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Transistor as a constant current source to variable DC input

AshwinNambiar

Nov 11, 2014
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I want to build a circuit using a transistor are a few light up my LEDs.Now,LEDs are a constant current source.This seems simple.However ,i have a situation:
1.The input DC voltage from the supply can vary from 16Volts to 32 Volts.
2.I want to build it using a single transistor,i.e., a linear circuit,"without switching"

The voltage can be any nominal voltage,say 20Volts..however a variation in the range of 16 to 32 or any abnormal voltage for that matter should not cause any change in the current to the LEDs..

I'll be glad to receive suggestions/inputs as to how to go about with this.
Thanks
 

BobK

Jan 5, 2010
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Here you go.

upload_2015-11-28_16-22-43.png

D2 is a 4.7V Zener diode.

This will regulate the current to approximately

I = (4.7 = 0.7) / R

In this example I used 20mA and, as the graph shows, the current varies from 19.59 to 19.89 mA.

At this current the transistor is dissapaing (32 - 4 - Vf) * I. For a red LED at Vf = 2.0V this would be 0.52W which is about the limit for a TO92 transistor, so it will get quite warm. At currents higher than this, you would have to use a power transistor.


Bob
 
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AshwinNambiar

Nov 11, 2014
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Here you go.

View attachment 23452

D2 is a 4.7V Zener diode.

This will regulate the current to approximately

I = (4.7 = 0.7) / R

In this example I used 20mA and, as the graph shows, the current varies from 19.59 to 19.89 mA.

At this current the LED is dissapaing (32 - 4 - Vf) * I. For a red LED at Vf = 2.0V this would be 0.52W which is about the limit for a TO92 transistor, so it will get quite warm. At currents higher than this, you would have to use a power transistor.


Bob
Thanks Bob,what i now want to know is,you are using a 4.7V zener diode,how are you taking care of the remaining voltage from the supply or rather how is it ensured that the zener is capable to dissipate the remaining power?Irrespective of any voltage from 16-32 ,the led should get a constant current.Well,this is clear from the explanation you have given but where i still have a doubt is ;how do we take care the varying voltage??
 

Harald Kapp

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how are you taking care of the remaining voltage from the supply
This will be seen across R1.
how is it ensured that the zener is capable to dissipate the remaining power
R1 needs to be schosen such that at minimum Vin the current through R1 is still sufficient to drive the base of Q1 for the desired outpt current plus the current need to keep the zener diode within its operating point.
At maximum Vin a proportionally higher current will flow through he zener diode which thereby defines the maximum power dissipation of said diode. You'l have to select the type of diode accordingly.
 

BobK

Jan 5, 2010
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The Zener will always drop about 4.7V at 32 V input the current through it is:

V = I * R
I = V / R

I = (32 - 4.7) / 1000 = 27mA

At this current it will dissipate 4.7 * 0.027 = 0.127 W or a little over 1/8 W. A 1/4 W Zener is appropriate.

The circuit works by controlling the voltage across R2. That will be the Zener voltage (4.7) minus the base to emitter voltage (Vbe) of the transistor which will be about 0.7. Thus we have 4.0V across the 200Ω R2 which gives us 20 mA for the current through R2. The current through the LED is nearly that same as this.

This is the classic, simple constant current source (actually sink) circuit which can be found in any elementary electronics book.

Bob
 

dorke

Jun 20, 2015
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The most important parameter is missing.
What is the LED current needed?
 
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