Flyback Converter diode is acting like a switch linking the transformer secondary and the load. When the primary switch is ON, the primary will store energy. This time the diode will be reversed biased. Meaning, the load side will be cut off from the transformer. On the other hand, when the primary switch is off, the stored energy will be transferred to the secondary. This time the diode will be forward biased and allow energy to be transferred to the load. Below is a common flyback converter schematic diagram.
Flyback Converter Diode Voltage
When the diode is forward biased, the voltage across it is very small. This is not relevant in most cases unless for those requirements that need to have a very tight voltage regulation to the output. The critical voltage on the diode will happen when the primary is storing energy. The voltage across the diode at this time is called peak inverse voltage (PIV) or peak reverse voltage (PRV). This is a destructive voltage and proper design is needed to avoid damaging the diode.
When the primary switch is on, the diode will be reversed biased and the circuit is can be drawn as below. The voltage across the diode at this time is can be computed using KVL.
Considering the loop direction as show above,
-Vsec – PIV – Vout = 0
PIV = -Vout – Vsec
The resulting PIV has negative sign which indicates an inverse voltage.
Vsec is the voltage on the secondary winding. It can be derived as:
Vpri / Vsec = Npri / Nsec
Vpri = Vin
Vsec = [ Nsec X Vin ] / Npri
Vsec – is the secondary winding voltage
Vpri – is the primary winding voltage that is equal to the input voltage Vin when the primary switch is on
Npri – is the primary winding number of turns
Nsec – is the secondary winding number of turns
Vin = 400V, Npri = 20, Nsec = 2, Vout = 12V, solve for the diode PIV and select the right diode.
1. Solve the Vsec
Vsec = [ Nsec X Vin ] / Npri = [2 X 400] / 20 = 40V
2. Solve PIV
PIV = -Vout – Vsec = – 12V – 40V = -52V
3. Select the Diode
The computed peak inverse voltage is 440V. now, what should be the voltage rating of the diode you are going to select? More than 440V. Well that is correct. But what level is enough and sure not to damage the diode?
The computed value is not yet accounting the effect of the parasitic of the circuit. In switching converter, ringing and voltage spikes are common and this can be seen on the diode as well. In my designs, I always add 30% for the voltage spike. This means I am anticipating an additional of 30% level. So, from 52V, I will consider now 52V + (0.3 X 52V) = 67.6V. Then, I need to consider a good design margin from this level. In my design, I only allow maximum voltage stress of 80%. Considering this, the diode voltage rating must be 67.6V / 0.8 = 84.5V. I will choose then a diode with a voltage rating of 100V (this is the standard value nearest) minimum. Why minimum? Because the 80% is supposed to be the maximum voltage stress and should only be reached when there is no other choice like parts availability and significant cost addition. I do not believe that a diode with a voltage rating of 200V is very much expensive to a 100V rated diode considering same current and package.
Flyback Converter Diode Current
The current will flow to the diode only when the primary switch is off. This is the flyback action. The current on the diode is the same to the current on the secondary winding.
For the secondary winding current derivation, refer to the article QR Flyback Secondary Current Derivation.
In my designs, I limit the current stress to just 80%. So, during diode selection, I used the criteria:
Idiode < 80% of the diode rating
For example, the computed current is 10A, I will select a diode that can handle a current of > 12.5A.
The 80% is the maximum limit I allowed in case there is no other option like parts availability, package issue or big cost adder. If ever I have many choices, I will not simply happy to select 12.5A diode to handle 10A current. I can select a diode that is rated 15A, 20A and so on. Per experience, I intentionally selected higher current rating like twice the actual current because the forward voltage will be lower.