How to limit current flowing through LM317

WBahn

Joined Mar 31, 2012
29,979
When there is 0.7V across the BE junction of the PNP, it is conducting, but that might mean that it is conducting nanoamps or microamps of current. We can essentially say that, in the limit, it is "ON" but that it just happens to be conducting zero current. That happens when Vbe = 0.7V and since there is zero collector current then there is zero base current and, hence, no voltage across the 5kΩ base resistor. So the 22Ω resistor has 0.7V across it and is conducting 31.4mA. Now, as soon as the transistor starts conducting anything more than zero current, there will be some base current which will result in more voltage across the 22Ω resistor and, hence, more current through it.
 

crutschow

Joined Mar 14, 2008
34,285
But now I have to ask, how does the PNP conduct with only 27.3ma, since the voltage across developed across the resistor has to be shared between the 5K resistor and the BE junction of the PNP, isnt it? So I guess we need more current through 22R
The voltage is shared but not the current. The current through the resistor (but not the PNP base current which goes into the LM317) is about 0.7V/22Ω = 33mA.
 

WBahn

Joined Mar 31, 2012
29,979
When there is 0.7V across the BE junction of the PNP, it is conducting, but that might mean that it is conducting nanoamps or microamps of current. We can essentially say that, in the limit, it is "ON" but that it just happens to be conducting zero current. That happens when Vbe = 0.7V and since there is zero collector current then there is zero base current and, hence, no voltage across the 5kΩ base resistor. So the 22Ω resistor has 0.7V across it and is conducting 31.4mA. Now, as soon as the transistor starts conducting anything more than zero current, there will be some base current which will result in more voltage across the 22Ω resistor and, hence, more current through it.
This may be information overkill, but we have been saying that the Vbe of the transistor is some fixed number, either 0.6V or 0.65V or 0.7V depending on who threw out the numbers for a particular example in the discussion. In reality, the Vbe increases as the base current increases. For a silicon transistor in the active region, the Vbe changes by about 60mV (at room temperature) for a factor of ten change in base (or collector) current. Looking at the 2N3905 datasheet, it says that it has a maximum base current of 50nA at a Vbe of 0.5V (and Vce=30V). If we assume that to be the actual value (and not an upper limit), then we would expect a base current of 0.5mA at Vbe=0.56V, 5mA at Vbe=0.62V, and 50mA at Vbe=0.68V. Since we've been talking about base currents in the PNP that are in the range of 5mA to 20mA, using anything between 0.6V and 0.7V is quite reasonable and the difference in the results of using one over the other is pretty negligible.
 

WBahn

Joined Mar 31, 2012
29,979
The voltage is shared but not the current. The current through the resistor (but not the PNP base current which goes into the LM317) is about 0.7V/22Ω = 33mA.
That's the current in the resistor only when the PNP is just barely on. If the PNP is conducting any significant current at all, then the current in the 22Ω resistor goes up considerably thanks to that 5kΩ base resistor. If the PNP has a collector current of just 30mA and a gain of 30, then there is 1mA of base current and an additional 5V of drop across the base resistor, which is mirror as an increase in voltage across the 22Ω resistor resulting in a current of 5.7V/22Ω=260mA.
 

Thread Starter

iinself

Joined Jan 18, 2013
98
Thanks for your responses. Please bear with me while I try to get this.
Can't I just use a diode to put the PNP in forward bias and thus make it easy for the regulator, for with the resistor the regulator input is varying based on the load?
 

WBahn

Joined Mar 31, 2012
29,979
Thanks for your responses. Please bear with me while I try to get this.
Can't I just use a diode to put the PNP in forward bias and thus make it easy for the regulator, for with the resistor the regulator input is varying based on the load?
You need the regulator input to vary based on load because you need the regulator to be able to control the current in the PNP (and, via that, the NPN) in response to changes in the output voltage. You can do that either using the output of the regulator or the input of the regulator. But we want the output of the regulator to not change, so that leaves us with changing something at the input of the regulator. The input resistor makes it so that the more current that is pulled through that resistor the more current will be sent through the bypass path.
 

