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A snubber circuit is essential for Flyback converter, to prevent the transistor from burning up. The overshoot on the transistor voltage, is due to the leakage inductance, $L_k$, of the transformer. ## Variables Edit

• $P_s$ - max power dissipated by the snubber resistor
• $R_s$ - snubber resistor
• $C_s$ - snubber capacitor
• $T_s$ - switching period
• $f_s$ - switching frequency
• $V_g$ - Input voltage to the converter
• $V_t$ - transistor max acceptable voltage
• $I$ - average input current
• $L_m$ - magnetizing inductance of the transformer
• $L_k$ - leakage inductance of the transformer
• $V_{t-peak}$ - transistor peak voltage, spec from datasheet

## Transistor snubber design Edit

Leakage inductance

It is not easy to calculate the leakage inductance of a transformer, but it can be measured after the transformer is built, or if a prebuilt transformer is used, it can be obtained from a datasheet. It can be assumed that the leakage inductance is 3% of the magnetizing inductance, $L_m$.

$L_k \approx 0.03 * L_m$

If a transformer is well designed, leakage inductance can be reduced to 1% of the magnetizing inductance.

### RCD snubber Edit

Snubber resistor

To calculate the snubber resistance, $R_s$, an acceptable max transistor voltage, $V_t$. You want to select a $V_t$ that has a wide margin from the peak transistor voltage rating specified in its datasheet. It still must be greater than the transistors blocking voltage, $V_g + V/n$
$V_{t-peak} > V_t > Vg + V/n$

Using this, you can calculate $R_s$, and $P_s$

$P_s = 1/2L_fI^2f_s$
$V_s = V_t - V_g$
$R_s = V_s^2/P_s$

Snubber capacitor

$C_s >> \frac{ 1 }{ f_s*R_s }$

Snubber diode

The diode voltage must be able to block voltage a high voltage, 1N4007 tends to work.

### Example Edit

flyback using the following specs. $V_g=311V, V = 30V, n = 0.2, f_s = 50kHz, L_m = 320uH, I = 5A, V_{t-peak} = 500V$

#### Calculations Edit

• $V_{t-peak} > V_t > Vg + V/n$
• $400V > V_t > 150 + 15/0.2$
• $400V > V_t > 225$
• Select: $V_t=325V$
• $L_k=0.03*L_m=(0.03)(0.001)=30uH$
• $P_s = 1/2L_fI^2f_s = (0.5)(30uH)(1.5A)^2(100kHz)=3.375W$
• $V_s = V_t - V_g = ?$
• $R_s = V_s^2/P_s = 9074\Omega$
• Select: $R_s = 10k\Omega$, 5W
• $C_s >> T_s/R_s = (10uS)/(10k\Omega) = 1nF$
• Select: $C_s = 47nF$, 500V

## Schottky snubber design Edit

If you choose to use a schottky diode its a good idea to have a snubber.

design to be added