Leakage saturation current Is for 2n2222

[hithesh123]

What is the leakage saturation current for 2n2222.
The Is i'm talking about is the one used in
Vbe = Vt ln(Ic/Is) equation.

 

[mister_rf]

Spice model for 2N2222
Is=14.34f Xti=3 Eg=1.11 Vaf=74.03 Bf=255.9 Ne=1.307+ Ise=14.34f Ikf=.2847 Xtb=1.5 Br=6.092 Nc=2 Isc=0 Ikr=0 Rc=1+ Cjc=7.306p Mjc=.3416Vjc=.75 Fc=.5 Cje=22.01p Mje=.377 Vje=.75+ Tr=46.91n Tf=411.1p Itf=.6 Vtf=1.7 Xtf=3 Rb=10

According to this Is=14.34fA = 14.34 * 10^-15 A

 

[dave9000]

it's a purely theoretical value, any value between 1fA and 1pA would do: the calculated Vbe changes by some Vt, i.e. it is unpredictable like in real life ;-)

 

[hithesh123]

it's a purely theoretical value, any value between 1fA and 1pA would do: the calculated Vbe changes by some Vt, i.e. it is unpredictable like in real life ;-)

Assuming Vt is 26mV, what is Is.

I am trying to calculate the bias values for base biased CE amplifier with emitter degeneration resistor.
My gain is 10. How do I calculate the values analytically. I can start with Vbe =0.8v.
Can't calculate Ib because Vre is not known.
The Is(14fA) in the post above does not give the correct value.

 

[FvM]

I agree about "theoretical value" in two regards:
- an analog designer reviews the Vbe value from the transistor datasheet rather than deriving it from an Is number
- the datasheet also gives an idea of the expectable Vbe type variation by specifying min and max values

Besides referring to SPICE parameters, that aren't always available for transistors, I would suggest the opposite way. Take a typical Vbe value from the datasheet and calculate Is from it, if you want to know it for any reason.

 

[dave9000]

"Is" is used just by simulators, usually you don't need it at all.
You usually must design your circuit in such a way to be independent (as much as possibile) on the variability of the device parameters.

About your circuit: the operating point is totally dependent on the hfe of Q1 --> it's a bad biasing method.
Supposing you know exactly the hfe (for that operating current) you can write:
hfe*(Vcc - (Vbe + Ic*R3))/R1 = (Vcc - Vc)/R2

But, once again, you have very little control of the operating point - try changing hfe from 50 to 500 to see what happen.

P.S. the Spice model from Microsemi says IS = 19.34n: see how relevant this parameter is in practice ;-)

 

[FvM]

About your circuit: the operating point is totally dependent on the hfe of Q1 --> it's a bad biasing method.

Yes, I didn't notice the schematic. If you don't need to achiecve an exact bias point, connecting R1 between base and collector can be a simple solution.

By the way, Is is not related to leakage current.

 

[mister_rf]

The manufacturing process itself has poor absolute tolerances. Key electrical parameters for the bipolar transistor can vary significantly because of manufacturing tolerances. For some parameters, the exact value is not important as long as it satisfies some minimum requirement.

 

[hithesh123]

About your circuit: the operating point is totally dependent on the hfe of Q1 --> it's a bad biasing method.
Supposing you know exactly the hfe (for that operating current) you can write:
hfe*(Vcc - (Vbe + Ic*R3))/R1 = (Vcc - Vc)/R2

Doesn't the emitter degeneration resistor help?

---------- Post added at 09:41 ---------- Previous post was at 09:39 ----------

Coming to the ckt,
How do I determine Ic ?

From the datasheet VBE ON vs Ic graph, I calculated Is as 9.4fA
Just Is is useless. I need Vbe. Should I assume Vbe?

 

[FvM]

Doesn't the emitter degeneration resistor help?

Not at all for the present circuit. It's only useful with a voltage divider bias circuit. But it reduces the voltage gain and increases linearity, which can be a sufficient reason to use it.

For your original circuit as well as for the suggested modified circuit, exact Vbe isn't an issue, assuming 0.7 V is sufficient. Current gain B however is critical and it's only loosely specified for 2N2222. If you require a specific bias point, e.g. Vce = 12 V, you should rather go for a voltage divider bias circuit.

 

[dave9000]

Ic is: (Vcc - Vc)/R2
If you are ANALYSING the circuit then just substitute the expression for Ic in the equation and solve for Vc.
If you are DESIGNING the circuit then:
[1] assume Vbe = 0.7V; the exact value doesn't matter because you will never have any control over it
[2] decide that the operating point will be Vc = V1/2 (maximize the output swing)
Now you know Ic (=V1/R2/2 =~24mA) and Vb (=Ic*R3+Vbe=~1.9V).
What you have to do is to find a way to "fix" Vb to 1.9V.
With your scheme you need to know the hfe, then: R1 = (V1-Vb)/(hfe*Ic).
For hfe=50 --> R1=46K; for hfe=300 --> R1=270k.
As you can see this method is not reliable at all!