Capacitors across rectifiers;why?

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It's interesting that I'm presently putting an article together on aspects of using 1N4007/UF4007 for valve amps, and I had just tried to measure the reverse leakage current through some sample 1N4004 diodes. 

I failed miserably when using the 1kVDC 2000 megohm range on a generic digital megohm meter - the leakage current was down below 0.5uA, even after heating up the sample diodes to what would have been well above 100C (using my quartz halogen desk lamp). 

I haven't got the incentive up yet to dust off a BPL RM175-LZ Mk2 megohm meter that's languishing with some other unrestored test equipment - that beast has the ability to read 5,000,000 megohm!
 
Caps across diodes are used to encourage equal (reverse) voltage sharing when diodes are connected in series.
For noise suppression you don't need a cap across each diode, you can just use a single cap across the transformer -the equivalent circuit is the same either way. For better results you need to add a resistor too, so you get proper snubbing.

 

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merlin said:
Caps across diodes are used to encourage equal (reverse) voltage sharing when diodes are connected in series.
For noise suppression you don't need a cap across each diode, you can just use a single cap across the transformer -the equivalent circuit is the same either way. For better results you need to add a resistor too, so you get proper snubbing.

Can you provide a link to the pic you posted please?

Cheers

Ian
 
merlin said:
It's from my HiFi book:
http://valvewizard.co.uk/Book3.html

LOL. Been meaning to buy myself a copy of that!

Can you confirm the vertical scales are all identical. What secondary AC voltage was used in the tests?

Cheers

Ian

PS. What was the transformer construction - EI or toroid?
 
ruffrecords said:
Can you confirm the vertical scales are all identical. What secondary AC voltage was used in the tests?
Yes the scales are all identical. It was a simple 12V transformer driving a full-wave rectifier. What you're looking at is the peak of the secondary AC waveform when the rectifier switches off, either with no cap, a cap across the secondary (slugger), or an RC (snubber) across the secondary.
 
merlin said:
Yes the scales are all identical. It was a simple 12V transformer driving a full-wave rectifier. What you're looking at is the peak of the secondary AC waveform when the rectifier switches off, either with no cap, a cap across the secondary (slugger), or an RC (snubber) across the secondary.

OK, understood. What was the load?

I wonder why every article I have seen about this topic uses a 12V transformer?? I have not had any audible effects of switch off transients in linear heater supplies (but I automatically include a snubber capacitor in all such supplies now). However, I have had problems with switch off in HT supplies interfering with the audio signal. I found a 100nF directly across the input to the bridge completely eliminated the spike. Of course the leakage inductance, dc resistance and winding capacitance in an HT winding is completely different to that of a 12V secondary so perhaps that explains the difference. Have you carried out any tests with HT supplies??

Cheers

Ian
 
The step in the  secondary voltage waveform in Merlin's cro screens is pretty much V = L.dI/dt, where L is the leakage inductance of that secondary winding and dI/dt is the dI/dt of the current waveform as the winding (ie. diode) current gets to zero.

Reducing the peak level of the diode forward current, and increasing the conduction duration of the diode current (ie. the ratio of effective winding resistance to load resistance), will lower the dI/dt at I=0.
 
trobbins said:
The step in the  secondary voltage waveform in Merlin's cro screens is pretty much V = L.dI/dt, where L is the leakage inductance of that secondary winding and dI/dt is the dI/dt of the current waveform as the winding (ie. diode) current gets to zero.

Reducing the peak level of the diode forward current, and increasing the conduction duration of the diode current (ie. the ratio of effective winding resistance to load resistance), will lower the dI/dt at I=0.

I do not understand the second sentence. The di/dt occurs when the diode turns off. At that point the diode current is very nearly the load current. The peak current occurs when the diode turns on.

Cheers

Ian
 
Yes I should have elaborated on what I meant by the I=0  point. 

During the diode's conduction duration, winding current has a sort of peaky half-sinusoid waveshape (for the common capacitor input filter circuit configuration).  At the end of that conduction duration, the current is falling at it's most rapid rate (highest magnitude of dI/dt) - that level of dI/dt then rapidly changes from a certain magnitude to dI/dt=0 at the point where I=0.

What happens when winding-diode current reaches zero is a mix of issues. 

That pre-existing dI/dt magnitude is a main issue, as it is generating a voltage across the secondary winding's leakage inductance.  But that dI/dt level is pretty low when compared to switchmode applications where much higher dI/dt at turn-off induces a much higher level of Qrr, that then induces the well know reverse recovery characteristic and waveforms.

