Bridge Rectifier & Heatsink Choice

Thread Starter

Warlord_1011

Joined Jan 8, 2013
7
HI all,

So i want to design a 12V 4A power supply. I am using the National Semiconductor Volt. Reg. Handbook from 1980, but section 8.3 isnt completely clear to me regarding the selection of the Bridge Rectifier.

It states that the RMS Capacitor current can be 2-3 times the load current, so in my case 8-12Arms. But it then goes on to say the Bridge Recfier should be rated for at least the full output current.

Im not sure if they mean the Output load current (in my case 4A) or the Capacitor Load current of 8-12Arms.

I have some spare GBJ2510 from a repair which i would like to use, they seem to be rated at 25A with heatsink or 4A without. So im not sure if i need to specifiy a heatsink, and the datasheet for this device just shows two power curves, with or without heatsink so i presume a 25A rating is with an Infinite Heatsink surely?
 

Thread Starter

Warlord_1011

Joined Jan 8, 2013
7
Hi MrChips, thanks for that, by all means i could do that, its just i am trying to get my head around the principles of it also.

Ive even dug out a copy of MicroElectronics by Sedra Smith but thats not quite helping either, using that I am calculating Idav as over 24A!

Im following the calcs in the book yet it doesnt seem to add up, im looking at a load power of 48W, ive opened old Linear Regulator Bench PSU's with smaller Rectifier diodes which are tiny compared to the Diodes i am calculating i need.
 

Thread Starter

Warlord_1011

Joined Jan 8, 2013
7
Ive tried calculating it a different way,

Vpk = 12*1.414 = 16.968
Vpk - 2*Vd = 15.268 (0.85V drop @ approx 4A on the GBJ2510)
Vrip = 1.418V (28,200uF Bank @ 50Hz full bridge so is 4A/2*50*0.0282 = 1.418V)
Vout Min = 13.85V
Iload = 4A
Equiv Load resistance 13.85/4 = 3.463Ohm

Using the Capacitor Voltage equation, Vt = Vmax*e^ (-t/RC)

Vt is 13.85, Vmax is 15.268, RC is 0.0976

Some rounding errors, but ive calculated that the time for the RC to discharge from Vpk to Vo is 9.518mS which at 100Hz rectified, means a diode conducts for 0.4824mS every 20mS (50Hz AC but 2 diodes conduct at a time) meaning i get a conduction duty cycle of 0.4824/20 = 2.412%

Q=CV and Q = It

So the 28200uF Capacitor bank, will have a ripple of 1.412V, which means a Change in charge of 0.04 Coulombs.

This will be delivered within the 0.4824mS Conduction period of the Diode, which means 0.04/0.4824mS = 82.89A Capacitor average charge Current.

82.89A + 4A = Average Diode current of 86A, this is ofcourse assuming the Diode conducts in a square wave pulse which it does not. it would be more like a spike with exponential decay. But worst case 86A

But this 86A is pulsed, per diode pair every 20mS, so that comes back to my DC of 2.412%, meaning an average current of 2.1A per diode every 20mS.

0.85V * 2.1A = 1.785W per Diode, so a total of 7.14W Average for the entire 4 diode Bridge Rectifier.


Does that seem about right?

I mean ill still bolt it to the side panel, but could probably get away with it in free Air?
 

LowQCab

Joined Nov 6, 2012
2,185
You're making this far more complex than it needs to be ........
Calculate the average total Input-Current to the device, pay no attention to peaks or transients.
To be safe, make sure the Bridge is rated for at least twice that average Current.
And substantially more than twice, if it has no Heat-Sink or Air-Flow.

To calculate Heat-Dissipation, look at the Data-Sheet for the Bridge and find the graph
that shows Current vs Forward-Voltage-Drop.
Multiply the Forward-Voltage-Drop by the Current, the result is Watts-of-Heat-Dissipation.

Then look at the graph for "Free-Air" Heat-Dissipation,
it will probably also include 3 or 4 curves for various Heat-Sink ratings as well.

The Heat-Sinks are rated in
Temperature-Rise-above-ambient, in degrees-C, per Watt of Heat-Dissipation.

There will be another graph that will show a "De-Rating-Curve" which will show
the Current-Rating of the device vs it's Temperature.
The Current-Rating will go down as the operating Temperature goes up.
.
.
.
 

MrSalts

Joined Apr 2, 2020
1,847
You calculate power to the bridge rectifier by the power lost to the component, not the power passed through the component. That is, the bridge likely has a voltage drop of 1.2v. Multiply that by your current. You have the approximate heat that must be dissipated because there is very little time that the drop is less than 1.2v (or what ever the datasheet says).
 

Slinger42

Joined Jul 24, 2020
1
The Diode pulse current you quote occurs every 10ms not 20ms by the way its a bridge so the output ripple frequency will be double the input frequency from the transformer., The diode bridge you reference (the KBPC25) has a IFsmax (Forward surge current rating of 300A over 8.3ms. Sothey will be fine. I have used this bridge running 12 @ 8A with 100,00uF without any problem with it bolted to a small heatsink without any failures over a 10 year period.
 

Thread Starter

Warlord_1011

Joined Jan 8, 2013
7
The Diode pulse current you quote occurs every 10ms not 20ms by the way its a bridge so the output ripple frequency will be double the input frequency from the transformer., The diode bridge you reference (the KBPC25) has a IFsmax (Forward surge current rating of 300A over 8.3ms. Sothey will be fine. I have used this bridge running 12 @ 8A with 100,00uF without any problem with it bolted to a small heatsink without any failures over a 10 year period.
Hi Slinger, no it is 20mS, because the first half of the 50Hz waveform, the "other pair" of the bridge rectifier is conducting, I was calculating it based on single diodes then at the end combining the power dissipations together to calc the total.

So i know the minimum rating, so will obviously double it.
 
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