# Transistor C.C

Discussion in 'General Electronics Chat' started by Shagas, Oct 21, 2013.

1. ### Shagas Thread Starter Active Member

May 13, 2013
802
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Hello !

Looking at these two circuits of a transistor constant current source circuit taken from talkingelectronics.com :

I have a few questions regarding this arrangement , but first let me try to explain what I understand is going on there:

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In the version with the two diodes the base is biased at 2*0.7 (1.4) volts.
VBE is 0.7 volts so that means that the emitter has to be at 0.7 in respect to ground which means that the transistor will push a magnitude of current through the emitter resistor so that there is 0.7V across it , therefore satisfying the condition that the emitter is at 0.7v IRT ground therefore giving us a constant current irregardless of the value of the collector resistor(if we have one and assuing VCC is large enough) , right ?

The circuit with the two transistors works in a similar fashion where if the emitter resistor drops more than 0.7 then it turns on the left transistor which turns off the right transistor therefore keeping 0.7 across the emitter resistor which gives us a constant current.

***

Now for some questions :

- Is this a good/accurate constant current source ?
-What is the diffrence between the two versions of the circuit in terms of performance?
- Are there any parameters that will cause my constant current to drift without me adjusting anything? perhaps VBE or something else?
- Is there any way that this circuit could be improved to provide constant current with a high frequency square wave input ? Say.. a few hundred khz or more assuming that the transistors are up for the job.

Tim

2. ### Brownout Well-Known Member

Jan 10, 2012
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Nope. The voltage across the resistor can be less than 0.7V. It cannot be more, however.

3. ### wayneh Expert

Sep 9, 2010
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For lighting an LED it should be fine. You need to define "good" and "accurate" if you need a better answer. With no personal experience, I'd say a ±10% fluctuation in current might be expected.
I'll speculate that since the 2-diode circuit adds one more diode drop, that maybe this introduces an additional source of drift and variation. Total wild guess.
Temperature for sure.
I can't answer. Do you mean constant average current, or constant while the pulse is on? Hmmm...I guess those are really the same thing.

Nov 30, 2010
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5. ### Shagas Thread Starter Active Member

May 13, 2013
802
74
Yes I guess that makes sense , thanks .

Thanks for the input , Wayneh

Well ... by good I mean a max of 1% deviation . but I guess without temperature stabilization etc that is not possible.

What I mean by this is that if i'm providing a 100khz square wave into the current source input then I want to see a faithfull representation of the 100khz square wave when I put my oscilloscope probes over the load resistor.
I guess i'm asking for alot from this simple circuit.

6. ### Shagas Thread Starter Active Member

May 13, 2013
802
74
Very nice #12 , I wanted to ask if the CC can be done with a jfet but it looks like my question is answered.
I'll be reading those posts now , thanks .

7. ### ScottWang Moderator

Aug 23, 2012
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These two circuits are not the constant current source, they are only one kind of current source has current limiting.

Why is that, becasue the real constant current source that it won't changed the current when the limiting resistor is changing, that's why it called constant current source.

The two bjts circuit is more useful than the two diodes circuit, because the protection bjt will shutdown the Base current of the bjt driver when the Vr raising up to 0.7V, and the bjt driver will turn off, but the two diodes that they just as a shunt diodes, when the Vr raising up to 0.7V the bjt driver will still working, the function is limiting the highest current, protecting the load to avoid to get over current.

8. ### Shagas Thread Starter Active Member

May 13, 2013
802
74
Also another question:

Why are there always inductors in laser diode drivers?
I know that inductors resist sudden changes in current , in other words their impedance rises with rising frequency .
Are they there to suppress sudden rushes of current that the transistor can't ?
If I remember reading a while ago that laser diodes are very very sensitive to overcurrent and that the diode can suffer micro-fractures when electrically 'shocked' (I read this quite a while ago so I'm not sure I recall it correctly ) . Is this why inductors are used in the drivers ?

9. ### Shagas Thread Starter Active Member

May 13, 2013
802
74
Scottwang , by limiting resistor do you mean the load resistor?

If so then this applies here .
I tried pulling a constant 10ma from this circuit (going through my DMM) and it pretty much holds at 10ma for a load resistance of zero to a few hundred ohms @ Vcc = 5V .

10. ### ScottWang Moderator

Aug 23, 2012
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Yes, you can changing the resistor, and to see how the current is changing.

Normally we using the current source including a limiting resistor is enough, because the real constant current source is more complicated.

A real constant current source can be in series many LEDs until the total voltages of the LEDs is great than the voltage source, and also don't care about the resistor.

11. ### Shagas Thread Starter Active Member

May 13, 2013
802
74
Well yes I understand what you are saying , but for now i'm just looking at simpler current sources which require voltages less than Vcc to deliver the current required

12. ### ScottWang Moderator

Aug 23, 2012
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Go check my posted. on #7 and choosing a properly resistor for the current.

