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7805 50V input, 5A output problem, pls help

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nimeshasilva

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Can anybody please help me with this prob?? I need a 5V supply which the input supply should be around 50VDC and should be able to withstand around 5A. I tried the circuit below and it fails just after exceeding 2A. is anything wrong I have done with the circuit?? or I'm well appreciate, if anybody have a better circuit..



Thank you...
 

How did you deduce that it fails after 2A? The output is approximately 5V with about 5A current flowing through. That's what the diagram shows.
 
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    IanP

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nimeshasilva said:
Can anybody please help me with this prob?? I need a 5V supply which the input supply should be around 50VDC and should be able to withstand around 5A. I tried the circuit below and it fails just after exceeding 2A. is anything wrong I have done with the circuit?? or I'm well appreciate, if anybody have a better circuit..
Thank you...
Click to expand...

2N3055 transistors have quite low current amplification factor, so a 1W 33V zener diode will definitely not be able to deliver enough current to its base to produce 5A of I[c] current ..
Simple calculations for, say, H[fe] <20 will show you how much base current will be required for 5000mA output …

:wink:
IanP
 

nimeshasilva said:
I need a 5V supply which the input supply should be around 50VDC and should be able to withstand around 5A.
Click to expand...
7805 input = 13V. Current delivery from 7805 = 550mA.
P=V*I=(13-5)*0.55=4.4W!
Is this just a schematic for educational purpose or it will be transformed into a real world circuit?
I think that the above calculation justifies a question like that.
 

Also, consider the overall circuit. Power loss = (50-5)*5 = 225W for an output power of 25W. Efficiency = (25/250)*100% = 10%

So, for a more efficient circuit, a switching regulator should be employed. There are switching regulators available for 5A current or more. One such is L4977 (7A output).

Hope this helps.
Tahmid.
 
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