How abruptly the diode stops conducting (ie. its dynamic resistance) is a main issue.  SS diodes like a 1N400x have low resistance (<<1 ohm) with low voltage across the device during  conduction, and only dynamically raise their resistance (>1ohm) as on-voltage falls (the change from on to off is <1V).  The dV/dt of the secondary winding is pretty slow for a mains waveform, so it does take some time for the coupled winding voltage to fall even 1V.

The effective step in dI/dt from some non-zero level, to a zero level, at the time when I reaches zero, forces the leakage inductance to find whatever path of least resistance to allow that voltage step to occur.  Winding shunt capacitance, diode capacitance, snubber capacitance are all paths for transient current flow at that time point.

That step in voltage across the leakage inductance forces a change in dV/dt at the diode anode, from a slowly changing -dV/dt occurring during diode conduction to a more negative level at I=0, effectively forcing the diode voltage to reverse a lot faster than if there was no leakage inductance.

As far as alleviating the impact of a given winding leakage inductance, the preferences would be:
- reduce dI/dt at on-set of diode turn-off arising from the charging pulse (as per earlier post).
- increase winding shunt capacitance if that was possible.
- add a snubber with low stray inductance as close to the winding terminals as possible.
- use a diode with as small a junction capacitance, and as low a residual Qrr as possible.

 
The di/dt during the conduction period is relatively small. At the point the diode turns off, the current drops from its present value (which as I mentioned is close to the load current) to zero in a very short time. di/dt is very large at this point simply because dt is very short. It is this di/dt that causes the spike as the magnetic field in the transformer secondary collapses.

This may explain why people see this more readily in 12VAC heater windings. In an HT circuit the load current is typically 100mA or so. In a heater circuit the load current is typically a couple of amps. Since the di is from the load current to zero it is clearly much larger in the case of heaters.

Cheers

Ian
 
Ian, are you referring to a choke input filter power supply?  That's the typical rectifier-filter arrangement that commutates the diode current from the load current level. 

A capacitor input filter arrangement causes the diode current to form a peaky half-sinusoid.
 
trobbins said:
Ian, are you referring to a choke input filter power supply?  That's the typical rectifier-filter arrangement that commutates the diode current from the load current level. 

A capacitor input filter arrangement causes the diode current to form a peaky half-sinusoid.

I am talking capacitor input filter.

Cheers

Ian
 
ruffrecords said:
The di/dt during the conduction period is relatively small...

Are you sure?

http://electronicmakeiteasy.blogspot.hr/2013/11/rectifier-circuit-diode-and-reservoir.html

 

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moamps said:
Are you sure?

http://electronicmakeiteasy.blogspot.hr/2013/11/rectifier-circuit-diode-and-reservoir.html

I suspect both trobbins and I are barking up the wrong tree. The diode current is a truncated half sine:

psu-f4.gif


di/dt is greatest at the start and end of the truncated sine where the waveform 'turns the corner', In fact they are equal and opposite which implies two spikes would be produced. The fact there is only a single spike indicates some other mechanism. like the discharge of diode junction capacitance at turn off, is responsible.

Edit: By the way, the link you posted is erroneous because it uses a zero ohm source. In reality there is no such thing and peak diode current is almost independent of capacitor values (except at switch on).

Cheers

Ian
 
Ian, the diode turn-on start of that half-sine current waveform certainly does induce a step in voltage across the leakage inductance - that can be seen in secondary winding voltage waveforms as well.

The concern about the diode turn-off end of the current waveform is that if the diode has significant junction stored charge then the diode current doesn't just stop at zero but reverses until the charge is depleted and the diode finally ends up blocking current, when another abrupt step in dI/dt occurs, which increases the magnitude of the voltage step compared to a diode that has noticeably less Qrr.  Merlin's excellent cro shots show that up well.
 
ruffrecords said:
The diode current is a truncated half sine
From my experience, isn't. It has much more harmonics.
di/dt is greatest at the start and end ....
So you have learnt that dI/dt isn't small at all. 
Edit: By the way, the link you posted is erroneous because it uses a zero ohm source. In reality there is no such thing and peak diode current is almost independent of capacitor values (except at switch on).

This works only if the power on the load is constant, which in article I posted isn't.

You should read Elliots article once again:
"While it may seem that a higher capacitance should draw larger peak currents, it must be understood that the larger values of capacitance discharge less between charge pulses, and ultimately require exactly the same 'top-up' energy as a smaller cap. This effect can be seen just by looking at the ripple voltage figures - with lower ripple voltage, there is less voltage change when the diodes conduct, so the peak current and waveform remain relatively constant.

