Current source, how to analyze?

I have this little circuit which runs off of 15V and supplies at point I a current to a load.
SE is the on/off switch for this circuit.

As an excercise I would like to understand the dynamics of this circuit, but other than the base voltage I am unable to calculate/predict anything here.

How does one approach this?

Uwe

Hello,


You have to think about several things when you look at this circuit:
1. Base emitter voltage Vbe
2. Base voltage from base to ground
3. Voltage across the 47ohm resistor VR11
4. Current through the base Ib
5. Voltage across the LED VLED, and current ILED
6. Collector current Ic
7. Bias current through the two 1k resistors IR1213
8. Current through the emitter Ie
9. Ie is equal to Ic+Ib

With light collector current Ic (large resistance collector to ground)
the transistor is biased by IR1213 and there is a little VR11. The base
emitter diode with R11 is in parallel to the LED so the LED gets only a
little voltage across it when Ic is light, not enough to light or maybe dim.
As the resistance from collector to ground is decreased the Ic
increases, which draws more current Ie which increases the voltage drop
across R11. As the Ic current is increased even more R11 drops even
more voltage until enough is dropped so that the LED starts to conduct
and light up a little. That current not only lights up the LED but
it also steals bias current from the base, making the base current Ib
less and less. This happens because Vb also decreases as the voltage
across R11 increases, and the LED starts to supply more and more of the
current through R12 and R13 that the base emitter used to supply.

As the base current decreases eventually a point is reached where the
beta of the transistor starts to limit the available current out of
the collector Ic, meaning the circuit limits the current as the LED
starts to get brighter and brighter.

Thank you very much, this has been exactly what I was hoping for and I am mulling your comments over.

I am running this from a 12V supply.
The load I am attaching is a tiny amplifier drawing 4.5 mA. The LED does not light up.
The voltage drop across the LED is 1.1V, I assume there is no current going thru the LED.
The voltage drop VR11 across R11 is 0.47V, that would indicate Ie=Ic=10mA.

That seems to contradict the measured current thru the load.
Also, if the LED is not conducting all current thru the two 1K resistors would be the base current only, maybe 2% of Ic or Ie.
With this base current he voltage at the base would have to be much smaller than it really is, I measure 11V.

I guess when I try to analyze this, I got the priorities wrong, the causes and the effects, and then things don't come out properly.

Uwe

Hi again,


I'm sorry, when i said that Ic is approximately equal to Ie that is
when the current in the collector is higher than what you are using.
A better way to look at this is to simply add the base current to the
collector current to get the emitter current:
Ie=Ib+Ic

Thus, with roughly 4ma load and 6ma bias (calculate that too) the
emitter current is 10ma.
Note that when (if) the LED starts to conduct that the base current
decreases, and this starts to limit the collector current as well as
make the LED light up.


I think this will clear up the discrepancies you are seeing.

Just wondering, do you have any data on the exact LED you are
using?

Your current source is made for a current of 23mA. The transistor is saturated and barely works a current source for only 4.5mA.

Hi again,


I hope i didnt imply that the circuit should be worked at 4ma.

The constant current level is easy to calculate. It is the
LED voltage minus the Vbe drop divided by R11, or:

Icc=(vLED-Vbe)/R11

which in the case of a red LED and R11=47 ohms comes out
to roughly:
Icc=1.1/R11=23ma approximate.

You can change R11 to a higher value if you need less current,
or change it to a lower value if you need more current.

If you change the LED to say a white LED then the drop will be
greater so then Icc=2.9/R11 very roughly.

I am using Digi Key #160-1698 as the LED. I think it requires 1.6V and 20 mA, not such a good choice if I want the LED to switch on, I guess.

I was one of my goals to modify the circuit in such a way that the LED would come on when I have the load plugged in, right now it doesn't do this.

And I am having problems with the explanations given. I do not understand how the base current could be comparable to the IC current as indicated in Audiogurus diagram with the low load. I was convinced that Ib<<Ic or Ie.

And where is a source for the formulas you use to calculate the design load?

Uwe

I deleted my post because I failed to read all of what you said so my point would be no help. I just got caught on a company computer, I have to delete my post again. Eff the boss he looking right at me.

The original circuit is a current regulator. The LED was used as a voltage reference. Frequently two series diodes are used instead of the LED.

audioguru said:

The original circuit is a current regulator. The LED was used as a voltage reference. Frequently two series diodes are used instead of the LED.

Click to expand...

I understand, but how come the relation between Ib and Ic (Ic=beta* Ib) does not hold??

Hi again Othello,


The reason is that transistors operate in more than one 'mode'.

The mode you are thinking about is called the 'active' mode, where
the transistor acts as a sort of linear device and Ic=Beta*Ib.

Another important mode is call the 'saturated' mode, and that is where
there can be much more base current and still the same collector current
because the transistor now looks sort of like a switch that has been
switched 'on' and that's the most that it can do...it acts almost like a short.
When this happens the transistor is said to have entered 'saturation', and
this occurs roughly when the collector base voltage is lower than the
emitter base voltage, although there are better definitions than that.
Typically the collector to emitter voltage goes down to 0.2 volts and
that's as far down as it can go. This is caused by enough base current
where the max Ic is less than Ib*Beta.
For example, if you have a 10k resistor on the collector to +10 volts
the max Ic would be 1ma, and if you drive the base with 1ma and the
Beta is 100 that would be enough base current to drive 100ma collector
current, but the 10k limits this to 1ma so the transistor goes into saturation.

If you would like to see some basic formulas for this circuit i can come back with
some a little later?

BTW, how much current do you need coming out of the collector in your
application?


P.S.
If you want the LED to come on i think you should be able to get that to happen
by increasing the 47 ohm resistor until it comes on. At that point the current
out of the collector will be limited to whatever that new value brings.

A current source is generally not a good power supply for an amplifier. You should probably be using a voltage source. There are ways to detect when a load is connected to a voltage source.

Thank you MrAl and Roff and the others.

After the sample calculation a few posts back I figured the value of the R11 I would need to supply the 4.5mA and to also switch on the LED. I have now a 150 Ohm resistor in there and that seems to work good.

Roff, I don't know why the author of this circuit chose a current source as a supply for the small amplifier (it is an onboard amp built into an electric guitar and the circuit we are discussing supplies that amp thru an extra wire in the input cable with juice. The LED, if it works, is a good status indicator that power is indeed available).

So now one of my goals has been met, the other would be to understand the circuit and the underlying principles better, and yes MrAl I would love to see some of the formulas which govern this, but a source would suffice, I can look for it myself, I just want to find the right ones...

Uwe