LM723, LM317, biasing the pass darlington?

Thread Starter

Hamlet

Joined Jun 10, 2015
264
I'm screwing around with this one on the workbench.
The 100k seems about right to get the full range of dim to bright at the bulb,
but I don't know if I need to bother with the 5k resistor.

Darlington  circiuit bias 4A bulb.png


Then, I want to apply it to a LM723 voltage regulator, like this:
Darlington  circiuit LM723 .png

I see this a lot. Usually, a BD139 into a 2N3055, no resistor bias, or whatever we'd call a
resistor on the base. (Frankly, I'm surprised that works with more than one 2n3055, why?)

Anyway, I'd like to try it with darlingtons. Do I need a resistor in there somewhere,
or does the puny 150ma output of the LM723 negate any worry about frying the bases off the
darlington transistors?

If I try it with a LM317, should a 100 ohm be enough to "bias" this darlington-to-darlington arraignment?

I have a TIP120 driving six TIP102s, all mounted to a heatsink and ready to go. I don't want to fry an afternoons work.
 

hp1729

Joined Nov 23, 2015
2,304
I'm screwing around with this one on the workbench.
The 100k seems about right to get the full range of dim to bright at the bulb,
but I don't know if I need to bother with the 5k resistor.

View attachment 95359


Then, I want to apply it to a LM723 voltage regulator, like this:
View attachment 95360

The TIP120 already has a built-in resistor. About 8K I think.


I see this a lot. Usually, a BD139 into a 2N3055, no resistor bias, or whatever we'd call a
resistor on the base. (Frankly, I'm surprised that works with more than one 2n3055, why?)

Anyway, I'd like to try it with darlingtons. Do I need a resistor in there somewhere,
or does the puny 150ma output of the LM723 negate any worry about frying the bases off the
darlington transistors?

If I try it with a LM317, should a 100 ohm be enough to "bias" this darlington-to-darlington arraignment?

I have a TIP120 driving six TIP102s, all mounted to a heatsink and ready to go. I don't want to fry an afternoons work.
 

hp1729

Joined Nov 23, 2015
2,304
Your taking the output from the wrong pin on the Lm723.

See fig 19 diagram on page 10 for series pass npn,


https://www.google.co.uk/url?sa=t&source=web&rct=j&url=http://www.ti.com/lit/ds/symlink/lm723.pdf&ved=0ahUKEwjwk-Lk663JAhVGVhoKHREMBCUQFggbMAA&usg=AFQjCNFjJyOyuBUcVcx3DiNZuHBiCa_lWg

re: wrong pin
:) For which package? For the DIP, pin 10 is the output. For the can, pin 6 is the output. Are you referencing the internal breakdown shown on the data sheet? That may be the can version depending on which data sheet you are looking at.
 

Thread Starter

Hamlet

Joined Jun 10, 2015
264
I think even at 100K ohm the light will still be on. The TIP120 and TIP102 both have a gain of about 1,000 so overall gain is 1,000,000. Even at 10,000,000 ohms you still have the light on, maybe dimly. Just use the TIP102?

Yes, yes! This is what I thought too. If I slap two, 2n2222 into a darlington configuration (say that fast) the florescent lighting in the room will trigger it weakly, just by touching my thumb to the base. This is not the case with the TIP120/TIP102. That's a four transistor chain. A guy in china with extensive dentalwork should be able to trigger it just by grinding his teeth on a clear evening, but no.
 

wayneh

Joined Sep 9, 2010
16,162
Two darlingtons to dim a lightbulb? That's huge overkill and I'd drop one of them. Then use a base resistor to protect the IC.

The 5K resistor is useful to limit the current when the pot gets inadvertently set to zero ohms.
 

Thread Starter

Hamlet

Joined Jun 10, 2015
264
Two darlingtons in series, I need 100k pot.
BD139 NPN, & one TIP102, in the darlington arrangement, 1M pot.

