Lm317

ErnieM

Joined Apr 24, 2011
8,377
Well... you can but...

Since there IS a potential difference between the outputs one output (with the higher voltage) will attempt to drive the entire load by itself, up to the point it reaches it’s current limit and starts to drop the output voltage. Then the other device picks up the rest of the load current. So one device is running at it’s max current, the other something less (but ultimately with NO potential difference between devices).

The scheme in Figure 13 pointed to by KMoffet works because a resistor is added in series with each output, which tends to equalize the current out of each device. A small difference in output voltage between devices now yields a small difference in output currents, so the devices all tend to share near equally instead of one device hogging all the current it can.
 
If one regulator is producing 5V and the other is say 4.
5V, then the potential difference is 500mV. Would this load down the regulator that is producing 5V?
 
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The scheme in Figure 13 pointed to by KMoffet works because a resistor is added in series with each output, which tends to equalize the current out of each device. A small difference in output voltage between devices now yields a small difference in output currents, so the devices all tend to share near equally instead of one device hogging all the current it can.
But you wouldn't though right? Bonkerz.
 

Lestraveled

Joined May 19, 2014
1,946
If one regulator is producing 5V and the other is say 4.
5V, then the potential difference is 500mV. Would this load down the regulator that is producing 5V?
The regulator outputting 5 volts would see all of the load. The regulator set at 4 volts would effectively see zero current.
 

KMoffett

Joined Dec 19, 2007
2,918
If one regulator is producing 5V and the other is say 4.
5V, then the potential difference is 500mV. Would this load down the regulator that is producing 5V?
The "4V regulator" would shut down because it's sensing 5V from the "5V regulator" output and will do what it needs to do to in an attempt to bring its sensed output down to 4V. It can't, so it shuts off its pass element. All the current then comes from the "5V regulator".

Ken
 

Lestraveled

Joined May 19, 2014
1,946
The LM317 is a voltage output device. It wants to control the voltage at the output pin to within millivolts of the set voltage. If the output pin is connected to another device that raises the voltage above the set point then the control circuit will reduce drive to the pass transistor until the output voltage reaches the set point. It only takes a small amount of millivolts to turn a LM317 off.

The normal method to parallel voltage devices is called "droop share". Resistance is put in series with the output of each voltage device. All regulators must be adjusted closely to each other. You gain current sharing, but you lose regulation.
 
???? Say that again.
Well the potential difference between the two or more regulators.
If one is putting out 5V and the other 4V, then there is 1V of potential difference. Now if the two regs were operating off different power sources, then there wpuld be an short between the regs. The reg putting out 4V would become an load.

Right?
 

KMoffett

Joined Dec 19, 2007
2,918
I'm thinking of two seperate power sources for an short.

Guranteed short in this case yeah?
Two batteries, different voltages..."short"...yes.
Two voltage regulated power supplies...no.
Same process...the lower voltage one will shut down trying to bring its shared output down to its set reference voltage.

Ken
 

Lestraveled

Joined May 19, 2014
1,946
Well the potential difference between the two or more regulators.
If one is putting out 5V and the other 4V, then there is 1V of potential difference. Now if the two regs were operating off different power sources, then there wpuld be an short between the regs. The reg putting out 4V would become an load.

Right?
Got it.

Generally the answer is no. In order for this to happen the 4 volt regulator would have to sink current as well as source current. That is not how they are built. Every simple linear regulator I know of can only source current. They can not act like a load unless some other parameter has been exceeded and they are in a "failure" mode.
 
Two batteries, different voltages..."short"...yes.
Two voltage regulated power supplies...no.
Same process...the lower voltage one will shut down trying to bring its shared output down to its set reference voltage.

Ken
I do a lot of battery projects so this explains it!

So if I have two mains powered lab power supplies operating off seperate isolation transformers, connect them together and set one at 5v and the other at 4v -- there will be an short though right?
 
Got it.

Generally the answer is no. In order for this to happen the 4 volt regulator would have to sink current as well as source current. That is not how they are built. Every simple linear regulator I know of can only source current. They can not act like a load unless some other parameter has been exceeded and they are in a "failure" mode.
Thanks Mark for the top answer.
 

KMoffett

Joined Dec 19, 2007
2,918
I do a lot of battery projects so this explains it!

So if I have two mains powered lab power supplies operating off seperate isolation transformers, connect them together and set one at 5v and the other at 4v -- there will be an short though right?
No. The two power supplies will behave exactly like the two LM317s.

Have you looked at the standard LM317 configuration? There two resistors in series across the output. The junction of these is fed back to an input of an internal differential opamp. The other input of the of opamp is an internal reference. The output of the opamp drives the internal pass transistor. If the output voltage is higher than the desired voltage, the voltage at the resistors' junction will be higher than the reference voltage and the opamp will reduce the drive to the pass transistor. This will continue until the voltage at the output matches the sensed voltage with the reference voltage. Lab supplies work the same. Since the 5v supply holds the top of the 4V regulator's series resistors at 5V, the 4V regulator reduces its pass transistor drive to cutoff in a futile attempt to sense 4V.

Ken
 
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