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24 V to 12 V with up to 30 Amp converter

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antoro

New Member
Dear all

I've 24V baterry want to step down to 12 V for my car audio system. I've try to using LM7812 connected to 6 TIP3055 (in paralel) by add 0.47 ohm at their emittor. This circuit working but i've got overheat in the TIP3055 (even after adding alumunium and fan cooler). It's happened when I give only a netbook (40W) as load. it's really weird because 6 TIP3055 must be can haddle 30 A (5 A for each transistor), means 30A x 12 volt = 360W. Is there anything I can do to reduce the heat? thanks before
 

MikeMl

Well-Known Member
Most Helpful Member
Get a huge heatsink. Power dissipation causes heating. It has to go somewhere. Use the car body/frame as the heatsink.

At your power levels, you will likely have to go to a switching (smps) converter, however, it might be difficult to prevent the switching noise from contaminating your audio.
 

unclejed613

Well-Known Member
Most Helpful Member
a schematic of what you have might be helpful. did you add input and output capacitors to the 7812, and on the output side of the 3055s? also, it wasn't very clear whether you had separate emitter resistors for the transistors, or just one. the collectors of the 3055 can be tied together, but the bases must each be fed with about a 100 ohm resistor, and each emitter has a 0.47 ohm emitter resistor. the base resistors are only tied together on the LM7812, and the emitter resistors are only tied together on the load side. there should be a 470uF cap from the input of the circuit, a 10uF cap from the 7812 output to ground, and a 1000uF cap from the regulator output to ground. the reason it's heating up on you could be from oscillation if you have no caps, or, if you have the emitters parallel without each one having it's own 0.47 ohm resistor, one or more devices are "current hogging". hopefully, you didn't run the regulator without having the 3055s heat sinked properly, or you could have one or more of them shorted. some of the heat is going to be the internal resistance of the transistor, because it's dropping half the power supply voltage inside the transistor. with a load of 6A, for instance, that's 1A for each transistor, which means each transistor is dissipating 12 watts. that's a total of 72 watts.
 

antoro

New Member
Dear
unclejed613
I already add 100 uf/35 volt in the input and output LM 7812, but there is no capasitor in the regulator output.
while the R0,47 ohm/5watt was also already installed in each emittor so there are 6 resistor working. yeah the output of LM7812 was connected straigh away to the bases of TIP3055. I's a good idea to add resistor to the bases. Iam a bit confuses with the 0.47 ohm/5 watts resistor. how much the current limit at 12 volt? I assume by increasing the resistor value at the emittor (for ex.2 ohm/5 watt) the current will not exceed 6A because 12V/2 Ohm = 6 Ampere. but I'am not sure it's gonna be work.
 

Cicero

Active Member
Was the Netbook the only thing attached? At 12V 40W, thats an avg current draw of 3.33A. Add anything else and its worse.

Bear in mind you've got a linear regulator, so your power dissipation required (just from the PSU as a whole) will be (Vin-Vout)*Iout = 40W to get rid of, around 6W each in the transistors! So whatever your total power draw is, you're losing double that because of your inefficient PSU. How big is your heatsink?
 
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