# Does Ic = (Hfe)Ib?

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#### Megamox

##### New Member
Hi All,

I've probably made a really silly mistake somewhere but I'm trying to work out an expression for the self biasing of this transistor circuit above so that the output always sits half way between the supply rail. I applied KCL to the node and got an expression above. I put the values in the expression which yielded a value for me to use. To test the expression I set up the circuit underneath with Vs = 9v and a 2n222 (Hfe = 200). However the simulation shows the values do not quite give an output which is half way below the rail. So to investigate further I worked out the base and collector currents in the circuit being simulated and see that Ic does not equal Hfe(Ib). Ic seems to be higher than it should be. Just wondering if I've missed something glaringly obvious? What's gone wrong!

Thanks,
Megamox

#### crutschow

##### Well-Known Member
You never want to bias a transistor with that circuit since it depends upon the value of Hfe, which varies with temperature and from unit to unit. The value of 200 is just a typical value. Better to add an emitter resistor to stabilize the bias point.

You simulation may be off because the simulator uses a different value for Hfe than you assumed.

#### MikeMl

##### Well-Known Member
Moreover, the expression Ic = (Hfe)Ib is only a crude approximation, that works for a narrow range of Vce, Ic, and temperature.

#### Megamox

##### New Member
Thanks guys. Yes absolutely, the Hfe dependence does make it limited. I just thought I'd play with the design and found it odd that I got a slight error. I've had a look at the circuit parameters and notice that Vbe is quoted as 0.75v which will obviously slightly effect calculations. But Hfe is quoted at 200.

Perhaps I need to go over some semiconductor theory to look at the original function of Ic and it's dependance on Ib which can be approximated to Ic = Hfe(Ib). The interesting thing is, If you look at the original biased circuit and were to assign a 'Hfe' to it, it would be about 208.8/0.987 =~ 211. If I rework the calculations for a Vs = 3v, the 'Hfe' comes out to 229. So not only is Hfe behaving as if it were above 200, but it's actually going up, the lower you make Vs. Ie the lower you make Vs, the more error introduced. I'm so used to applying Ic = hfe(Ib) that I guess I forget it's only an approximation! I guess sometimes you really cant design everything on paper first, you've got to tweak things in practise.

Megamox

#### crutschow

##### Well-Known Member
And you never design a circuit based upon the "typical" value of Hfe. It needs to be able to operate properly with the minimum value.

#### Megamox

##### New Member
That's right, I think the art of circuit design in some respects is designing circuitry with predictable output, which involves making it independent of component parameters and characteristics.

Megamox

#### Winterstone

##### Banned
Quote Chrutschow:You never want to bias a transistor with that circuit since it depends upon the value of Hfe, which varies with temperature and from unit to unit. The value of 200 is just a typical value. Better to add an emitter resistor to stabilize the bias point.

I am not sure if it is appropriate to be so rigorous against megamoxs first circuit (with one single bias resistor). Rather, I think to select a suitable bias scheme strongly depends on the application and on the consequences the bias circuitry has (input resistance, lower frequency limit, linear amplification range,...). Simulations show that this bias method is not as bad as it looks at first sight.

Quote megamox: That's right, I think the art of circuit design in some respects is designing circuitry with predictable output, which involves making it independent of component parameters and characteristics.

I am sure, you know that it is not possible to design a transistor stage that independent of component parameters and characteristics .
The art of good circuit design is always a trade-off between requirements, performance and expense (cost).

#### Winterstone

##### Banned
Sorry - I couldnt figure out why this post appears twice.

Quote Chrutschow:You never want to bias a transistor with that circuit since it depends upon the value of Hfe, which varies with temperature and from unit to unit. The value of 200 is just a typical value. Better to add an emitter resistor to stabilize the bias point.

I am not sure if it is appropriate to be so rigorous against megamoxs first circuit (with one single bias resistor). Rather, I think to select a suitable bias scheme strongly depends on the application and on the consequences the bias circuitry has (input resistance, lower frequency limit, linear amplification range,...). Simulations show that this bias method is not as bad as it looks at first sight.

Quote megamox: That's right, I think the art of circuit design in some respects is designing circuitry with predictable output, which involves making it independent of component parameters and characteristics.

I am sure, you know that it is not possible to design a transistor stage that independent of component parameters and characteristics .
The art of good circuit design is always a trade-off between requirements, performance and expense (cost).

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#### Winterstone

##### Banned
Moreover, the expression Ic = (Hfe)Ib is only a crude approximation, that works for a narrow range of Vce, Ic, and temperature.

