Resistors purpose.

[sureshot]

Found this schematic on the web, but unsure of the 10 ohm 3 watt resistor on the output of the lm317. It's between the emitter, base and volts out. Any ideas would be really appreciated as to the roll this resistor plays. And do you think this circuit can deliver 5 Amps ? I'm not so sure.
Thanks for any help.

[Relayer]

Hello sureshot,
The 10 resistor is there to supply enough base current to
the 2N3055 pass transistor, though its usually placed on the input side
of the LM317 regulator.
With enough heat-sinking and provided the transformer can deliver more
than 5 Amps, I can't see why it wouldn't work.
Though I would put at least a 6800uF or a 4700 and a 1000uF in place of
the 3300uF that the schematic shows. Its a tad low for my liking.
Regards,
Relayer

[rstofer]

The 10 Ohm resistor provides the base-emiter voltage for the 2N3055.  As more current flows to the load through the resistor, the voltage between the base and emitter increases, increasing the amount of current flowing through the transistor.

Take 35V AC and rectify it to get about 50 VDC.  Now, consider what happens when you try for an output of 1.2V at 5A.  The transistor has to drop about (50V - 1.2V) or 48.8 V @ 5A or around 244 Watts.  Somehow I don't see this turning out well.

I know, the numbers aren't exact.  I didn't account for the 1.4V drop across the bridge rectifier and I didn't account for the current contributed by the voltage regulator but it's close enough to show that it isn't likely to work at low voltages.

At 30V it still won't work.  (50-30) or 20V @ 5A is about 100 Watts.  I don't think so.  See Figure 1 and the following table where it talks about 1.52 deg C per Watt temperature increase:

http://www.onsemi.com/pub/Collateral/2N3055-D.PDF

Take 100 W and see from Figure 1 that the case temperature can't be higher than about 48 deg C.

See Maximum Rating of 115W @ 25 deg C - that's a Maximum and it depends on the heatsink keeping the device at 25 deg C.

I probably messed up the thermal numbers but they are close enough.  This thing isn't close to working.

In my view, the transformer is seriously over-voltage.  Even then, it is good to have a tapped secondary so that lower voltages can be selected when lower output voltage is desired.
 

[bson]

The output resistor doesn't need to be 3W; at around 0.7V the 2N3055 will fully bypass the regulator and resistor.  P=U²/R = 0.7*0.7/10 = 50mW = 1/21W.  So even an 0402 1/16W resistor is perfectly fine here, or any 1/8W or 1/4W TH garden variety.

The main problem with this circuit though is it uses a regulator instead of an op amp, and the regulator will be managing the output voltage, which is the base voltage - not the actual output. Replace the regulator with an op amp and the 10ohm resistor with a 1k or 2k one and use the actual output for feedback, and now it controls the output by adjusting the base voltage (and for very small loads drives it directly with the transistor off, through the resistor) rather than controlling the base.

[JS]

The output resistor doesn't need to be 3W; at around 0.7V the 2N3055 will fully bypass the regulator and resistor.  P=U²/R = 0.7*0.7/10 = 50mW = 1/21W.  So even an 0402 1/16W resistor is perfectly fine here, or any 1/8W or 1/4W TH garden variety.

The main problem with this circuit though is it uses a regulator instead of an op amp, and the regulator will be managing the output voltage, which is the base voltage - not the actual output. Replace the regulator with an op amp and the 10ohm resistor with a 1k or 2k one and use the actual output for feedback, and now it controls the output by adjusting the base voltage (and for very small loads drives it directly with the transistor off, through the resistor) rather than controlling the base.

And you are likely to have a very unstable circuit, not to consider that the opamp needs to provide quite a high current to drive the 2N3055 base. Then, a darlington, but the stability problem is still there which should be addressed with a more suitable feedback loop.

JS

[romons]

The resistor is there to create a Vbe for the pass transistor. As the output current rises, the Vbe drop increase until the outboard pass transistor turns on and starts diverting current around the regulator.

The real problem is that the regulator won't really be able to keep regulation. As the current draw increases, the output will droop until the pass transistor is turned on. At that point, it'll stay pretty much at 0.6V below whatever the regulator is set for. There is no current limiting on the pass transistor, so it'll just pass more and more current until it burns up. The irony is that a much larger resistor will prevent the droop by turning on immediately. That'll allow you to control the output voltage with the two pots (although it will still not be the best regulation)

If you move the 10 ohm resistor and base of the transistor to BEFORE the regulator, then it'll regulate much better. It still doesn't have any current limiting.