Thread Starter

iinself

Joined Jan 18, 2013
98
You need the regulator input to vary based on load because you need the regulator to be able to control the current in the PNP (and, via that, the NPN) in response to changes in the output voltage. You can do that either using the output of the regulator or the input of the regulator. But we want the output of the regulator to not change, so that leaves us with changing something at the input of the regulator. The input resistor makes it so that the more current that is pulled through that resistor the more current will be sent through the bypass path.
Thanks, got it, so it is our feedback element ?
 

WBahn

Joined Mar 31, 2012
29,979
Hi, I have this transistor at hand - 2SA1490 - it is PNP and can handle 8A, can I just use this and skip the NPN transistor on top.

http://pdf1.alldatasheet.com/datasheet-pdf/view/33555/WINGS/2SA1490.html
I would not use that transistor. It is a TO-92 case, which means that heat sinking it is going to be a lot more difficult. Also, the 8A is the absolute max current before they won't say that it won't be destroyed. If you look at the saturation numbers, with Vcesat of 1.5V it is saturated with a 300mA base current and a 3A collector current. So to get 7A through the thing you will probably need a lot more voltage headroom than the 10V you have to work with. Even if we allow for the beta being 50 (and at the currents you need it won't be), that would still require a base current of 140 mA in addition to the current in the input resistor, which will probably put you in the range of needing to heatsink the regulator anyway.

Keep in mind that you are going to have to heat sink something as long as you are using a linear regulator. You need to dissipate 70W (10V drop at 7A) and that doesn't happen (successfully) without planning.

Edit: Originally said that it is in a TO-220 case and what I meant to say was that finding a transistor in a TO-220 case would be a better choice.
 
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Thread Starter

iinself

Joined Jan 18, 2013
98
I would not use that transistor. It is a TO-220 case, which means that heat sinking it is going to be a lot more difficult. Also, the 8A is the absolute max current before they won't say that it won't be destroyed. If you look at the saturation numbers, with Vcesat of 1.5V it is saturated with a 300mA base current and a 3A collector current. So to get 7A through the thing you will probably need a lot more voltage headroom than the 10V you have to work with. Even if we allow for the beta being 50 (and at the currents you need it won't be), that would still require a base current of 140 mA in addition to the current in the input resistor, which will probably put you in the range of needing to heatsink the regulator anyway.

Keep in mind that you are going to have to heat sink something as long as you are using a linear regulator. You need to dissipate 70W (10V drop at 7A) and that doesn't happen (successfully) without planning.
Thanks
 

Thread Starter

iinself

Joined Jan 18, 2013
98
Hi,
would the attach circuit work. I am trying to work with the components I have, so this attempt.

I will use the first LM317 to bring down voltage to 50V (from 60V) and then follow it with another LM317 as a current limiter. It will limit the current to 10mA. This will drive the base of 2 TIP120 (Ic Max = 5A and hfe = 1000). Looking for about 7A at 47V.
 

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WBahn

Joined Mar 31, 2012
29,979
Your specs keep changing. Before you wanted to maintain an output voltage of 50V. Not it appears you want 47V. What are your actual requirements?

That circuit is going to give you all kinds of problems.

First, The input to the second LM317 is regulated to 50V. You need a couple volts of overhead so the most you can reliably count on at the output is 48V. Then you've got another 1.25V across the sense resistor so now you are down to about 47V. Then you've got about 2.5V for the Vbe of the TIP120 bringing you down to about 44V to 45V.

Worse, the goal of the second LM317 is to produce an output voltage that will drive 10mA of current into the base of the TIP120. You then appear to be relying on the hfe of the transistor being 1000, which would nominally equate to a current of 10A. But what if the load does need 10A? The output will go as high as it can in order to slam the TIP120s on as hard as possible to try to get that 10A into the load. Is that what you want? Also, you never count on the hfe being a specific value or even being constant. If you look at the chart in the Fairchild datasheet you will see that the typical hfe, with Vce=4V, varies from ~450 at Ic=~0.1A to a peak of ~4500 at Ic=~2.5A before dropping back to about 2000 at Ic=~5A. And those are the "typical" values. The actual values can vary a lot from device to device and they will also vary considerably with temperature and Vce. Circuits that rely on a specific value of beta are almost always a bad idea.

