# Question about current sources - Not getting the correct current through load

Discussion in 'General Electronics Discussion' started by Robert Hill, Jul 6, 2015.

1. ### Robert Hill

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Mar 5, 2015
Hi all,

I've been learning about current sources and tried to make a simple one as a practice.

I followed the design 2.21 on page 72 of the 2nd edition of the art of electronics.

Essentially I have a 5v VCC going into a standard LED (the load) and then the negative lead of the LED goes into the collector of an NPN transistor. The base of the transistor is biased by a 3.3v supply through a 10ohm resistor. The emitter of the transistor goes to ground through a 1k resistor. By my calculation I should get a current of 2.7miliamps through the LED ( 2.7v/1000Ohms =0.0027 amps)
This is verified when I build the circuit in the free software 'circuit simulator'.

However when I built the circuit in real life I got 3.4 miliamps through the LED. This seemed to be basically the same as if I ran the LED from 5v supply in series with a couple of hundred ohm resistor.

Can anyone help me understand why this is or if I've gone wrong somewhere? It is to do with the unreliability of components such as resistors having 5% inaccuracy in their ratings or perhaps it is the transistor I am using?

Any help appreciated.
Thanks,

Last edited by a moderator: Jul 6, 2015

5,165
1,087
Dec 18, 2013
Hi Robert
I think your right, it's the tolerance of real life components. The current that will pass through the LED will be the voltage appearing across the emitter resistor divided by the emitter resistance. If you actually measure the voltage and resistor value you should see that they match your measurements.

3. ### davennModerator

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Sep 5, 2009
Hi Robert

where did you get the values to put into you calculations ?

real life components are not going to have idea values or properties

the LED, the collector-emitter resistance of the transistor, the base resistor and the emitter resistor

you are only looking at 0.7 mA difference and that would easily be a result in the difference in tolerance values of the 1k emitter resistor
even before you take into account the transistor and LED variations

your 1k resistor, say it has a tolerance of 10% ( pretty standard unless a 1% metal film type)
its value could be anything from 900 Ohms to 1,100 Ohms

see your problem ?

Dave

4. ### Gryd3

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Jun 25, 2014
My electronics teacher was a stickler for doing the work... but I figured I was familiar enough with it, so I did the math... Got all the right answers, but they were too exact so he gave me a 0 on the work.
When you are dealing with real-life components, the ideal values you learn about in books is almost never observed unless you buy precision components... and that's no problem because many of the components will operate as long as they are within a given 'range'.

So, moral of the story. Fudge the numbers if you going to use a calculator, and never expect your carefully planned out circuit to give you an exact output!

*Think about this too... perhaps your circuit IS perfect... your multi-meter certainly is not... simply connecting a meter or probe to your circuit will change it's behaviour. The effect is incredibly slight, but remember your multi-meter can be looked at like a resistor! A very large value resistor when used to measure voltage, and a very low value when measuring current.

5. ### Robert Hill

112
12
Mar 5, 2015
Nice one chaps, thanks,

Gryd3 - sounds like your teacher probably quite enjoyed giving you a 0.

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5,165
1,087
Dec 18, 2013
We had a guy just out of Uni, green as a cucumber. Designing a simple transistor switch. He asked me for a 1.589879 K resistor. I said they are out the back next to the 100.23587 uH inductors, just bellow the 100.325478 pF capacitors

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7. ### BobK

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Jan 5, 2010
Most likely, it is the 0.6V you are using for Vbe. This is a reasonable typical value, but is not accurate at all. Vbe will vary quite a bit with base current.

Bob

8. ### Colin Mitchell

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Aug 31, 2014
You have all missed the obvious.
The LED could be red, yellow, blue,green indio or violet.
They all have different characteristic voltage drops and the voltage across each varies slightly due to the different manufacturer and brightness and if it is cheap, junk, good quality, high brightness, low current, super-bright, or high-bright.
The 10 ohm resistor turns the transistor into an emitter-follower and this changes the whole aspect enormously.

9. ### Colin Mitchell

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Aug 31, 2014
The problem is this: The text book states a RESISTOR LOAD. You have used a LED.
And secondly the voltages you have used, create a conflict, due to the characteristic voltage across a LED.
You need what we call a BUFFER and this consists of a resistor in series with the LED. Next you need to choose a higher supply voltage to remove the conflict.

