What is the impact of the beta dependence on a BJT with voltage divider bias?

[likephysics]

For a BJT with voltage divider bias (but without emitter deneration resistor), what is the beta dependence?
Ib is (Vth-Vbe)/Rth [where, Vth and Rth are thevenin equivalent of the divider)

Ic = β * Ib

The above is almost same as just base bias, where a resistor from base is collected to supply.

The design guideline is usually, the current flowing thru the divider should be 10x the base current. How does that help in this case?

 

[yungman]

It is not good to use voltage divider bias with emitter resistor.You cannot use Rth for calculation. The Vbe is a logarithmic device where

$$V_{BE}=V_T ln (\frac {I_C}{I_S})\;\Rightarrow\; Δ V_{BE}=V_T ln (\frac {I_{C_1}}{I_{C_2}})$$

where Vbe increase about 26mV every time you double the current. You'll burn the transistor before you get the Vbe to even 1V.

Base bias is very different concept. You use a very high value resistor as the base resistor Rb to V+. So the Ib≈ V+/Rb. Then you can get Ie=βIb.

 

[NascentOxygen]

It is not good to use voltage divider bias with emitter resistor.

ITYM "without".

You cannot use Rth for calculation.

He could still use Rth in the calculation. It might be impractical to use a really high value for RB, but feasible to use a potential divider with significant Rth. It may not be quite as stable as a single RB, but it's a poor design anyway so can't be made much worse. Just don't design for the pot divider to waste 10 times the base current, it serves no purpose here, and may still need to include a third resistor to the base.

 

[fleem]

agree with above but throwing my wording in...

For any typical BJT ckt, you almost certainly want a fairly constant bias current going in the base. It wouldn't do for that current to vary wildly with, for example, temperature. When you have a NPN (for example) BJT with grounded emitter, the only reasonable way to bias it is with a single pull-up resistor. This acts as a fairly good current source because we know the drop across that resistor will always be a little less than Vcc. You can also add a little negative DC feedback to help compensate for part variation by connecting that pull up to the bottom of the collector's load, but that's a different story. Anyway, we don't want a voltage divider biasing the base in this ckt because a voltage divider is used to produce more of a voltage source than a current source--that's its purpose-- and we don't want a voltage source we want a current source. Attempting to control the base current in this ckt by trying to manage a constant voltage on the base is bad design because the base-emitter junction voltage varies a lot with temperature. So a constant base voltage will cause base current that varies with temperature, and you have an extra unneeded component, to boot (namely the pull-down resistor).

In any case if you've been asked to solve that ckt, you'll have to know the base-emitter voltage drop for the current temperature quite accurately, especially if the voltage divider's unloaded output voltage is barely above the base-emitter drop voltage. If you've been told the base-emitter drop is a constant (it never is), then you know the voltages on the resistors so you know the current through those resistors so you know how much of the current in the pull-up resistor is going through the pull-down resistor, and whatever is left over is what's going through the base.

 

[yungman]

ITYM "without".

Opps! Finger is getting too old!
He could still use Rth in the calculation. It might be impractical to use a really high value for RB, but feasible to use a potential divider with significant Rth. It may not be quite as stable as a single RB, but it's a poor design anyway so can't be made much worse. Just don't design for the pot divider to waste 10 times the base current, it serves no purpose here, and may still need to include a third resistor to the base.

It's never a good practice to just grounded the emitter and try to bias the BJT anyway, You always want to have emitter resistor and voltage divider at the base to set up a stable bias, then use a capacitor to connect the emitter to the ground and get a AC ground.

It is not good practice to deal with Rth. Refer back to the formula I posted the Vbe change with the log of the current ratio. It is not linear. You can only calculate Rth at a set current.