If the capacitor is smaller than optimum, then there will be very large differences between various values. Smaller than optimum is absolutely not recommended, and peak to peak ripple voltage should be no more than 10% of the total supply voltage for best results."

 
moamps said:
From my experience, isn't. It has much more harmonics.So you have learnt that dI/dt isn't small at all. 
This works only if the power on the load is constant, which in article I posted isn't.

You should read Elliots article once again:
"While it may seem that a higher capacitance should draw larger peak currents, it must be understood that the larger values of capacitance discharge less between charge pulses, and ultimately require exactly the same 'top-up' energy as a smaller cap. This effect can be seen just by looking at the ripple voltage figures - with lower ripple voltage, there is less voltage change when the diodes conduct, so the peak current and waveform remain relatively constant.

If the capacitor is smaller than optimum, then there will be very large differences between various values. Smaller than optimum is absolutely not recommended, and peak to peak ripple voltage should be no more than 10% of the total supply voltage for best results."

I have read Elliot's article, that is where I got the truncated half sine pics from.. Elliot also debunks the bigger caps have bigger peak currents myth the article you linked to was proposing:

Major Myth Regarding Capacitance
I only heard about this myth recently, and while I can imagine how it came about, it's completely bogus. Some people claim that as the capacitance is increased for a given sized transformer, the peak current is also increased. There are conflicting additional claims that the RMS input current to the transformer either A) does, or B) does not increase as well. Added to this is a further claim that the transformer will overheat because the current is higher.

In essence, this is all complete rubbish. Incorrect measurement techniques or bad simulation practices may lead one to believe that this is the case, but it is not. The important thing is that we can only examine the steady state current - inrush current will quite obviously be greater with larger capacitance, but this is a transient event. Because transient events are just that - transient - there is no point analysing them and making absolute claims, because every transient will be different. Transformers can survive massive short term overloads without any harm, and a soft start circuit will tame the transient currents to something less scary.

The steady-state conditions are applicable to most power supplies within about 100ms after power is applied. If one were to use a 2 Farad capacitor on a 15VA transformer, this time will be extended considerably, but this would be silly, and we are not interested in the effects of silly combinations.

If we use the transformer/rectifier circuit described above as an example, we can either measure or simulate the effects of using a much larger than normal capacitor. As shown in Figure 2, the selected capacitor is 4,700µF and the load current is 1.44A - all fairly normal. The transformer secondary current is 2.7A RMS, so a 120VA transformer is well within its ratings. Even overloads are not a problem - if they are infrequent, the transformer will be perfectly happy as long as it has a chance to cool down so its maximum temperature is never exceeded. A fan can be used to increase the VA rating of most transformers, albeit with some variability.

No problems so far. However, many audiophile expectations will demand that the capacitance be at least 10,000µF, around 50,000µF for passable performance, but (of course) 100,000µF would be much better. This is (IMO) rather pointless. I won't argue with 10,000µF, but any more is really wasted and not necessary.

Now, according to the myth (sorry - 'theory'), this extra capacitance will cause the transformer's RMS current to increase, accompanied by a dramatic increase (or not) of the peak current - all during steady state conditions. It simply doesn't happen that way.

Adding more capacitance will ...

Decrease the ripple voltage
Increase the average DC voltage very slightly
Increase the inrush current (dramatically for larger capacitance values)
Barely affect the steady state RMS current
Have almost zero effect on the steady state peak current
Not cause the transformer to overheat, provided sensible limits are placed on the cap value
What is sensible? As with all things, it depends on the context. For a 25V transformer providing a worst case rectified and smoothed current of 1.44A into a 20 ohm load (as described above), a sensible upper limit would be perhaps 50,000µF, although even 100,000µF will cause no harm. Sensible values are those that consider the law of diminishing returns, where, after a certain point is reached further increases yield little additional benefit.

Provided you choose a cap that gives a ripple of 10% or less of the output voltage you will be fine.

But this is all off topic. What we are talking about is spikes caused by rectifiers. I was pointing out that there are two significant di/dt events in the rectifier current but only one spike so perhaps we are not talking about the correct mechanism.

Cheers

Ian
 
ruffrecords said:
What was the load?
I think it was a 1000uF cap and a couple hundred milliamp resistive load.

I wonder why every article I have seen about this topic uses a 12V transformer??
Because it's a lot easier to centre the scope on a high-res image of the tiny switching point when it's riding on only a small AC waveform. But the design principles are the same whatever the voltage.
 
>   Merlin's excellent cro shots

You (and I) are going to have to change our lingo, now that 'scopes mostly do not have Cathode Rays in them.
 
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