains hum.

[Mechatrommer]

is it possible that this very high inrush, although its for very very short time, is enough to yield the diode in another shorter time afterward of operation?

[nixxon]

LTSPICE the circuit! After all this thinking and calculations. It would take less than 3 minutes for someone familiar with the software to simulate it.

I tried, but for some reason failed to import diode 1N4007 characteristics.

I can't figure out how to alter a general diode using SPICE directive the way described here:
http://www.electro-tech-online.com/general-electronics-chat/92131-ltspice-issue-2.html

.MODEL 1N4007 D(Rs=0.010 Tt=2u Cjo=15p Vj=0.6 BV=1000 Ibv=5u)

C'mon:, SPICE it!

[madires]

is it possible that this very high inrush, although its for very very short time, is enough to yield the diode in another shorter time afterward of operation?

If the current is large enough to do so some minor damage, I would guess so.

Another thought about the arc. If the 33uF cap is charged to 330V and the polarity of the mains voltage changes to -330V you got 660V across the diode.

[Mechatrommer]

Another thought about the arc. If the 33uF cap is charged to 330V and the polarity of the mains voltage changes to -330V you got 660V across the diode.

i believe thats not possible since the diode's leads are separated far away, or if you are talking reverse voltage, 6A10 is capable of 1000V reversed.

[lewis]

1. Watch the voltage rating of all the resistors. Pairs in series is better for voltages above 250V DC.
2. It's possible you have counterfeit diodes, they might actually be lower voltage types
3. 220-470R series resistor would help

[IanB]

You can minimize fireworks if you use a low energy supply for testing. Put two small transformers back to back to make an isolation transformer. Then even if you have an accidental short there won't be enough power available to make flashes or bangs.

Secondly, test the circuit on lower voltages first. Try 24 V, 60 V, 120 V then 240 V in sequence. A small 6-0-6 and a small 12-0-12 pair of transformers will let you generate these voltages safely.

[TerraHertz]

Argh! You cannot use veroboard for mains voltage circuits! Not without a copper-free clearance of at least two hole spacings (preferably three!) between all circuit elements. But I guess you learned that already.
Also be aware that at these voltages, the paint on resistors, the plastic sheath on electros, etc are NOT rated to withstand breakdown. On the back of that small copper-backed board, you've got a 33M resistor with the metal endcap bridging a gap between copper areas. Bad!

Anyway, either the diode or the electro, or both, are now dead. Possible causes:
* Salvaged diodes, got overheated during desoldering, either outright dead or reduced reverse voltage margin.
* Very high brief inrush current when you closed the switch while mains was at positive peak and cap was completely discharged. Damage to either the cap, or diode. If the diode, then the cap gets 240VAC applied and expires rapidly due to high AC current, heating, and possible internal shorts or opens.

If you want to diagnose its operation, and don't have a variac, then try this - remove the 500K (470K?) discharge resistor. Put that resistor in series with the mains input.
Now if the components are all OK, the cap will charge up _slowly_. Multimeter will do for observing behavior.
Does the cap charge? To what voltage? Does it hold charge when mains removed?

[Obligatory warning, about working with circuits that will kill you if you touch them.]