Just so we're on the same page, we're talking about the transformer + rectifier + capacitor circuit that has been used for linear power supplies for decades, right? Something like this:
+--+----+-----+
| | | |
d1- -d2 | |
^ ^ | |
+---\ /-R1---+ | | |
120VAC T1 24VAC | | | (load)
+---/ \---------+ | |
| | | |
d3- -d4 ---c1 |
^ ^ --- |
| | | |
+--+----+-----+-- GND
I'm assuming what you're really trying to find out is: What VA rating should I specify when I buy a transformer for my system?
The I^2R heating of a coil inside that transformer is proportional to the RMS current through that coil.
If you keep that RMS current low enough, the manufacturer guarantees that the transformer will not overheat and fail.
(Most manufacturers specify that max RMS current indirectly, implying it from the VA rating of the transformer.)
quick, conservative calculation
Say you already know the peak number of electrons per second flowing through some diode (Id_max) and what fraction of the full 1/60 second cycle that diode has nonzero flow (D).
Then I can get a quick estimate of the VA rating required for the transformer with
estimated_I_RMS = 2 * D * Id_max^2.
So, for example, if you somehow know that any one diode is conducting 1/17th of a full cycle -- in other words, d1 conducts 2/17 of one half cycle, then it has zero current while d2 conducts 2/17 of the next half cycle, and so nonzero current is flowing through the transformer 2/17 of the time.
Say, for example, you also know that Id_max is 2 A.
At any instant, whenever (nonzero) electrons are flowing through any diode, exactly the same number of electrons per second are flowing through the transformer.
So the maximum electrons per second through any diode (Id_max) is the same as the maximum electrons per second through the transformer (Itx_max).
Then I estimate the RMS current through the transformer as
2 * (1/17) * (2 A)^2 = about 0.47 A_RMS
so for a 24 VAC output transformer, I would need to specify
estimated_VA = Vrms * estimated_I_RMS = 24 VAC * 0.47 A_RMS = about 11.3 VA.
Of course, no one sells transformers that are exactly 11.3 VA, so I'd round up to a 12 VA or a 15 VA or a 20 VA transformer -- whatever my suppliers have in stock at some reasonable cost.
This is a conservative estimate -- the actual RMS current through the transformer is somewhat less than this estimate, but more than the RMS current through the load.
more details
To more accurately calculate the actual RMS current flowing through the transformer,
I could divide up the complete cycle into 6 or so time slices,
estimate the current flowing during each time slice --
that's pretty easy when it's zero --
and then do the root-mean-square (RMS) calculation:
square each current, average each of those squared values, weighted by the time that current was flowing, and then that the square root of that average.
It might be quicker and more accurate to run a simulation with thousands of time-slices than to work it out by hand.
There are many techniques for reducing the RMS current through the transformer while supplying exactly the same power to the load.
Electric power companies love those techniques,
because their customers are just as happy (the load gets exactly the same power),
they get paid the same amount of money (for customers that pay per kWh),
and they can spend less money for transformers and long power lines (because higher RMS currents require bigger, heavier, more expensive transformers and power lines).
Those techniques go by the general name of "power factor correction".
Related:
Perhaps the simplest such technique is the "R1" resistor in the above diagram.
Some systems use a more complicated "valley fill" circuit -- see Serial capacitors in electronic ballast of a fluorescent lamp .
And many systems -- such as most computer power supplies -- use an even more complicated "active power correction" system.
Flogging the FREDs
Voltage fed converters with transformer isolation will exhibit ringing in the secondary. Ringing is caused by parasitic inductances and capacitances in the circuit, with the dominant elements will being the transformer leakage inductance (\$ L_ {\text {Lk}}\$) and junction capacitance ( \$ C_j\$)of the bridge diodes. The diode data sheet shows \$ C_j\$ of 32pF. I'm going to make a naive guess at \$ L_ {\text {Lk}}\$ of 500nH, but it will have to be measured to really know. So, an LC of 500nH and 32pF is what must be snubbed.
Spike amplitude without snubbing will be \$ 2 n V_ {\text {in}}\$, where \$ n \$ is transformer turns ratio and the factor of 2 is what you get for a high Q resonance.