Next, if you are planning to put two TIP120s in parallel, then you are asking for thermal runaway to happen unless you design the circuit in such a way as to prevent it.
 

#12

Joined Nov 30, 2010
18,224
That should work, right after you add the capacitors that stop the 317 chips from oscillating, as per the datasheet.
I assume you left them out just to keep the drawing simple.
I'm also assuming that static load that you drew.
 
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Thread Starter

iinself

Joined Jan 18, 2013
98
Your specs keep changing. Before you wanted to maintain an output voltage of 50V. Not it appears you want 47V. What are your actual requirements?

That circuit is going to give you all kinds of problems.

First, The input to the second LM317 is regulated to 50V. You need a couple volts of overhead so the most you can reliably count on at the output is 48V. Then you've got another 1.25V across the sense resistor so now you are down to about 47V. Then you've got about 2.5V for the Vbe of the TIP120 bringing you down to about 44V to 45V.

Worse, the goal of the second LM317 is to produce an output voltage that will drive 10mA of current into the base of the TIP120. You then appear to be relying on the hfe of the transistor being 1000, which would nominally equate to a current of 10A. But what if the load does need 10A? The output will go as high as it can in order to slam the TIP120s on as hard as possible to try to get that 10A into the load. Is that what you want? Also, you never count on the hfe being a specific value or even being constant. If you look at the chart in the Fairchild datasheet you will see that the typical hfe, with Vce=4V, varies from ~450 at Ic=~0.1A to a peak of ~4500 at Ic=~2.5A before dropping back to about 2000 at Ic=~5A. And those are the "typical" values. The actual values can vary a lot from device to device and they will also vary considerably with temperature and Vce. Circuits that rely on a specific value of beta are almost always a bad idea.

Next, if you are planning to put two TIP120s in parallel, then you are asking for thermal runaway to happen unless you design the circuit in such a way as to prevent it.
Sorry about that. I can live with something as low as 45V, it is to drive a chipamp, I can compromise on the output watts of the amp. I am looking for about 7 A. I used an hfe of 700. Anyway thanks for you inputs.
 

WBahn

Joined Mar 31, 2012
29,979
That should work, right after you add the capacitors that stop the 317 chips from oscillating, as per the datasheet.
I assume you left them out just to keep the drawing simple.
How is that going to work?

What will the circuit do if a load of, say, 100Ω is present? How about if 10Ω is present?
 

Thread Starter

iinself

Joined Jan 18, 2013
98
That should work, right after you add the capacitors that stop the 317 chips from oscillating, as per the datasheet.
I assume you left them out just to keep the drawing simple.
Yes, I did leave out the caps for the sake of the diagram.
 

WBahn

Joined Mar 31, 2012
29,979
Sorry about that. I can live with something as low as 45V, it is to drive a chipamp, I can compromise on the output watts of the amp. I am looking for about 7 A. I used an hfe of 700. Anyway thanks for you inputs.
What determines the current? When you say you are looking for 7A, is that under all conditions? Or is that just the max current at the minimum value of the load impedance?

What is the maximum voltage you can live with at the output of the circuit?
 

#12

Joined Nov 30, 2010
18,224
How is that going to work?

What will the circuit do if a load of, say, 100Ω is present? How about if 10Ω is present?

I had to go in and edit, "for THAT static load that you drew".
Now I'm seriously doubting myself.
 

Thread Starter

iinself

Joined Jan 18, 2013
98
How is that going to work?

What will the circuit do if a load of, say, 100Ω is present? How about if 10Ω is present?
I was hoping the change in the emitter/collector current will cause the base current to adjust to maintain the hfe ?
 
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