All of this is explained in detail in Art of Electronics. I have the 2nd edition. If you can find it, look at figure 6.36 and 6.37. The section is named "Outboard pass transistors". Figure 6.37 shows a couple of current limited uses of pass transistors, which they call 'outboard transistor boosters'

Regards

[Zero999]

It will work without the resistor, which is there to allow the LM317 to take some of the current.

This is a bad circuit. Use the standard PNP pass transistor configuration.

https://www.eevblog.com/forum/beginners/building-a-13-8-volt-linear-power-supply/msg1281765/#msg1281765

[Ian.M]

.... and if all you've got is NPN high power transistors, use a Sziklai pair with a medium power PNP to get a composite high power high gain PNP equivalent pass device.

A base resistor is advisable so it doesn't fuse the base:emitter junction on sudden load increases.

If you add an emitter resistor to the pass transistor, and put it and the LM317 on the same heatsink, you can get reasonable over-current and thermal overload protection.  As the current through the pass transistor increases, its base drive decreases, so the pass transistor emitter resistor sets the maximum current through the pass transistor when the LM317 enters current limiting (at about 2.5A at 25 deg C - its not well specified and due to its thermal protection, it can only reach 2.5A transiently).

[sureshot]

Thank you for the replys. I have put together a few variants of the pnp versions of this circuit with single, and up to four transistors with good results. I've been looking for an npn variety of the same circuit with limited parts count (That's the attraction to it for me) I've not got round to the LM723 circuits yet, but will in the near future. From what I've seen so far, the best option for an npn final is driven by a more modest pnp transistor. Is there a circuit out there that can use an npn as the current supplier, whilst retaining reasonable regulation ? Any protection i add on as a final, normally a thyristor crow bar circuit. Thanks again for all the replys. I do try to keep the margin for input to output as close as possible to avoid excessive heat. Generally the pnp variaty work ok. So the 10 ohm 3 watt resistor is of little use ? Thank you again for the help.

[sureshot]

Sometimes things come back to me of circuits I've seen on the web. I'm not 100% sure, but i can vaguely remember seeing a circuit where an npn power transistor was on the negative volts side of the supply rails. As said i can't be certain, but is that an option worth exploring ? Just a thought.

[Ian.M]

Sure, use a LM337 negative regulator and you can use a NPN pass transistor.   Just take the classic circuit Hero999 posted above and flip it for negative Vin and negative Vout.

However, simplicity comes at a price.   None of the circuits above compare favourably to a LM723 based regulator.

[Zero999]

.... and if all you've got is NPN high power transistors, use a Sziklai pair with a medium power PNP to get a composite high power high gain PNP equivalent pass device.

A base resistor is advisable so it doesn't fuse the base:emitter junction on sudden load increases.

I agree.

If you add an emitter resistor to the pass transistor, and put it and the LM317 on the same heatsink, you can get reasonable over-current and thermal overload protection.  As the current through the pass transistor increases, its base drive decreases, so the pass transistor emitter resistor sets the maximum current through the pass transistor when the LM317 enters current limiting (at about 2.5A at 25 deg C - its not well specified and due to its thermal protection, it can only reach 2.5A transiently).

Here's another idea which uses sense resistors to detect when the LM317 is current limiting and cuts the drive to the pass transistor. It also shows how the current can be spread across multiple pass transistors.

When the LM317 is not current limiting and the current is low, the volt drop across R3 is too low to turn on Tr1 and it operates as an ordinary LM317 circuit.

If the current through the LM317 exceeds about 600mA, the voltage drop on R3 becomes high enough to turn on Tr1, which turns on Tr2, causing current to be diverted around the LM317 circuit. Any increase in current increases the voltage drop on R3, causing Tr1 to turn on harder and more current to flow around it.

When the LM317 starts to current limit, the voltage across R3 stops increasing, which keeps the voltage across it constant, therefore the base voltage on Tr1 the same, so the current through R3 also is fixed. The voltage drop across R4 is approximately double that of R3 and is also stable, thus keeping the voltage on Tr2's base and the voltage drop on R_E the constant. Because the voltage across R_E is constant, the current is also constant.


https://www.eevblog.com/forum/beginners/building-a-13-8-volt-linear-power-supply/msg1281994/#msg1281994

[sureshot]

So as i want a positive output, the negative rails not the way to go. I was thinking it could some how be posative output with an npn transistor on the negative rail of a positive voltage regulator. Thanks for posting those schematics, they look very interesting. That base resistor is quite a large resistance for TR1 does it need to be that high ?

