# Current detection (darlington) question

#### TsAmE

Joined Apr 19, 2010
72
Calculate the input current Iin for which the Darlington transistor D will begin to turn ON, and estimate a minimum value of Iin which guarantees that D will saturate. (here, the saturation voltage can be ignored; you may assume that in saturation VC = 0)

My attempt:

Iin = Vin / R
= 1.2 / 1000
= 0.0012mA

Ic = 15 / 15
= 1mA

Iin (sat) : Iin = Ic / 1000
= 1 / 1000
= 0.001mA

This was my final answer and I was right up to here, but the answer for Iin (sat) was actually 2.2μA. I dont get why Iin (sat) is the sum of the 2 Iin s that were worked out.

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#### Jony130

Joined Feb 17, 2009
5,539
Don't forgot that I_in = Ib + Vbe/1MΩ = 1μA + 1.2μA

#### TsAmE

Joined Apr 19, 2010
72
Don't forgot that I_in = Ib + Vbe/1MΩ = 1μA + 1.2μA
Wouldnt you use the supply voltage as opposed to Vbe? (since Vbe = 0 when the transistor is saturated).

Would Iin = Ie (emitter), which is why you add Ib and Ic?

#### Georacer

Joined Nov 25, 2009
5,182
$$V_{\small{be}}$$ is never 0 when the transisor allows current through it. Think of the BE part of the transistor as a simple diode. If current is to flow through it you must have a voltage difference of 0.7 volts, give or take, depending on how high is Ic. $$V_{\small{be}}$$ is 0 only when the transistor is cutoff.

#### Jony130

Joined Feb 17, 2009
5,539
Wouldnt you use the supply voltage as opposed to Vbe?
Not in this case, in this case Vin must be equal Vbe because we don't have Rb resistor in series with base.

since Vbe = 0 when the transistor is saturated.
If Vbe = 0V the BJT is cut-off.

Would Iin = Ie (emitter), which is why you add Ib and Ic?
No, Iin is not equal Ie.
Ie = Ib + Ic = (β +1 ) *Ib (this equation always hold even in saturation)

In your circuit Iin is equal the sum of base current (Ib) and current that is flow through R1 resistor(1MΩ).
Don't forget that Kirchhoff's first law must hold.

Iin
= IR1 + Ib

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#### TsAmE

Joined Apr 19, 2010
72
Thanks a lot for all the diagrams with the info, cleared a lot of doubts. There is another thing which I would like to know (refer to new attachment), why is it that this setup wont work?

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#### Georacer

Joined Nov 25, 2009
5,182
I hope that you don't plan to use this circuit as an amplifier, as it is highly unstable. There are much better assemblies for that purpose. But, for the sake of the argument, in order for the transistor to operate in the active region, Vbb should be marginaly higher than 0.7 and quite lower than 0.7+R/beta*10^-3. In a common transistor with Icsat in the region of some mamps, Ib should be under a mamp to avoid saturation. Some specifications on the circuit would allow for a deeper analysis. It seems though a good assembly for a limited current switch.

#### Ghar

Joined Mar 8, 2010
655
I hope that you don't plan to use this circuit as an amplifier, as it is highly unstable.
I would say "unpredictable" or something as the word stability has a very specific definition in electronics.

#### Georacer

Joined Nov 25, 2009
5,182
Guess you are right. What I wanted to say is that the gain of the amplifier will vary greatly with the input signal width and ambient temperature.

#### TsAmE

Joined Apr 19, 2010
72
Guess you are right. What I wanted to say is that the gain of the amplifier will vary greatly with the input signal width and ambient temperature.
Why would the gain vary greatly? I thought each transistor had a partically gain witch is fixed (β).

To me, it seems like you could say:

Iin (IB) = Vin - 1.2 / R

and then use that to calculate IC.

#### Jony130

Joined Feb 17, 2009
5,539
Thanks a lot for all the diagrams with the info, cleared a lot of doubts. There is another thing which I would like to know (refer to new attachment), why is it that this setup wont work?
Becaues in your first circuit R1 is connect parallel to base-emitter junction.

And in this circuit

Resistor R is connect in series with base-emitter junction.
So you should already know that current in series stays the same.
Because we have only one current which flow from Input voltage source through R resistor then base-emitter and then back to input voltage source.
Iin = IR = Ib
And that's why to find Ib we must write:
Ib = (Vin - Vbe) /R (KVL)

Why would the gain vary greatly? I thought each transistor had a partically gain witch is fixed (β).

To me, it seems like you could say:

Iin (IB) = Vin - 1.2 / R

and then use that to calculate IC.
Only in theory β is fixed.
In real world β change with Ic and temperature.
Look at the data-sheets and figure 2 (page 3)

http://www.iele.polsl.pl/elenota/Philips/bc516_4.pdf

And voltage gain is equal Au = 40 * Ic * Rc
And Ic change with β and with Ib so gain also must change.
And almost all BJT parameters are temperature dependent.

#### Georacer

Joined Nov 25, 2009
5,182
For the reasons Johny130 mentioned, a good transistor amplifier should be assembled in a way that minimizes the effect of those changes of performance on the output. I am no expert on the subject, maybe someone could give you an easy solution if you specified your needs, but a look on the Sedra/Smith handbook couldn't possibly hurt. It contains many schematics and assembly types that compensate for variable b and temperature.

#### TsAmE

Joined Apr 19, 2010
72
So basically connecting R in series with base-emitter junction:

*Would let a very high current to flow, possibly destroying the darlington.

While connecting R parallel to the base-emitter junction:

*Part of the current will be grounded, and a reduced current will travel through the base.

Is this right?