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12V to 5V Convertor

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Suraj143

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I need to convert 12V to 5V & needs 2A of current. Going to use the below circuit & need to know the transistor part number. Planning to use TIP125. Is it ok?
 

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ronsimpson

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The transistor you chose TIP125 is a Darlington so it takes 1.3 volts Base to Emitter to get it turned on. You show a 1 ohm resistor so 1.3 amps in the LM7805 before the transistor turns on.

I think you want a non Darlington. It will turn on at 0.65 volts to turn on. (0.6 amps)
upload_2018-2-26_21-59-34.png
I think you want a 2 to 5 amp transistor. The voltage rating could be as low as 20V.
 

Suraj143

Active Member
Hi I built the circuit & checked. The 7805 is getting too much heat with 1Ohms resistor. How to overcome this?
 

Pommie

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You have 7V drop and 700mA so it's dissipating nearly 5W. No wonder it's getting hot.

Mike.
 

MikeMl

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Hi I built the circuit & checked. The 7805 is getting too much heat with 1Ohms resistor. How to overcome this?
The higher resistance from emitter to base of the PNP, the sooner it will turn on.

When the load current is 2A, and the volt difference between the input (12V) and the load (5V), the total power wasted is (12-5)*2 = 14W, which requires a very large heatsink. Depending on which package your 7805 is in, you may need to select the resistor so that about 13W is dissipated in the PNP and about 1W is dissipated in the reg.

68.png

This shows the current in the PNP (green) and current in the regulator (yellow) vs the value of R1. It also shows power dissipated in Q1 (red) and the regulator (blue).
 
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AnalogKid

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At 2 A output current, you are dissipating 14 W in the regulator circuit. Changing the value of R1 changes the proportion of how much heat is dissipated in the regulator and how much is dissipated in the transistor, but the total always will be 14 W. Assuming a 50/50 split, 7 W in a TO-220 package with no heatsink will melt fingerprints.

ak
 

Suraj143

Active Member
Thank you for the replies.

You all correct. The power dissipation is always 14W. That is too much heat.

At 2 A output current, you are dissipating 14 W in the regulator circuit. Changing the value of R1 changes the proportion of how much heat is dissipated in the regulator and how much is dissipated in the transistor, but the total always will be 14 W. Assuming a 50/50 split, 7 W in a TO-220 package with no heatsink will melt fingerprints.

ak
You got it, I wanted to mount those giants without a heat sink. I think I have to go for a buck module :(
 

cowboybob

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MikeMl

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Word of warning: the referenced buck converter generates horrible RFI. I cannot use them in any radio-related application. They are also difficult to tame if the application has low-level audio or analog data aquisition for ADC.
 

Externet

Active Member
LT1185 and similar others do perform well. 7805 is not the only 5V regulator out there. There is 3, 5...Amperes ones. Use a low DC input voltage for less dissipation.
 

AnalogKid

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The LT 1185 is a linear regulator. As such, it will dissipate the same 14 W as any other linear regulator in this application. It does eliminate the need for an external current boost transistor, but all that does is concentrate the entire 14 W in one device.

Note - in post #1, the input voltage is fixed at 12 V.

ak
 
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