There are different types of voltage snubbers; Clamping, Energy transfer resonant, and Dissipative. The clamping and resonant types require more parts and some involvement of active switches which I think make them impractical for this case. So, I am only going to cover dissipative snubbers because they are the most simple and work well with passive switches (like diodes or synchronous rectifiers).
The form of dissipative snubber that I will cover is a series RC placed in parallel with each bridge diode.
Some facts about RC dampening snubbers:
- They are all about impedance matching. You don't get to choose the snubber resistor value \$ R_d\$. The parasitic LC determines that for you by characteristic impedance Zo.
- You do get to choose the value of the snubber cap \$ C_d\$. That's important since the cap value sets the snubber loss (\$ P_ {\text {Rd}}\$)as \$ C_d F V^2\$ . Where V is the pedestal voltage and F is switching frequency. The snubber cap must provide a low impedance at the LC resonance of the parasitics, so it needs to be several times \$ C_j\$.
Some guidelines, and what to expect with RC dampening snubbers:
For \$ L_ {\text {Lk}}\$ of 500nH and \$ C_j\$ of 32pF, Zo will be 125Ohms. So, \$ R_d\$ would be 125 to match Zo. You may have to fine tune this a little since \$ C_j\$ is non-linear and falls off with reverse voltage.
Choosing the snubber cap \$ C_d\$ : Choose \$ 3 C_j\leq C_d\leq 10 C_j \$ . Higher values in the range do provide better dampening. For example, \$
C_d\$ of \$ 3 C_j\$ will result in a peak diode voltage of \$ 1.5 n V_ {\text
{in}}\$, while \$ C_d\$ of \$ 10 C_j\$ will result in a peak diode voltage of
\$ 1.2 n V_ {\text {in}}\$.
Dissipative snubber performance will not improve for \$ C_d\$ values
greater than \$ 10 C_j\$.
Power loss \$ P_ {\text {Rd}}\$, with a pedestal voltage of 1250V and F of 50KHz.
- If \$ C_d\$ is \$ 3 C_j\$ or 100pF, \$ P_ {\text {Rd}}\$ = \$ C_d F V^2\$ or 7.8W.
- If \$ C_d\$ is \$ 10 C_j\$ or 330pF, \$ P_ {\text {Rd}}\$ = \$ C_d F V^2\$ or 25.8W.
\$ C_d\$ of \$ 10 C_j\$ gives the best dampening with peak voltage of 1.2 time the pedestal voltage, but you can save some power with smaller snubbing caps if you can stand the higher peak voltage.
Best Answer
Edit: As Andy pointed out, I didnt quite pick up on the fact that you are trying to feed the primary of the transformer with DC. This will in fact be very damaging to all components involved, and potentially hazardous as well. please do not do this! You will blow something up, load or no load! The only reason power supplies can even operate like this, is because it is rapidly switching on/off the incoming HV DC power towards the tranformer in order to simulate some kind of AC. A diode bridge straight into a transformer is no good. Disregard everything I said in the comments about the transformer, as this is not applicable to your situation. The answer @Andy aka provided goes into more detail about why some power supplies can be made to work while having a rectiefied input. My apologies!
Your reasoning is almost on point. Diodes can happily be connected to mains voltage depending on how you use them. With power supplies this is indeed very common as you pointed out. If you had 1000V of forward voltage on a diode, that would indeed be strange for the scenario's you are describing. The forward voltage of a diode however isn't determined by just the voltage on one if its 'legs', its measured across the device. And this voltage is developed based on the current that is flowing trough it. https://www.farnell.com/datasheets/639187.pdf
This picture is directly taken out of the 1N4007 datasheet. Perhaps this will make it easier to understand the forward current and forward voltage relationship. The remainder of the supply (mains?) voltage is dropped by the thing the rectifier is powering.
As other people are pointing out as well, the only way to develop such a high voltage with a forward biased diode (conducting diode), is to force the current/voltage trough/across it by shorting it, bypassing the load or essentially having the diode itself as the only load. If you connect a diode across mains like that it will fail very quickly and violently.