I was reading an article on power supplies by Rod Elliott when something caught my attention. He basically says that diodes used in a full-wave center tapped rectifier must have a Vrrm higher than twice the peak of AC voltage, even giving an example where rectifying 25-0-25 volts of AC you need diodes rated for 200V reverse voltage.
That would mean that I need at least 300Vrrm diodes in the power supplies that I use for my ESP P101 amps as I have a 50-0-50 V transformer (around 140 volts peak, double that as stated by Rod), but I had already used BY229-200 (200Vrrm), later replaced with MBR40250 Schottkys (250Vrrm) with no ill effects... Am I pushing these diodes far beyond their specs or is there a mistake in Rod's article? (I for one cannot see a moment when a diode is reverse-biased with more than 140V)
Here it is, for reference (check the "Rectifier Diodes" paragraph):
http://sound.westhost.com/power-supplies.htm
That would mean that I need at least 300Vrrm diodes in the power supplies that I use for my ESP P101 amps as I have a 50-0-50 V transformer (around 140 volts peak, double that as stated by Rod), but I had already used BY229-200 (200Vrrm), later replaced with MBR40250 Schottkys (250Vrrm) with no ill effects... Am I pushing these diodes far beyond their specs or is there a mistake in Rod's article? (I for one cannot see a moment when a diode is reverse-biased with more than 140V)
Here it is, for reference (check the "Rectifier Diodes" paragraph):
http://sound.westhost.com/power-supplies.htm
For a 60Hz transformer I use the following.
Vfd=Forward Voltage Drop of the output rectifier Diode
Vs= Secondary Side Vac RMS Voltage on the Transformer
Vrr=Reverse voltage on the rectifier.
The 1.18 multiplier compensates for the difference between the nominal line voltage and the maximum line voltage the AC mains should see.
The Sqrt(2) term gives you the peak voltage of a sin wave.
The 2* multipliier is the voltage seen across a diode when the transformer winding is full negative related to the center tap.
The 1.2 term derates the Diode by 20% for safety margin.
(2*(Vs*1.18*Sqrt(2))+Vfd)*1.2=Vrr
So a 25-0-25 venter tapped transformer output winding would need a 102V rectifier for a sin wave transformer. You could use a 100V diode if you had too but I would probably use a 150V diode just to make sure I had margin.
For a PWM controller such as a flyback then then the equation is different since you have to account for the leakage inductance spike and differences in the duty cycle of the PWQM signal.
I hope this helps.
Tony
Vfd=Forward Voltage Drop of the output rectifier Diode
Vs= Secondary Side Vac RMS Voltage on the Transformer
Vrr=Reverse voltage on the rectifier.
The 1.18 multiplier compensates for the difference between the nominal line voltage and the maximum line voltage the AC mains should see.
The Sqrt(2) term gives you the peak voltage of a sin wave.
The 2* multipliier is the voltage seen across a diode when the transformer winding is full negative related to the center tap.
The 1.2 term derates the Diode by 20% for safety margin.
(2*(Vs*1.18*Sqrt(2))+Vfd)*1.2=Vrr
So a 25-0-25 venter tapped transformer output winding would need a 102V rectifier for a sin wave transformer. You could use a 100V diode if you had too but I would probably use a 150V diode just to make sure I had margin.
For a PWM controller such as a flyback then then the equation is different since you have to account for the leakage inductance spike and differences in the duty cycle of the PWQM signal.
I hope this helps.
Tony
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