Electronic – Is paralleling diodes a bad idea

Either can work correctly if designed properly. If you have a dumb rectifier supply feeding a 7805, then all the rectifier part needs to do is guarantee the minimum input voltage to the 7805 is met.

The problem is that such a power supply only charges up the input cap at the line cycle peaks, then the 7805 will drain it between the peaks. This means the cap needs to be big enough to still supply the minimum 7805 input voltage at the worst case current drain for the maximum time between the peaks.

The advantage of a full wave rectifier is that both the positive and negative peaks are used. This means the cap is charged up twice as often. Since the maximum time since the last peak is less, the cap can be less to support the same maximum current draw. The downside of a full wave rectifier is that it takes 4 diodes instead of 1, and one more diode drop of voltage is lost. Diodes are cheap and small, so most of the time a full wave rectifier makes more sense. Another way to make a full wave rectifier is with a center tapped transformer secondary. The center is connected to ground and there is one diode from each end to the raw positive supply. This full wave rectifies with only one diode drop in the path, but requires a heavier and more expensive transformer.

A advantage of a half wave rectifier is that one side of the AC input can be directly connected to the same ground as the DC output. That doesn't matter when the AC input is a transformer secondary, but it can be a issue if the AC is already ground-referenced.

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.