How to limit current flowing through LM317

Thread Starter

iinself

Joined Jan 18, 2013
98
Hi,
I am to reduce Vin to Vout but my current requirements are higher than the 1.5A that LM317 can supply. Hence I am using a transistor to provide this high current (8A) based on the high current circuit in the datasheet. Since the current is being provided by the transistor I do not want to put a large heat sink on the LM317 also and thus would like to reduce the current through it to a bare minimum enough t drive the base of T1. Will the resistor RL do the job?

Thanks
 

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WBahn

Joined Mar 31, 2012
29,979
That topology isn't going to give you very good voltage regulation. Your circuit is simply going to hold the voltage at the input to the base resistor at a set level and as your circuit draws more current the amount of base current will increase and the output voltage will drop by any additional drop across the base resistor (plus the small amount of increase in Vbe of the transistor).

You want a circuit in which changes in output voltage are servo'ed to increase or decrease the base drive voltage of the bypass transistor.
 

ian field

Joined Oct 27, 2012
6,536
Hi,
I am to reduce Vin to Vout but my current requirements are higher than the 1.5A that LM317 can supply. Hence I am using a transistor to provide this high current (8A) based on the high current circuit in the datasheet. Since the current is being provided by the transistor I do not want to put a large heat sink on the LM317 also and thus would like to reduce the current through it to a bare minimum enough t drive the base of T1. Will the resistor RL do the job?

Thanks
You've got the external transistor in the wrong place!

You need to develop the Vbe bias across a resistor in series with the 317 input, the collector is in parallel with the 317 output. The orientation of the transistor means it needs a PNP to boost a positive regulator.

Its fully described in various appnotes and application examples in some of the datasheets.
 

Thread Starter

iinself

Joined Jan 18, 2013
98
That topology isn't going to give you very good voltage regulation. Your circuit is simply going to hold the voltage at the input to the base resistor at a set level and as your circuit draws more current the amount of base current will increase and the output voltage will drop by any additional drop across the base resistor (plus the small amount of increase in Vbe of the transistor).

You want a circuit in which changes in output voltage are servo'ed to increase or decrease the base drive voltage of the bypass transistor.
I was hoping that since T1 is forward biased, it was going to act as closed switch and all current to the load will be drawn through it. So will this circuit work. Is it right to assume that most of the current is coming from T1 now ?
 

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ian field

Joined Oct 27, 2012
6,536
I was hoping that since T1 is forward biased, it was going to act as closed switch and all current to the load will be drawn through it. So will this circuit work. Is it right to assume that most of the current is coming from T1 now ?
Do it like in the appnote or the regulation will be useless.
 

Thread Starter

iinself

Joined Jan 18, 2013
98
You've got the external transistor in the wrong place!

You need to develop the Vbe bias across a resistor in series with the 317 input, the collector is in parallel with the 317 output. The orientation of the transistor means it needs a PNP to boost a positive regulator.

Its fully described in various appnotes and application examples in some of the datasheets.
Currently I have a circuit that this that is working to supply in excess of 1.5 A, I found this circuit here - http://www.reuk.co.uk/LM317-High-Current-Voltage-Regulator.htm. The 220R is in the wrong place but with that correction it is working fine. I don't understand the need for 6a4 diode though?
 

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WBahn

Joined Mar 31, 2012
29,979
I was hoping that since T1 is forward biased, it was going to act as closed switch and all current to the load will be drawn through it. So will this circuit work. Is it right to assume that most of the current is coming from T1 now ?
The voltage at the load will be the voltage at the LM317 output minus the IR drop across the base resistor minus the Vbe of the pass transistor. The base current will be the load current divided by the transistor beta. So you have:

\(V_{load} \; = \; V_{317} - \frac{I_{load}}{\beta}R_b - V_{be}
\)

Hence as the load current changes, so will the load voltage. If beta is 100 and Rb is 1kΩ, then a 10mA change in load current will result in a 0.1V change in load voltage.
 

WBahn

Joined Mar 31, 2012
29,979
There is a way to do load sharing in multiple transistors - but not all the appnotes show it.
There are lots of ways to do it. The simplest is to just use ballast resistors in the emitter paths before they join up with the other transistors.
 

WBahn

Joined Mar 31, 2012
29,979

ronv

Joined Nov 12, 2008
3,770
You know how it is.. One bad internet circuit leads to another. You may notice there is nothing to turn on T1. :rolleyes:

I think the diodes are a misguided attempt to discharge the input cap and the adjust pin cap if the input or output is shorted to ground.
 

WBahn

Joined Mar 31, 2012
29,979
You know how it is.. One bad internet circuit leads to another. You may notice there is nothing to turn on T1. :rolleyes:

I think the diodes are a misguided attempt to discharge the input cap and the adjust pin cap if the input or output is shorted to ground.
I didn't even pay attention to the schematic with the problem of not being able to turn on T1 since I was looking for a schematic that had that 6A4 diode in it and that one doesn't.

I don't think it's for temperature compensation. I can't find the data, but I don't see anything that indicates that that 6A4 diode has a positive tempco.
 

crutschow

Joined Mar 14, 2008
34,285
Any scheme with a NPN follower at the LM317 output will have poor regulation and temperature stability. There's no good reason to even consider them.
Use one of the schemes shown in the data sheet using a PNP transistor booster for a stable output voltage.
 

Thread Starter

iinself

Joined Jan 18, 2013
98
The voltage at the load will be the voltage at the LM317 output minus the IR drop across the base resistor minus the Vbe of the pass transistor. The base current will be the load current divided by the transistor beta. So you have:

\(
V_{load} \; = \; V_{317} - \frac{I_{load}}{\beta}R_b - V_{be}
\)

Hence as the load current changes, so will the load voltage. If beta is 100 and Rb is 1kΩ, then a 10mA change in load current will result in a 0.1V change in load voltage.
Thanks that helps a lot.
 
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