Calculating Vc and Ic for darlington

[TsAmE]

D is a Darlington transistor (whose base current may be neglected).
Estimate the collector voltage VC and the collector current iC.

Attempt:

Vin = V x R2 / R1 + R2
= 6 x 10 / 20 + 10
= 2V

IB = Vin - 1.2 / RB
= 2 - 1.2 / 20
= 0.04mA

Ic = IB x β
= 0.04 x 1000
= 40mA

V = Ic x Rc
= 40 x 2
= 80V (saturated)

Therefore Vc = 0.7V

I know my answer is wrong, but I dont get what I did wrong.

 

[OMID-313]

Hi friend

I think the calculation of the Ic is wrong.
Ib × β is the collector's current , and not the resistor's. Did you get it !?
Therefore , you must first calculate the resistor's current ( which is more than collector's current ) , and then calculate it's voltage.

Good Luck
Hope I've helped !

 

[TsAmE]

I think the calculation of the Ic is wrong.
Ib × β is the collector's current , and not the resistor's. Did you get it !?
Therefore , you must first calculate the resistor's current ( which is more than collector's current ) , and then calculate it's voltage.

I thought that Ic was the same current for the resistor since this is the amplified current? How can there be 2 different Ic s?

 

[LvW]

Quote:
Vin = V x R2 / R1 + R2
= 6 x 10 / 20 + 10
= 2V

You should first define all voltages (what is V, Vin, RB ?)
Nevertheless, your equation implies that V=Vc and Vin=Vbase, correct?
In this case V cannot be identical to 6 V.

 

[TsAmE]

I made a mistake in my initial post.

Vin = V x R2 / R1 + R2
= 6 x 10 / 20 + 10
= 2V

Vin is supposed to be:

Vin = V x R2 + R3 / R1 + R2 + R3
= 6 x (20 + 10 / 2 + 20 + 20)
= 4.2V

IB will then = 0.15mA

and IC = 150mA

and V across the resistor will = 300V, which still leads to my answer of Vc = 0.7.

V = 6V
Vin = 4.2V
V across RB = 6 - 4.2 = 1.8V

Ignore my first part of my post, cause I made a mistake.

 

[LvW]

TsAmE,

I am not motivated to check your equation without knowing the definitions of R1, R2 and R3. Is it so hard to be exact?
More than that, don't ignore the correct way of writing equations (brackets !). Otherwise, it's not possible to read the equations.

 

[TsAmE]

TsAmE,

I am not motivated to check your equation without knowing the definitions of R1, R2 and R3. Is it so hard to be exact?
More than that, don't ignore the correct way of writing equations (brackets !). Otherwise, it's not possible to read the equations.

Oh ok I am really sorry. Here are the corrections:

R1 = 2kΩ
R2 = 20kΩ
R3 = 10kΩ

Vin = V x ( (R2 + R3) / (R1 + R2 + R3) )
= 6 x ( (20 + 10) / (2 + 20 + 20) )
= 4.2V

 

[LvW]

Up to now, unfortunately, you did not tell us what Vin shall be!
However, according to your formula it is the voltage at the collector node
under the condition Ic=0 (current only through 3 resistors in series).

Is that your goal? I doubt.

Added after 23 minutes:

I think, the circuit cannot be analyzed without an additional info.
For a single transistor, in such a case, one assumes app. 0.6 or 0.7 volts at the base node; likewise for a Darlington app. 1.2...1.4 volts.

 

[TsAmE]

Up to now, unfortunately, you did not tell us what Vin shall be!
However, according to your formula it is the voltage at the collector node
under the condition Ic=0 (current only through 3 resistors in series).

Is that your goal? I doubt.

Added after 23 minutes:

I think, the circuit cannot be analyzed without an additional info.
For a single transistor, in such a case, one assumes app. 0.6 or 0.7 volts at the base node; likewise for a Darlington app. 1.2...1.4 volts.

I did specify Vin in the calculation in my previous post (Vin = 4.2V). If you mean the particular Vin that turns on the transistor, then since this is a darlington it is 1.2V.

My goal is to estimate the collector voltage VC and the collector current iC.

 

[LvW]

I did specify Vin in the calculation in my previous post (Vin = 4.2V). If you mean the particular Vin that turns on the transistor, then since this is a darlington it is 1.2V.
My goal is to estimate the collector voltage VC and the collector current iC.

I can only guess that you mean Vin=VB (at the base), right?
And if it is so, why did you specify thius value? How exactly looks you darlington combination? Any resistors inside?

In any case, if VB is really known, it shouldn't be a problem for you to calculate the current through R2 and R3. And then, of course, you have Vc and Ic. You only need Ohm's rule.

 

[FvM]

Referring to your initial post, your calculation doesn't follow the said prerequisites. If it's said, Ib should ne neglected, why don't you?

Vbe must be known, of course, it's the key for determing the bias point (For a reasonable exercise problem, it must be either given, or taken as granted). If it's 1.2V, than Vce is 3.6V and IC is 1.2 mA - 0.12 mA, that's all.

 

[srizbf]

FvM has given the answer .
elaborating the same:

in the circuit figure shown(attached):
assume component names asR1=10k R2=20K and Rc=2K

if Vbe of darlington is 1.2v then neglecting base current ,
Vr1=1.2v.

and Vr2=2.4v(since R2 is 2 times R1)

So, Vc=2.4+1.2=3.6v

supply V=6v

hence Ic=(V-Vc)/Rc
therefore Ic=(6-3.6)/2k
which is 1.2mA.

So Vc is 3.6v and Ic is 1.2mA
(what FvM gave in short)

srizbf
19thjune2010

 

[FvM]

LvW has given the answer too, leaving the calculation to the original poster...

 

[TsAmE]

Thanks a lot for the help. Im just curious about one thing:

*For Vb (Vin) cant you calculate it using voltage division from the +6V supply rail to the earth of the 10k resistor to get a specific Vin, rather than 1.2V? Which is what I attempted. (I know you use Vin = 1.2V when not enough info is given).

 

[srizbf]

if you start solving for Vin(Vbe) then you have to write many loop equations and solve.

since base current is very small and Vbe of darlington is 1.2v(typical)
it is better to go approximate method.

even if you write the equations you will get the values close to the approximate ones.

srizbf
20thjune2010

what software did you use to draw your png circuit file attached?

 

[TsAmE]

I didnt use software, was an example from my online notes, but Yenka is a pretty good app for drawing circuits.