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.
You can make some simplifying assumptions. They will add less noise to the result than the fluctuation of line voltage this circuit will encounter, so don't worry about it too much.
Assume that the capacitor is charged at the peak of each line cycle, and then discharges thru the resistor the rest of the time. The cap therefore has to not drop more than 2 V during 17 ms. You say your acceptable output voltage is 14 to 16 volts, so assume the cap gets charged to 16 volts. Discharging from 16 to 14 volts would take .134 time constants, since .134 time constants is 17 ms, you know that a full time constant is 125 ms, which is R x C. 125 ms / 150 Ω = 830 µF.
Of course in practise you don't want to cut it that tight. At the very least, you want a 1 mF capacitor, but I'd probably use 1.5 mF or 2 mF. That will be electrolytic at this capacitance and voltage level. It should be rated for at least 20 V, although 25 V would provide longer life.
Note that the current provided by the transformer will come is large and short spikes, and their level will be directly proportional to the input line voltage. For robustness, you should design this circuit to operate properly from at least 105 V to 125 V AC. This gets tricky. There is a reason we use switchers and regulation nowadays.
Best Answer
It may seem that conduction begins at 0 volts, but it does not; it depends on the LR time constant and the frequency of the AC source.
The LTspice circuit file for these two circuits is here, so you can simulate them simultaneously and see the differences between the start of conduction.
Notice particularly that for the circuit on the left, conduction begins at a voltage more negative than zero volts.