Electronic – Dual active bridge converter

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.

After discussing this at university here's the answer: no, it's not possible.

Because if the external diode has a higher V_F value, it will be short-circuited by the low-V_F high-Q_rr internal body diode of the MOSFET. Therefore at turn of the internal body diode will have the same switching losses, since the external diode practically never goes in conduction mode.

The only "solution" seems to be to get a hand of a bit higher (~10-20%) R_DS,on MOSFET which has a better diode (10-20 times lower Q_rr) and eventually put some more MOSFETs in parallel in order to lower conduction losses.