[sureshot]

So with the pnp to npn drive circuit, are R1 to R5 0.5 watt resistors, I'm familiar with R1 and R2 or a potentiometer.  But are all other resistors half watt values. In the circuit you posted I've posted below.

[Zero999]

So as i want a positive output, the negative rails not the way to go. I was thinking it could some how be posative output with an npn transistor on the negative rail of a positive voltage regulator. Thanks for posting those schematics, they look very interesting. That base resistor is quite a large resistance for TR1 does it need to be that high ?

The current through R5 is small, so the voltage drop is low. If Tr2 has an Hfe of 20 and the output current is 3A, the current through Tr1 will be about 150mA and if it has an Hfe of 200, the base current will be 750µA, giving a volt drop of 750mV across R5.  It can be reduced, to 100R, if the extra volt drop is unacceptable.

So with the pnp to npn drive circuit, are R1 to R5 0.5 watt resistors, I'm familiar with R1 and R2 or a potentiometer.  But are all other resistors half watt values. In the circuit you posted I've posted below.

Yes, 0.5W is more than enough for the resistors in that circuit.

R2 doesn't have to be a potentiometer. It can be a fixed resistor. The formula for calculating R1 and R2 can be found on the LM317 data sheet.
http://www.ti.com/lit/ds/symlink/lm317.pdf

[StillTrying]

For a bit less overshoot I think the 2N3055's base-emitter turn-off resistor should be about 1k, rather than 10k, and definitely not 0.47R as in reply#11.

[bson]

And you are likely to have a very unstable circuit,

This is true for capacitive loads, and the easiest is to use external compensation to stabilize an op amp based regulator, but then it's no longer a beginner project.

[Zero999]

For a bit less overshoot I think the 2N3055's base-emitter turn-off resistor should be about 1k, rather than 10k,

I agree.

and definitely not 0.47R as in reply#11.

There was a reason for the extremely low value in that circuit. Please read the description more carefully and study the schematic more. Hint: it's part of the current limiting circuitry.

[wraper]

The main problem with this circuit though is it uses a regulator instead of an op amp, and the regulator will be managing the output voltage, which is the base voltage - not the actual output.

The problem is that 220 ohm resistor is placed incorrectly. It should be connected after 2N3055, not directly at the output of LM317. No opamps needed.

[sureshot]

Thank you for the explanations, I'm trying to make sense of it. I don't see a labelled 220 ohm resistor, I guess it's R1 I normally use 120 ohm resistor for R1. Yes I've had a play with an lm317 / 338 calculator on the REUK website. There's a big difference between R4 being 0.47 ohm and 1K. So RE resistors are ballast resistors, something like 0.1 ohm 5 watts ? I'm only assuming they are.

[borghese]

The maximum input-output differential voltage of the LM317 is 40V so the output voltage must never fall below 10V assuming an input voltage of 50V

[sureshot]

Correction, yes I see the 220 ohm resistor in the picture I posted opening post. I normally use 120 ohm for that, the voltage doesn't drift as much.

[David Hess]

For a bit less overshoot I think the 2N3055's base-emitter turn-off resistor should be about 1k, rather than 10k, and definitely not 0.47R as in reply#11.

10k or even 1k is way too high.

0.47 ohms is low but it depends on how much current should be passed around the 2N3055.  If the ratio of current between the pass transistor and regulator is adjusted correctly, then the regulator's current limiting and thermal protection can also protect the pass transistor as discussed below.

[sureshot]

Thanks for the replys. So looking at that it would be 75 ohm for R4 ? I've been looking at 100 ma versions of the LM317, and L7812 voltage regulators. But to use a T092 package devise calls for little work load from the regulator. Say something like 0.030 mA or there about. I've not used an 78L12 or LM317L yet, but plan to try these lower power regulators.

[David Hess]

Thanks for the replys. So looking at that it would be 75 ohm for R4 ? I've been looking at 100 ma versions of the LM317, and L7812 voltage regulators. But to use a T092 package devise calls for little work load from the regulator. Say something like 0.030 mA or there about. I've not used an 78L12 or LM317L yet, but plan to try these lower power regulators.

It depends a little on the drive requirements.  If drive is limited, I usually figure on diverting 1/10th of the current away from the base-emitter junction so with 100mA from a 78L12 or LM317LZ, I might go with 0.6V Vbe / 10mA or 60 ohms which is close enough to 75 ohms.  In practice up to 4 times that or 300 ohms would work just as well; the value is not that critical.

In small signal applications, 1 to 10k or even higher could be suitable.