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    Bridge diode thermal problem

    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 http://www.mouser.com/datasheet/2/34...s607m-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!
    Click image for larger version. 

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    Hasan

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    Re: Bridge diode thermal problem

    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.



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    Re: Bridge diode thermal problem

    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
    A best design is easily achieved with good test specs™
    A better question deserves a better answer. ™
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    Re: Bridge diode thermal problem

    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.

    Click image for larger version. 

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    Re: Bridge diode thermal problem

    Quote Originally Posted by BigBoss View Post
    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 !
    Hasan



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    Re: Bridge diode thermal problem

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

    Schottky's should work well at these voltages.



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    Re: Bridge diode thermal problem

    Quote Originally Posted by BradtheRad View Post
    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.

    Click image for larger version. 

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    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 - - -

    Quote Originally Posted by SunnySkyguy View Post
    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.
    Hasan



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    Re: Bridge diode thermal problem

    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
    http://tinyurl.com/y5z7q4fq
    Last edited by SunnySkyguy; 7th October 2019 at 15:49.
    A best design is easily achieved with good test specs™
    A better question deserves a better answer. ™
    ... so include all your acceptance criteria ( values, % tolerance) and assumptions in your question or any design.

    ... Tony Stewart EE since 1975
    - slightly north of Toronto



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    Re: Bridge diode thermal problem

    Quote Originally Posted by Hasan2017 View Post
    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.



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    Re: Bridge diode thermal problem

    Quote Originally Posted by SunnySkyguy View Post
    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
    http://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.
    Hasan



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    Re: Bridge diode thermal problem

    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 ...



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  12. #12
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    Re: Bridge diode thermal problem

    Quote Originally Posted by Easy peasy View Post
    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.
    Hasan



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    Re: Bridge diode thermal problem

    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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