Darlington  circiuit bias 4A bulb3.png
What gives? With the BD139, I can pick up China. All I have to to
is touch the base of the BD139, and my lamp starts to glow. It's much
more sensitive.

A TIP102 isn't very sensitive by itself, despite being a "Darlington". It can
handle a lot of current though, and its what I had in my junk box, and I put
six of 'em on the heatsink. (This isn't about just dimming a bulb. In my previous
post I show them as the output of a LM723/)
 

ronv

Joined Nov 12, 2008
3,770
You might want to talk about what you are actually trying to do.
What you have shown here is not very well regulated.
Anyway, without the 5 k to ground the circuit has no voltage reference and relies only on base current and the gain of the transistors.
The darlington has a current gain from 1ooo to 20,000 and the BD139 from 40 to 250. Put them together and you may have a gain of 40,000 to 5,000,000. So very little current is needed to light your bulb.
If you are just dimming a bulb, why not PWM the darlington or better yet a fet?
But if you are going to build a heater like this, make sure you add some resistance in the emitters of the TIP102's to balance them.
In theory you don't need a resistor in the base if you never short the emitter, but I would add a little anyway just to protect the driver. Probably will still blow the TIP102.
If the regulator can output 150 ma. the extra transistor should not be needed. (.150 X 1000 = 150 amps)
 

Thread Starter

Hamlet

Joined Jun 10, 2015
264
I'm trying to design the output stage for a voltage regulated supply.
I don't care about voltage regulation in my test apparatus. I just
wanted to discover how much current gain I could get. There wasn't
enough current gain using just the TIP102. So, I added a stage, the TIP120.
There should have be enormous gain, but no, it was almost the same.
So I replaced the TIP120 with a BD139 and got my enormous current gain.

I have not tried this with an LM317 or LM723, in an actual regulation scenario.
150ma divided by 6 is 25ma. I don't think that's enough current to light up the
six TIP102 darlingtons. I could be wrong, but when I drove their bases with a pot
straight from the power rail, I didn't get much gain.

I don't have enough experience or nomenclature to explain it better than that.
All I can do at this point is to poke around at the workbench and discover what
works, and then guess why.
 

Dodgydave

Joined Jun 22, 2012
8,677
Why dont you start by looking at the datasheets of some regulator ics, they have circuits that work so it will give you help, like the one posted earlier on, some chip numbers LM338, Lm723, LM317, L200C,Lm2596
 
Last edited:

ronv

Joined Nov 12, 2008
3,770
I'm trying to design the output stage for a voltage regulated supply.
I don't care about voltage regulation in my test apparatus. I just
wanted to discover how much current gain I could get. There wasn't
enough current gain using just the TIP102. So, I added a stage, the TIP120.
There should have be enormous gain, but no, it was almost the same.
So I replaced the TIP120 with a BD139 and got my enormous current gain.

I have not tried this with an LM317 or LM723, in an actual regulation scenario.
150ma divided by 6 is 25ma. I don't think that's enough current to light up the
six TIP102 darlingtons. I could be wrong, but when I drove their bases with a pot
straight from the power rail, I didn't get much gain.

I don't have enough experience or nomenclature to explain it better than that.
All I can do at this point is to poke around at the workbench and discover what
works, and then guess why.
Ahh, I see what's bothering you.
The darlington has some internal resistors to help it turn off.
upload_2015-11-27_11-40-21.png
So you have to drive it with enough current to get above the 2 voltage drops from base to emitter. This takes some current before the transistor turns on.
With the BD139 there are no internal resistors that have to be overcome to turn it on so the gain seems higher.
You can check Hfe or current gain in the data sheet to see the gain of the transistors.
 

Thread Starter

Hamlet

Joined Jun 10, 2015
264
Yes, those darn resistors! This seems plausible. I do read data sheets until
my eyes bleed. I don't understand everything on them, but I get Hfe. The resistors
of the TIP120 are something like R1=5k, R2=150. The Hfe is high, but the initial
current demand to overcome is even higher.

I'll give ronv the hat tip. Thank you.

Now I need to do some reading.
 
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