10. ### Colin Mitchell

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Aug 31, 2014
I agree, the text-book uses far-too-complex language to explain this complex concept and it is no wonder you don't understand it.
I will explain in slow-motion how the circuit works.

The transistor is basically an emitter-follower and this means the voltage delivered to the base must have the capability of providing a current.
The current does not have to be large but it must be about say 5% or 10% of the current you want to deliver through the load.
You will not need this current in the final stage, but it is needed to get the stage to start to operate.
This point was explained as: the base voltage can be delivered via a voltage divider but the voltage divider must be STIFF.
In other words, the voltage divider must be able to deliver a current without the voltage it is delivering DROPPING.
The text also states the voltage can be delivered via a zener.

But it never states the transistor is an EMITTER FOLLOWER.

To see how the circuit works we start with just a transistor and emitter resistor - NO LOAD

The circuit starts by delivering a voltage to the base and the transistor rises and the emitter follows but it is 0.7v less than the base.
This means current flows though the emitter resistor and this current is say 10mA.
This current is supplied by the BASE.
The next fact to realise it this: The transistor is FULLY TURNED ON.
Even though there is no load, the transistor is in a fully turned on condition.
The transistor is fully turned ON and current will flow through the LOAD and this current will also flow though the emitter resistor and the voltage across the emitter resistor will increase.
This means the voltage on the emitter of the transistor will increase but the voltage on the base will remain constant.
This means the voltage across the base-emitter terminals of the transistor will decrease and the transistor will start to turn OFF.
This means the current that was originally supplied by the base to produce the voltage across the emitter resistor, will now be supplied by the LOAD.
Suppose the supply voltage increases.
This will cause more current to flow through the LOAD and also through the emitter resistor.
This will raise the emitter voltage and the transistor will turn OFF slightly.
This will increase the voltage between the collector and emitter and if the supply rises by say 3v, the voltage across the collector and emitter will rise by 3v and the current through the circuit will not change.
In other words the circuit is a constant current circuit.
It’s that simple.
It’s a pity the text book did not explain it this way.

11. ### AnalogKid

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Jun 10, 2015
What is obvious is ten posts and no schematic.

ak

12. ### davennModerator

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Sep 5, 2009
it didn't need a schematic

nor did it need Colin's multiple posts

The question was answered in the 3 posts following the OP

13. ### Gryd3

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Jun 25, 2014
But this has been the most detailed post from Colin I've seen in a while. Although it's a little.... rough, it cover plenty more details than many of the partial single sentences I've seen in the past.

14. ### davennModerator

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Sep 5, 2009
LOL it is but a bit late and much more detail than needed
probably confused the OP

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15. ### Martaine2005

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May 12, 2015
I like Colin's post..
I can smell the keyboard smoke from the UK.

Martin

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16. ### LvW

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Apr 12, 2014
My comments are in bold.

I am sorry, but this text has produced more questions to me than giving answers.
Colin - perhaps I misunderstood something, but I must admit that I am really confused. It is such a simple circuit which exploits the fact that the BJT is nothing else than a voltage-controlled (non-ideal) current source (Ic=f(Vbe)). This can be easily prooved having a look on the output characteristics of the BJT. The small slope is an indication of a large output resistance r,out. And the emitter resistor RE causes a further increase of this resistance - thereby improving the quality of the current source. I think, the keyword "emitter follower" cannot help at all to understand the working principle of the circuit.

Last edited: Jul 8, 2015
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17. ### AnalogKid

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Jun 10, 2015
My point was that there are many on this forum who do not have the referenced book, so this whole thread has the feel of an in-group/out-group dialog. I don't think it is too much to expect that anyone asking or describing anything on an electronics forum should pitch in a schematic. It is a simple courtesy that shows a bit of awareness, something I've pointed out to Colin more than once.

ak

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18. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Here is the circuit:

Sorry about the large image.

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19. ### AnalogKid

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Jun 10, 2015
Thanks. Some of the posts now make more sense.

ak

20. ### LvW

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Apr 12, 2014
Yes - it is nothing else than a simple common emitter stage (of course, not an emitter follower) with RE feedback - and the current Ic through the "load" is considered to be the output of the current source.
No magic behind the circuit.

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