13. ### WBahn Moderator

Mar 31, 2012
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My guess is that the primary focus of either of these two circuits is to make the current largely immune to variations in the signal providing the control input. Note that the supply is indicated as being anything from 6V to 15V. If the input signal is derived from that same supply, then there could be significant variations in it's "on" value.

This circuit is almost certainly meant to be operated with the input voltage noticeably more than one Vbe drop above the voltage at the top of the resistor. The intent of the feedback transistor or the two diodes is to shunt the excess current in the base resistor around the drive transistor to make this happen.

Let's assume that the input signal could vary from, say, 5V to 15V. At 5V there would be about 360μA of current in the 10kΩ resistor and at 15V there would be about 3.6mA in it. Now let's assume that the transistor has a β of 100 and that the resistor is chosen to yield an LED current of 10mA. That means that we need 100uA of base current, so we need to shunt 260μA to 3500μA around the transistor. That's over an order of magnitude difference.

The performance, in terms of clamping the current at a fixed value and not drifting with temperature, should be considerably better with the two-transistor circuit than with the two-diode circuit.

In the two-diode circuit, the voltage across the resistor will be 2Vd-Vbe. The voltage across the diodes will vary with the amount of current that has to be bypassed. In going from 260μA to 3.6mA, at room temperature we can expect the forward voltage across each diode to increase by a bit over 60mA. That's a total increase of about 135mV. If the nominal voltage across the resistor is 0.7V, then that would amount to an increase in current of about 20%. There would only be an increase in Vbe of about 5mV, so that would have negligible restoring effect.

In the two-transistor case, the increase in Vbe of the feedback transistor would only be about 65mV so the net increase in resistor current would be just under 10%.

In the first circuit, the only parameter of real significance (with regards to drift) is Vbe of the feedback transistor. In the two-diode circuit, you have the drops of both diodes and the drive transistor that will vary with temperature.

14. ### Shagas Thread Starter Active Member

May 13, 2013
802
74
Thank you for your analysis WBahn I was hoping you would join .

Question:
How did you get 360uA of current in the 10k at 5v and 3.6mA at 15V?
We don't even know the RE so how did you calculate the current?

- I understand that that temperature drift will cause the VBE to chance therefore changing current etc , is there any simple way of countering this in this circuit using a few more biasing components ?

Also , what do you think of this one ?

Would the zener be a more stable reference?

15. ### ScottWang Moderator

Aug 23, 2012
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>the input signal could vary from, say, 5V to 15V
Hey, I found a little bug, maybe you forgot to wear your glasses, my daughter always told me that I didn't bring my eyeballs with me when I can't see something clearly.

So do you think if the input voltage vary from 6~15V then using the logical level MOSFET to replace the bjt, will it help?

16. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
What bug?

Originally I had 6V to 15V, but dropped it to 5V to allow for the possibility of some drop from the supply in the originating signal. I thought about putting a similar drop for the upper voltage, but decided to keep the range wide for the purpose of doing worst case kind of stuff.

17. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
To calculate the current in the base resistor we just need to know the voltage at both ends, right? The voltage at the left end is Vin and the voltage at the right end it 2Vbe or 2Vd. In either case, it is about 1.4V. But I did make a math error and the current for the upper end should be 1.36mA and not 3.6mA. That actually raised a red flag when I typed it and I wondered, "Does it make sense that the increase is exactly an order of magnitude?" I didn't think it did, but noticed that the increase in voltage was 10V and then went muddle-minded and immediately convinced myself that this explained it because the difference after the two diode drops must narrow the range enough. Of course, this is completely wrong. Maybe this is the bug Scott was referring to.

Before we talk about how to achieve something, let's first establish what it is we are trying to achieve. Otherwise you'll end up buying a battleship with a popgun and a rowboat would have sufficed.

I'll have to look at this later.

18. ### ScottWang Moderator

Aug 23, 2012
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The little bug had turned to a beautiful butterfly.

19. ### Shagas Thread Starter Active Member

May 13, 2013
802
74
Yes I see what you mean .

Well i'm just looking at transistors in more detail in general so I don't have a specific purpose at the moment for the current source but i'm gathering information for when I do have a specific project .

For now we can just assume that i'm trying to drive a load with say... a 1-3% accuracy of the desired set current.

Also as I meantioned earlier my first goal was to input a square wave from a microcontroller at 0-5V logic levels and when putting my oscilloscope over the load resistor I would like to get an exact representation of the square wave (voltage) across the resistor.
And my second question is how I could improve the speed of this circuit to say...1Mhz or more.
With these circuits that I showed at the beginning with just two transistors and an emitter resistor the square wave (voltage) over the load resistor starts to curve on the edges at higher frequencies . I would like to improve the transient response of this circuit (I hope I'm using the correct term here)

20. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
The edges of your squarewave represent very high order harmonics and you might well be getting into the region where parasitic capacitances are introducing poles into the response.

What does it look like when you probe the squarewave directly?

I don't understand what you mean by an exact representation of the squarewave. You have a circuit that is going to clamp the waveform at the resistor once the input gets above a certain level, so how will having that level be stable within 1% give you any more fidelity to the squarewave you are trying to represent?