base resistor to reduce sensitivity of Ic vs. Vb

I am trying to manually check how adding a base resistor can reduce Ic sensitivty regarding to base terminal bias voltage Vb.

Since it is a simple common emitter BJT transistor (emitter directly to ground, collector directly to Vcc),

So Ic=K*exp(Vbe/Vt)

where Vbe=Vb-(Ic/beta)*Rb

Therefore, Ic=K*exp{[Vb-(Ic/beta)*Rb]/Vt}.

Now Ic (which is Y) vs. Vb (which is X) is a transcendetal equation that cannot be easily solved.

How do I evaluate the Ic sensitivity with regard to Vb, without resorting to a simulation tool?

Another question is how much Rb is really needed to stabilize Ic?

Any comments or insight is much appreciated.

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

RF_Jim said:

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

Click to expand...

Thanks Jim. I am interested in both aspects. I haven't looked at other ways to stabilize the Ib for a desired Ic. Any recommendation is very much appreciated!

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

jeffrey samuel said:

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

Click to expand...

Is potential divider the resistor divider approach (from Vcc, with large R values)? not sure what do you mean by base feedback?

Base feed back is just using a feed back resistance to feed the collector current through a high resistance to the base junction

And potential divider bias Fromm Vcc using couple of resistors one large and another small in series connection the base junction is connected to the point where the two resistors meet

For a visual aid, here is a demonstration of 3 biasing methods, with scope traces.



The supply ranges between 2 and 16 V. The upper range of the sweep would dissipate a lot of power and make the transistor heat up.

Thank you Brad for the detailed design info!

Could anyone comment on the original questions I asked?

newbie_rf said:

Thank you Brad for the detailed design info!

Could anyone comment on the original questions I asked?

Click to expand...

I am trying to manually check how adding a base resistor can reduce Ic sensitivty regarding to base terminal bias voltage Vb.

Click to expand...

The typical B-E operating range is just a few tenths of a volt.

Example, if you apply 0.35 V to the bias (referenced to emitter) you'll get miniscule current (through C-E)...

whereas if you apply 0.75 V, you'll get very high current.

newbie_rf said:

Could anyone comment on the original questions I asked?

Click to expand...

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.

LvW said:

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.

Click to expand...

You actually answered my 1st question. The denominator is always>1, so it is the reduction factor. In semiconductor, there are often some bare devices for process monitoring. When testing these, direct base voltage is used. Knowing its current vs. Vb will help to reduce the chance to burn the device. So base or emitter resistance is needed sometimes. Hope this is clear to you as to why I was asking those questions. Resistor biasing network is nice, but it is not always desirable for low noise, or process monitoring.

newbie_rf said:

You actually answered my 1st question. The denominator is always>1, so it is the reduction factor.

Click to expand...

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.

LvW said:

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.

Click to expand...

no, it is not a simple linear relationship. Note that when numerator increases by 10%, denominator will increase to a certain percentage too. Compared to the case where Rb=0, there is quite some reduction of sensitivity.
Thank you for your involvement in this discussion.

Newbie_rf - OK, now I know what you mean and you are right. There was a misunderstanding since in your first post you spoke about the "base terminal bias voltage Vb".
In fact, Vb is not the voltage at the base terminal but the bias voltage applied at the left side of Rb, right?

- - - Updated - - -

The answer to your original question is twofold: Are you interested in the "absolute" or in the "relative" sensitivity ?

* Absolute sensitivity: Your qualitative assumption is realistic: A variation of let`s say (+-) 10mV around Vb has less influence on Ic if Vb is large because of a large value of Rb (between Vb and the base node).
* Relative sensitivity: This is a classical quality figure for many circuits (in particular for active filters to see how much parts tolerances influence the pole data).
It is defined (for our case) as:

(dIc)/Ic=S(Ic,Vb)*dVb/Vb with S=Sensitivity of Ic versus Vb.
In words: The figure S determines how much a relative change in Vb will cause a relative change in Ic.

Thus: S(Ic,Vb)=(Vb/Ic)*dIc/dVb

It is not a problem to differentiate the expression for Ic - and the result is

S(Ic,Vb)=Vb/Vt.

As you can see, you have the choice which sensitivity (absolute or relative) is important for you.

dIc/dVb=1? my math is getting rusty, could you show why please?
Thank you very much.

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

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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)

LvW said:

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)

Click to expand...

I really like your update and remark. I went through what you did and was able to verify its correctness.

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.
For the Rb=0 case, even relative sensitivity is dependent on Vb. As Vb increases, Ic is still sort of following the exponential curve. Therefore, sensitivity is larger in the higher Vb region than the lower Vb region.
With Rb added, the relatively sensitivy is reduced compared to no Rb case for the same value of far end external Vb.

The inverse calculation idea is just brilliant.

newbie_rf said:

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.

Click to expand...

I think, the sensitivity concept and the definition of the sensitivity figure S makes sense for a constant Ic only (since we are interested in the influence of Vb fluctuations on a certain value of Ic).
That means: If Vb is increased the value of Rb must be increased at the same time (to keep Ib and Ic constant). Thus, there will be no danger of damage.
Finally, for Rb and Vb approaching infinity we arrive at an idealized current source Ib - and the sensitivity figure S(Ic,Vb) merges into S(Ic,Ib), which proves to be unity for beta=constant.