# constant current circuit analysis.

Discussion in 'General Electronics Chat' started by unlv007, Apr 21, 2008.

1. ### unlv007 Thread Starter Active Member

Apr 5, 2008
44
0
An excellent constant current circuit has been posted by Bill Marsden. The circuit diagram is attached as a pdf. I am explaing how the circuit acts as constant current source in the steps below, please check my steps and confirm if i am right.

Ie=Ib + Ic
Ib << Ie,Ic Ie=Ic=Iout
Ie= (Ve-Vcc)/R1 (please see location of points a, b, c in the diagram attached) ...................(1)

Ve=Veb + Vrs.....................(2)
Let the pot between q and s have resistance R3 in top half and R4 in bottom half as shown.

Vrs=(R4/R3)(Vcc-Vd)................(3)
By combining (1),(2),(3) and assuming Veb=Vd=0.6V
Ie=(Vcc-Vd)(R4/R3-1)/R1....(4)

Ideally we will seek an expression like Ie=(Vcc){ R4/(R3* R1)}
Now the extra diode put does not cancel the Vbe drop of transistor as in (4). I guess the reason for diode addition is to nullify the Vbe drop of transistor. So, is my analysis correct?

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2. ### gotumal Active Member

Mar 24, 2008
99
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Hi,

Actually transistors have NTC. As temerature increases, resistance decreases increasing the current. In this circuit the current will increase for the given position of pot for temp raise. To compensate this they might have added diode. Since it has the same temp characteristics of transistor, the equivalent current increases with temp raise increaing the voltage at base and thus compensating the output current.

Please tell me if I am wrong.

3. ### Caveman Senior Member

Apr 15, 2008
471
1
Your equation 1 should be Ie = (Vcc-Ve)/R1 to match the diagram.
Eq 2 is right.
Eq 3 is wrong. First you should explicitely state that you are assuming that Ib is negligible. Then
Vrs = (Vcc-Vd)*(R4/(R3+R4)).

But....all this math is just not doing it for me. You should learn to look at circuits more intuitively. Then put the math to work to get exact results.
If you realize that Vd is basically the same as Veb, then you can see that there are two paths from Vcc to r. Each includes a pn junction drop plus a resistance:
R1 and Veb
Vd and R3

Now, since Veb and Vd are both pn junction drops, we will assume them to be the same. That means that the voltages across R1 and R3 are the same. So, if we know the current through R3, we know its voltage, and therefore we know the current through R1:
I(R1)= I(R3)*R3/R1

Now we can see that Vcc, Vd, and the total of R3+R4 will define a current through R3, right? This current is pretty much independent of the rest of the circuit. The base current is there, but we are assuming it is really small in comparison.

So, the current though R1 equals Ie = Ic.
So, the approximate answer with our assumptions is:
I(out) = (Vcc-Vd)/(R3+R4)*R3/R1

Now another interesting point of this circuit is the temperature compensation. It is very well defined that PN junction drops will decrease their voltage 2mV/degree C. Since the diode and the transistor are paired in the way they are, they will both drop together and keep our assumptions basically true.

However, Vd does define the current directly and there is no compensation for this, so looking at the output equation, the current will increase slightly as temperature goes up.

4. ### Caveman Senior Member

Apr 15, 2008
471
1
I have to eat crow on this one. I was thinking about it, and I realized that R4 is providing something here.

The previous analysis is all correct if you assume that R4 is 0 ohms.
But with R4 there is an effect that helps temperature compensation.

Look at it this way. As the temperature goes up, Vd reduces which tends to increase the current through R3. Since V(R3) = V(R1), the output current will increase somewhat. However, if I(e) increases, then I(b) will increase. This current will go into R4, which will reduce the voltage through R3, cancelling the original increase in current.

This all of course happens simultaneously so you never see the actual increase in current, and the proportionality of the different values matters, but it does provide compensation. Cool circuit.

5. ### Wendy Moderator

Mar 24, 2008
21,779
3,024
It's a pretty classic circuit, and in the eBook. Took me forever to wrap my head around a bipolar transistor being a constant current source when I was trying to understand transistors way back when, but when I got it it stuck.

I would agree with your analysis, just remember some of the negligable factors can rise up and bite you in ways you didn't anticipate, like the base current. Temperature also emulates base current to a degree, but it can be compensated for as Caveman has explained.

The diode can be a minor compensation as is, but if you tie it across the transistor heat sink where it has the same temperature, it gets better.

6. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
218
I wonder if it would be better to use a current mirror composed of two matched PNP transistors to implement this circuit.

hgmjr

7. ### Caveman Senior Member

Apr 15, 2008
471
1
Yeah, I think the best would be just like you said. Replace the the diode with the same pnp transistor type and the base connected to the collector.

Even better, you could get a dual transistor package that has two pnp's on the same silicon. They would be exactly matched then. Rohm makes these. The only issue is that they are in a small SMD package.

