First question: Is this academic exercise or for design?
2nd question: If this is for practical design, have you looked at ways to 'stabilize' or control the Ib bias for a desired Ic collector current?
RF_Jim
There are more stabler methods like potential divider method and base feed back method
The potential feed back has theoretically very high stability in real time appliances
Thank you Brad for the detailed design info!
Could anyone comment on the original questions I asked?
I am trying to manually check how adding a base resistor can reduce Ic sensitivty regarding to base terminal bias voltage Vb.
Could anyone comment on the original questions I asked?
Here comes your original question
I am trying to manually check how adding a base resistor can reduce Ic sensitivty regarding to base terminal bias voltage Vb.
Ic=K*exp{[Vb-(Ic/beta)*Rb]/Vt}.
Now my question:
Why do you think that Rb can reduce the Ic sensitivity "regarding to Vb" ?
If you rewrite the last equation you get the following:
Ic=K*exp{Vb/Vt}/exp{(Ic/beta)*Rb/Vt}
I don`t see any sensitivity reduction.
More than that - why do you want such a reduction? More important is a reduction in sensitivity with respect to parameter uncertainties of the transistor.
And that`s what negative feedback can do - either using an emitter resistor Re or a base voltage divider that is driven by the collector voltage rather than the supply voltage.
You actually answered my 1st question. The denominator is always>1, so it is the reduction factor.
Hi Newbie,
up to now I was of the opinion that your desire is a reduction in SENSITIVITY.
Of course, the denominator>1 reduces Ic (for constant Vb). That`s no big surprise because of the voltage drop across Rb.
However - does it reduce the sensitivity of Ic against Vb changes? Increase the numerator in the rewritten equation (post#10) by 10% - and Ic will increase by 10% also.
Hi,
yes, something wrong with my calculation. In case Rb and the voltage drop Ib*Rb exist the calculation of the differential dIc/dVb is not correct since Ib depends also on Ic. It was too simple.
I`ll try a correction.
LvW
- - - Updated - - -
Here comes the correction (I hope no further errors):
S(Ic,Vb)=Vb/(Vt+Rb*Ib).
Please note that for Rb=0 this expression equals the sensitivity figure as given in my former posting#14.
With: Vb=Rb*Ib+0.65V (approximation) the above expression can be simplified to
S(Ic,Vb)=Vb/(Vt+Vb-0.65V).
As can be seen, the sensitivity S approaches unity for very large voltages Vb.
Remark: The simplest way to calculate dIc/dVb was to start with the inverse, which is dVb/dIc=.....= (Vt/Ic)+(Rb/beta)
if we increase Vb to a large number, surely the limit of S will be 1. In reality, Vb is not able to go too far before the device is damaged.
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