In this context, it is perhaps helpful to give a short explanation to my claims in post #7:

*It is true that the current gain Hfe (and it´s uncertainties) enters directly into the calculation of the bias resistor. At the same time, the base voltage Vbe plays only a minor role (if compared with the major influence of Vce).
* The situation regarding Hfe apparently looks better for the classical bias scheme with a voltage divider at the base and an emitter resistor Re. Now the value of Hfe plays a minor role only during calculation of the base resistors. However, now the voltage Vbe (and its uncertainties) enters directly into the calculation . And in this context, one shouldnt forget that the value of Vbe also is only a rough estimate (0.7V...0.75V...0.8V) and the real value, which is not known, depends also on Hfe.
Remember: The Ic=f(Vbe) characteristic is derived from the input characteristic Ib=f(Vbe) simply by multiplication with Hfe - and the function also is temperature dependent.

W.

#### Megamox

##### New Member
Yes Indeed that's a far better description Winterstone! Having seen this design being used in various circuits, I thought it had some advantages and I certainly wouldn't mind using it for some things. It would probably be an entirely incorrect choice for other applications too. I think it might be helpful for me to look up how and why the h-parameters are defined for the BJT so I can use them more appropriately. It does indeed look like h-parameters are defined for a subset of environmental and operating conditions and need to be recalculated for new configurations. Whatever the standard 'environment' is for a transistors hfe, or the window of where the relationships hold to be true, I assume my multimeter implements it as it generally gives the same ball park figure for the gain as quoted on datasheets.

Megamox

Edit: I wonder if a hot multimeter gives a different value for hfe than a cold one?

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#### Winterstone

##### Banned
It does indeed look like h-parameters are defined for a subset of environmental and operating conditions and need to be recalculated for new configurations.

Yes - of course. The h-parameters are given for the common emitter configuration and a specific operating point (mostly Vce=5V, Ic=2mA) for normal ambient temperatures.
There is a set of curves,which allow the adaptation to another collector current and/or another voltage Vce.
But one shouldn`t forget that these parameters are connected with rather large tolerances.

#### MOR_AL

##### New Member
Hfe is equal to Ic/Ib by definition (if Icb0 = 0).
Your estimate Hfe isn’t the right value. For your first circuit a more appropriate Vbe value would be something like 0,6V.
For low currents you have to include Icb0 (Ic = Hfe Ib + (Hfe + 1) Icb0), but normally I think it isn’t necessary.
If an emitter resistor isn’t appropriate, you can use your first circuit. It depends of Vbe, but all other circuits also depend of Vbe on some degree.
The collector to base resistor includes a negative feedback as emitter resistor does. Imagine if Vce reduces. Ib also reduces and Ic tends to maintain. Its variation is less than if Rb was connected to Vcc.
As you see, Hfe isn’t a stable value with temperature and also has a spread value. Those calculations are useful to get the approximate value of your circuit. In practice you have to consider a difference of 10% to 15% from calculated to practical value. If you want to reduce this difference using transistors you have to include temperature effect and consider equations as dVcc/dT (dVcc/dIc * dIc/dVbe * dVbe/dT) (T is temperature) and dIc/dT (dIc/dVbe * dVbe/dT). You will see that those variations will be reduced by a factor of Hfe + 1.

dIc/dVbe = - Hfe / (Rb + Re(Hfe + 1)) (Rb to Vcc).
dIc/dIcb0 = (Rb + Re) / [Re + Rb (1 – alfa)] (alfa = Hfe / (Hfe + 1) = = 1 and Rb to Vcc).
dVbe/dT from datasheet.

From your first circuit: Rb from base to collector.

dIc/dVbe = Hfe / [(Hfe + 1) Rc + Rb + (Hfe +1) Re] (If you don’t include Re, make Re = 0)
dVbe/dT from datasheet.

Finally.
I calculate those equations (in 1974) and I am quite sure they are correct. You have to calculate as an exercise.

Sorry my English.

MOR_AL

#### Claude Abraham

##### Member
Ic = alpha*Ie is the best description of bjt operation. A circuit should be designed around alpha value, not beta. The beta value is related to alpha per alpha = beta/(beta+1). But there are cases where we can force a value of Ib & get a predictable Ic value. When the bjt is used as a saturated switch we drive the base with a value around one tenth the desired collector current. If Ib = Ic/10, the bjt is saturated and Ic is predictable & fixed for any bjt with a beta value well above 10.

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#### audioguru

##### Well-Known Member
Cheep (cluck, cluck) circuits use a single resistor to bias the transistor that has no emitter resistor. Then they simply throw away all the circuits that DO NOT WORK!

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