8. ### Wendy Moderator

Mar 24, 2008
21,779
3,024
It is possible to overdesign a circuit. Remember KISS. If it works, even if it isn't perfect it can be good enough. Hmmm, I think I've seen that somewhere.

Here is a redraw of the circuit to make the classic design more obvious for beginners...

9. ### Caveman Senior Member

Apr 15, 2008
471
1
Hey, it's less parts. That's simpler, right?

10. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
218
Exactly!

Right, again.

And there is an improvement in headroom for the output. You don't lose the voltage drop across the emitter resistor because there is no emitter resistor.

hgmjr

11. ### Wendy Moderator

Mar 24, 2008
21,779
3,024
How much do these parts cost, and how available are they? I spent less than a buck for a TIP105 last time I bought one at Tanners.

The other question is how adjustable a current mirror is. Schematics please?

12. ### Caveman Senior Member

Apr 15, 2008
471
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\$0.46 for single quantity IMX8 (dual npn) or \$0.43 for IMT4 (dual pnp). In stock digikey (\$25 min order) and Mouser (no min order).

Oh, and Bill, I'm in Austin and it is probably true in Dallas as well. Since Mouser is based in Texas, I can order UPS ground and get it the next day from them! It costs like \$3 total for most shipping. Almost makes it not worth checking out the places in town.

13. ### Wendy Moderator

Mar 24, 2008
21,779
3,024
Ever use BG Micro? Another mail order place, but this one has my zip code. I go in there occasionally. About every one or two weeks I drive to Tanners, which is an old fashioned electronics parts outlet, with kits and other stuff on teh side. Think candy store for technicians.

I've heard of current mirrors, but I'm not familiar with them. Care to point some examples out? I'm really interested how they would achieve the variability a basic transistor has.

14. ### Caveman Senior Member

Apr 15, 2008
471
1
We have M.C. Howards which is local.

Go to http://www.4qdtec.com/csm.html
It has tons of current mirrors and sources with general explanations. It is just a topology of setting up a circuit. #5 is the classic current mirror.

15. ### Wendy Moderator

Mar 24, 2008
21,779
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OK, I understand. By having identical junctions what happens to one happens to the other, almost like quantum entanglement. Thanks.

I have seen this in various analogue IC chips for constant current sources without really understanding it. The temperature issue would be a killer though.

If the transistors were merely close, could the circuit work (even partly), or would it completely fall apart? I've been looking for some low voltage dropping current sources for LEDs and whatnot, matter of fact I can think of a project that could have used it for some high current LEDs not to long ago in another thread. The drop from an LM317 circuit was killing his design.

I'd seen matched transistor designs from way back when, but thought they were primarily for differential amps, this is a totally new application to me.

16. ### SgtWookie Expert

Jul 17, 2007
22,201
1,809
The thermal coupling between the two transistors is very important.

Remember the other day when I mentioned the non-linear gain response and I would've attached curves but my keyboard was messed up? Well, my keyboard is repaired and I've attached the curves for a 2N3904. (ETA:added 2n3906 curves as well) The typical pulsed gain current vs collector current (upper left plot) shows the non-linearity I was talking about, but the thing to look at here is the difference in hFE over temperature. You can see that even if two transistors are basically matched before insertion into a circuit, if there is a difference in temperature between the two, they will have different gains.

You could slap some epoxy or heat sink compound on a pair of TO-92's and mechanically couple them together. However, you're just not going to get that thermal coupling as good as if they were on the same substrate. Besides, curve tracers seem to be a bit hard to come by nowadays, unless you build one yourself or maybe find one at a ham radio swapmeet. Far more time and cost effective to just buy a matched pair that were made just for that purpose.

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17. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
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Don't overlook the availability of reasonably priced matched transistors in the same IC package. It is hard to get better thermal communication between two transistors than to stick them together in the same package.

hgmjr

18. ### Caveman Senior Member

Apr 15, 2008
471
1
If you go back and look at some old circuits from my idols, Bob Pease and Jim Williams, you will see that they either tested them for matching and then glued them together, or they used these MATxxx parts which was a matched dual transistor package. However, when I looked them up, they were like \$10 or something. That's why the Rohm parts are so good(except for the package which is smaller than a SOT6). 50 cents (single qty) is a bargain.

19. ### SgtWookie Expert

Jul 17, 2007
22,201
1,809
Caveman,
I agree, they're a good deal. There are even less expensive matched NPN and PNP pairs out there. On Mouser's site, I spotted a dual NPN unit by Fairchild for \$0.14/each if you buy them singly.
http://www.mouser.com/Search/ProductDetail.aspx?qs=UMEuL5FsraCAVR1yzjolNA==

At those kinds of prices, one could hardly justify trying to jury-rig something from discretes. I mean - you'd spend more in heat sink compound than you'd save by buying discretes, even if the discretes WERE all pre-matched - and you still wouldn't be getting the thermal coupling nearly as good as the IC.