# how to analyse this circuit?

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#### hayowazzup

##### Advanced Member level 4
Hi, I just found this circuit on wikipedia and i'm trying to figure how it work.

Can someone tell me how the diode work as a reference in the circuit? how does it depend on the current through R1?

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#### rolupusoru

##### Newbie level 3
Hi,

R1 and DZ1 act as a voltage divider. DZ1 will change it's reverse resistance so that it's dropout voltage is equal to the breakdown voltage. R1 will acts as a current limiter. This is true for currents in the ordr of tens of mA.The whole circuit looks to me as a voltage stabilizer, where the voltage output of Q1 will be Vdz1 - Vbe.So R2 will have a dropout voltage of DZ1-0.7v.

#### sherazi

##### Banned
exactly As rolupusoru said! the diode is a zener diode and its used here to regulate teh volatge to the level VDZ1 -Vbe....

#### kak111

##### Advanced Member level 4
Yes exactly so,

but i will call this circuit as constant current sink.

When voltage Vs fluctuates , current I load has near constant value.

##### Advanced Member level 2
I have only ever used this circuit for a slightly different purpose....

To me it is a constant current source. As the load resistance changes, the current will remain constant, but the voltage across the load will vary. (I am assuming Vs+ remains constant the way I use it.)

Alternatively it could be a current limiter.... say the load is a motor, and the motor stalls, this circuit would automatically prevent excessive current flowing and hopefully prevent motor burn out.

All of the replys in this thread are correct and valid (at least as far as I can tell with my limited understanding). It is just that the circuit can be used in different ways, and viewed in different ways.

#### Syncopator

##### Full Member level 6
It's a constant current source.

The voltage across the zener is fixed. The voltage across the emitter resistor is equal to the zener voltage minus Vbe (0.7V).

The current through the emitter resistor is the emitter voltage divided by the the value of R2.

Since the base current is so small it is ignored; the emitter and collector currents are considered to be equal.

The current through the load is, therefore, Ve/Re.

So long as the supply voltage is sufficient, its actual value is irrelevant.

#### lovexxnu

##### Junior Member level 1
yes, this is constant current sink.
the current depends on the Re

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