Hi, i was looking for a 5A regulator that is accessible to me in my region. I found only the LM350 IC, but it has a maximum amp delivery capability of 3A, in his datasheet there is a technique to cascade it with a OPAMP, but i'm not entirely sure what the OPAMP is doing and some 0.1 resistors. Let me show:



Okay, there are 0.1 Ω resistors inserted before the ic's what are the purpose of them ?

And, what is the OPAMP is doing exactly in this circuit ? Can it be changed to a jellybean LM358 ?

 

The purpose of the op-amp and 0.1 ohm resistors is to make sure that the current is equally divided between the two LM350s.

If you simply paralleled two LM350 one of them (that has the slightly higher output voltage) would do all the work and supply all the current.

LM358 will not be OK in your circuit because the inputs need to operate up to the +ve supply voltage. You could probably use the LM358 if you use a resistor divider on each input. Or have a separate supply for the op-amp that was a few volts higher than the input voltage.

 

Adding an external pass transistor would be simpler:

This circuit protects the pass transistor during short circuit/overload conditions if it's connected to the same heat sink as the regulator.

R1 and R2 determine current division, D matches the BE junction of Q1. You want the regulator current to be at it's maximum under full load.

 

Yes, i tried to build the circuit in Proteus, but it's demonstrating weird behaviors, like when the voltage goes low, one of the LM350 conducts all the current and the other practically nothing. My goal would be to able to conduct almost 10A, i was going to start small by understanding these circuits that are in LM350 datasheet, but it seems to be not a trivial thing to do.

I intend to try to build one of these, in the meantime, any of you care to explain to me the next circuit ?



It's the usage of three in parallel, but the circuit changes a bit.

In the 6A circuit, the R3(2k) is doing what in the adjust pin ? In my view, it's only limitting the current.
Also the output is only chosen by R4 and R5 ?

In the 10A circuit, what's the resistive 5k divider network doing ? Also the Vcc pin of the opamp is connected
differently, not sure i understood this connection.

 

My goal would be to able to conduct almost 10A, i was going to start small by understanding these circuits that are in LM350 datasheet, but it seems to be not a trivial thing to do.

Why not consider the circuit I posted. One regulator, one power transistor, a diode, a few resistors, and you're done.

Just set the current ratio so 3A goes through the regulator and 7A goes through the pass transistor.

 

I managed to recover my old account, so i'll be answering on this one. @dl324 , i didn't try your circuit, i was doing some research on the circuits that used the opamp.

I wish to understand them, but yours seem to work fine, i'll give a try.

 

i was doing some research on the circuits that used the opamp.

Using a simulator for such a simple circuit doesn't make sense to me. It's kind of like using a calculator for a simple multiplication.

 

For me it isnt :/. What does R3 does in your circuit ? Also the diode in BE junction implicates a 0.7v drop to the IN pin of the regulator right ? I running on 12v, and i need 10.5v on the output, can't afford that drop.

 

What does R3 does in your circuit ?

That resistor determines when the pass transistor starts turning on.

Also the diode in BE junction implicates a 0.7v drop to the IN pin of the regulator right ? I running on 12v, and i need 10.5v on the output, can't afford that drop.

Then you can't use an LM350. You need a Low Drop Out (LDO) regulator.

 

Yes i am starting to realize i aint gonna get 10.5 on the output.