# Is it possible? constant current with 2 NPN BJTs

Discussion in 'General Electronics Chat' started by Tired guy, Apr 17, 2012.

1. ### Tired guy Thread Starter New Member

Apr 17, 2012
11
0
Hello people
My knowledge of electronics is... well, barley existent.
I have a battery, So, if I know anything about electronics, It means DC, which makes my life easier.
In the schematics for a constant current source with NPN BJT, I'm pretty sure there's ground, but with a battery-powered device, I don't expect I'll be connected to the ground...
So I can understand only very little of the designs I see

Here's what I got: A battery, some resistors, 2 NPN BJTs, and 2 electrodes which will be in things with different resistances.
Can I use these stuff to make sure a constant current will pass through whatever will be between the electrodes?

I don't need to know the math if it's complicated, i'll figure it out. I just want to know if it's possible, and if it is, how will it look
(Many years ago, I saw this website, and then I lost it, and now I've found it again, but there's collage - and I don't have the time to study all this stuff, I have to wait for a vacation... but I want to know now if I should even try to figure it out or just buy a voltage regulator. don't want to wait too long for it to arrive...).

Thank you very much!

2. ### jimkeith Active Member

Oct 26, 2011
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Try this circuit

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3. ### Tired guy Thread Starter New Member

Apr 17, 2012
11
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thanks... I'm missing 1 NPN BJT for this, so I guess this means I'll need a voltage regulator.
This looks interesting though, gotta figure out how this works (and for that I need to wait for a vacation from college to learn all about circuits...)

4. ### #12 Expert

Nov 30, 2010
16,346
6,831
The 2 transistors on the left could be replaced with a zener diode...or a base emitter junction connected backwards so it breaks down like a zener diode at about 5 to 6 volts, or 2 diodes in forward bias condition.

All the resistor values would be different.
The zener method would require more supply voltage than the example presented.

5. ### chrisw1990 Active Member

Oct 22, 2011
543
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you could have a single p-fet and a pnp transistor.. =]

6. ### #12 Expert

Nov 30, 2010
16,346
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A single j-fet makes a good current source all by itself...p type OR n type. Problem is they aren't adjustable over a very wide range and you don't have any.

If you had any, adding a j-fet to a bjt would make a good, adjustable, current source (as chrisw said).

Dec 26, 2010
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There is a very well-known current limiter circuit, albeit somewhat crude, which uses just two BJTs and a couple of resistors. A slightly better version could use an FET as the output device, and there are also a whole bunch of rather more accurate versions based on voltage regulator ICs ( and some explicitly current regulating ones ). Unfortunately I have only a mobile device here and can't draw any of these things up, but you should be able to find them easily enough.

The voltage regulator hack is particularly easy to understand. All you need is a suitably low-voltage output regulator loaded into a fixed resistance, and the current is defined ( at that value plus the regulator's own consumption, which sets a minimum to the practical value you can obtain for a given device).

All well and good, but for your system to stand any chance you need to know the limits of the resistance you are feeding, so that you can provide an appropriate input voltage to the system. Remember that the current regulator circuit will actually drop some voltage in itself, maybe a couple of volts lost in the simple two-transistor type, but several volts for a typical regulator-and-resistor setup.

8. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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Here is a 2 tranny current sink. If you need a source, we can think about that.

• ###### 2 tranny current sink.png
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jimkeith likes this.

Dec 26, 2010
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Good point that. I was thinking of sinks too, though of course I suppose it ondepends which polarity of parts would be used. I tend to think of NPNs and positive voltages (regulators etc.) by default.

Perhaps it would help if the original poster state his actual requirements in terms of how much current ( and range of driving voltage ) is needed, and in what direction.

I'm a bit puzzled too about the reference to a problem due to the lack of a ground connection. Could that be explained too - what is the issue?

10. ### Ron H AAC Fanatic!

Apr 14, 2005
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Ground in a circuit generally does not refer to earthing. Ground is just a reference node in a circuit against which all other nodes are measured.

11. ### atferrari AAC Fanatic!

Jan 6, 2004
2,652
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I expected both transistors diode-connected (base to collector).

12. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
It sounds to me like he might be able to use a sink or a source.

Meanwhile, I simulated a cool 4-transistor, 2-terminal floatable source/sink which has a simulated output impedance of over 20 Megohms over the range of about 3V minimum across the current source/sink to Vmax, which is determined by transistor breakdown voltages.
I'm attaching the schematic and sim results for the Zout test. The slope at the right side is due to capacitance. I got the same Zout results when I configured it as a sink.

PS I just did this to satisfy my own curiosity.

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• ###### 2 terminal floating current source-sink dynamic range.asc
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Last edited: Apr 17, 2012
13. ### Tired guy Thread Starter New Member

Apr 17, 2012
11
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well, thanks everyone don't have a zener diode, and only 2 NPN BJTs, and since you said it won't be very accurate... I guess I'll buy a voltage regulator

I have lots of stuff I took apart laying around, but my soldering iron is very old and I can't seem to get anything out of the PCBs...

14. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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What do you need to drive with a constant current?

Dec 26, 2010
2,147
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If your soldering iron is so bad that you really can't de-solder with it, then that should be the first thing on your list to fix. Very likely it just needs a new tip, or even a thorough clean. I find it handy to keep one of those little tins of tip cleaner handy - the bit needs to be kept bright and properly tinned or nothing will solder right.

If your iron really is no good, get a new one, or you will just end up getting frustrated and probably give up on the whole project.

Personally, I would try the simple circuit right away to get some practice, it's not as if a couple of transistors cost much, wherever you are you can buy them on line for ( pence / cents ) each, provided we are not looking for much power.

By the way, what sort of current / voltage do you need?

16. ### WBahn Moderator

Mar 31, 2012
17,777
4,805
I'm confused by some of the seeming contradictions in the original post. You say you only have 2 NPN transistors (and a few other things). That makes it sound like an assignment of some kind (yet you talk about having to do it in your free time). Also, you then talk about the alternative being to buy a voltage regulator. So why couldn't you buy a third transistor (cheaper than the voltage regulator). Also, your problem is that you want to regulate the current, so how would you use the voltage regulator to achieve this?

To answer the basic question you asked about whether it could be done, the answer is, "Yes, within reasonable limits." If you explain what it is you are trying to achieve, including the range of values it needs to work over and how "constant" your constant current needs to be, we could help you a lot better.

17. ### WBahn Moderator

Mar 31, 2012
17,777
4,805
I can't imagine why you want both transistors to be diode connected. Generally in current mirror circuits, you want one to be diode connected so that the current flowing it is establishes a corresponding base-emitter voltage and then you apply that base-emitter voltage to the second transistor and that one you need to have NOT be diode connected so that it can act like an active device.

The classic current mirror has the two transistors connected base-to-base and emitter-to-emitter. The accuracy depends entirely on the two devices being well-matched and the performance is not too good; namely, the output impedance is pretty low. However, there are a number of other current mirror/source designs, such as the one you were responding to, that relax and/or remove the need for well-matched devices and/or increase the output impedance considerably.

I have found that when I have some overhead voltage to spare, I can get a very good current mirror just by adding ballast resistors to the emitters of each transistor.

18. ### atferrari AAC Fanatic!

Jan 6, 2004
2,652
766
In this post, the circuit i was alluding to, is a constant current source, the subject of the thread.

Two diode-connected NPN BJT keep the base somewhat biased, right?

Last edited: Apr 19, 2012

Apr 17, 2012
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20. ### Ron H AAC Fanatic!

Apr 14, 2005
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You're correct. The schematic contains a couple of "typos". The bases should be connected to the collectors.