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Bridge diode thermal problem

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Hasan2017

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Hi there!

Hope this post find you well.
Like my other post, you might think its silly!

I am talking about a basic bridge recrifier made of https://www.mouser.com/datasheet/2/345/rs601m-rs607m-5532.pdf
Take a look on my circuit that is been used for 20V/5A AC application, problem is diode and caps temparature is raising upto 52 and 45 degree respectively..

My senior said " add more 2 similar capacitor to reduce ripple" then "heating can be minimized"

Dont you think its a wrong idea!
bridge_diode.PNG
 

RS605M is not an appropriate choice for this kind of application.It's used in HV applications rather than Low Voltage/High Current.
2200uF/per Amper is a good rule of thumb for 50Hz applications.But 45 degree is -thought- coming from rectifier diode ( by radiation ) if they are close.But nevertheless 45 degree is not a serious issue for electrolytic caps.
 

Caps must be have specified ripple current to reduce temp. Derate 30% min. Choose low ESR types.

Bridge: clamp 10W heatsink with thermal grease or use 30A bridge as %Ipk = 1/%V ripple and I^2R increases but shorter RC=T duration is linear
 

Capacitive filtering draws short bursts of current. To get 5A output you must supply bursts of current on the order of 30A to 60 A. I*I*R losses are high. Heating effect is high, through diodes and capacitors. Your transformer may or may not limit these extremes to a reasonable level (depending on internal impedance, saturation ratings, etc.).

However if you add a choke filter it lengthens those bursts and reduces their amplitude. A mere 4mH is sufficient for a draw of several Amperes as your supply must handle.

A full diode bridge is not essential. Two diodes are sufficient. Ground is one end of the secondary. The split volt levels stay better balanced even if loads are unequal.

bipolar supply fm single sec 4mH choke 2 diodes 2 caps 200W.png
 

RS605M is not an appropriate choice for this kind of application.It's used in HV applications rather than Low Voltage/High Current.
2200uF/per Amper is a good rule of thumb for 50Hz applications.But 45 degree is -thought- coming from rectifier diode ( by radiation ) if they are close.But nevertheless 45 degree is not a serious issue for electrolytic caps.

The connector J1 comes from a Power Transformer's (220V,3~5A,60Hz) secondary output. It is low voltage/high current application. Dont you think adding more component dissipates power when facing DC !
 

Don't use a 1000V rectifier for 20V...

Schottky's should work well at these voltages.
 

Capacitive filtering draws short bursts of current. To get 5A output you must supply bursts of current on the order of 30A to 60 A. I*I*R losses are high. Heating effect is high, through diodes and capacitors. Your transformer may or may not limit these extremes to a reasonable level (depending on internal impedance, saturation ratings, etc.).

However if you add a choke filter it lengthens those bursts and reduces their amplitude. A mere 4mH is sufficient for a draw of several Amperes as your supply must handle.

A full diode bridge is not essential. Two diodes are sufficient. Ground is one end of the secondary. The split volt levels stay better balanced even if loads are unequal.

View attachment 155915


Not sure whats the benifit to use such design. For a full cycle conduction, how is your signal appears to load. Without load could you measure output voltage . During circuit analysis does LC filter act for anything!

- - - Updated - - -

Caps must be have specified ripple current to reduce temp. Derate 30% min. Choose low ESR types.

Bridge: clamp 10W heatsink with thermal grease or use 30A bridge as %Ipk = 1/%V ripple and I^2R increases but shorter RC=T duration is linear

Impressive answer!
What you have suggested for " bridge " can be understandable with any circuit or graph.
Looking at your " RC" selection, you want me to make small " time constant". It should be explaind.
 

When you choose low ESR caps , the Tau=ESR*C will be < 10us as a result will also be able to handle higher Irms ripple current from I^2*ESR=Pd

- from self heating Pd*Rja =T rise due to thermal properties.

So you dont need to choose lower RC , but by evaluating specs of Ripple current limit or (ESR+Rdiode+Rsource)=Req*C =Tc charge time and Rload*C=Td discharge time and thus Vripple/Vin = +/- Req/Rload and self heating I^2*R * Rja (thermal junction to ambient resistance)

- - - Updated - - -

This Falstad Web simulation allows you to add any ESR and see the effects or gang diodes
https://tinyurl.com/y5z7q4fq
 
Last edited:

Without load could you measure output voltage . During circuit analysis does LC filter act for anything!

With load absent, an unregulated supply tends to rise to the maximum amplitude coming from the transformer. This is true whether filtering is choke type or capacitor type.

You must install a voltage regulator (perhaps two) if you want to keep voltage below a certain level.
 

When you choose low ESR caps , the Tau=ESR*C will be < 10us as a result will also be able to handle higher Irms ripple current from I^2*ESR=Pd

- from self heating Pd*Rja =T rise due to thermal properties.

So you dont need to choose lower RC , but by evaluating specs of Ripple current limit or (ESR+Rdiode+Rsource)=Req*C =Tc charge time and Rload*C=Td discharge time and thus Vripple/Vin = +/- Req/Rload and self heating I^2*R * Rja (thermal junction to ambient resistance)

- - - Updated - - -

This Falstad Web simulation allows you to add any ESR and see the effects or gang diodes
https://tinyurl.com/y5z7q4fq

Looks interesting, even though back dated issue!
Dont you mean number of component can be minimize !
What you have said "gang diode" action is not clear here.
 

bigger caps = shorter current pulse = higher rms current draw = more heating in diodes and wires ...

45V, 10A schottky a good idea if pk volts out of Tx are < 35V ... 5W zener clamps for schottky diode protection ...
 

bigger caps = shorter current pulse = higher rms current draw = more heating in diodes and wires ...

45V, 10A schottky a good idea if pk volts out of Tx are < 35V ... 5W zener clamps for schottky diode protection ...

Kindly disscuss with these issue with a suitable circuit diagram and equation.
 

Hi,

Since you are the designer - at least I guess so - you should draw the "suitable" schematic on your own ... within a simulation tool.
Then run the simulation and check power dissipation.

Free simulation tools are available.
But you could simply use Excel.

Klaus
 

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