In the partial pressure of the bias circuit in dc circuit, the book says the current on the bias res-CodePudding

In the partial pressure of the bias circuit in dc circuit, the book says the current on the bias resistors R2 I2 is far greater than the base current IB, namely I2 & gt;> Could you tell me how this is concluded, the IB?
My understanding is triode body resistance is small, I1 shunt should be mostly distributed to the Ib, didn't want to know why I1=I2 + Ib, can be roughly as I1=I2,

For bosses to reassure

CodePudding user response:

Ic=beta Ib, beta are large

CodePudding user response:

refer to 1st floor core electrical world response:

Ic=beta Ib, beta is generally much

But this discussion is I1 shunt under static condition,,
Want to know is

Ic=beta Ib, this no problem,
Calculated according to I1=I2,
I1, I2==133 ua, Ub=1.3 V, Ue=0.6 V, the Ie=0.6/1=0.6 k mA
And static conducting Ic=[Vcc - Ue]/Rc=7.2 mA,
So it is not in conformity with the Ie=(1 + beta) Ib,
And suppose beta for 100 times, back out of the Ib uA level,
So, I want to know why I1=I2 + Ib can approximate the Ib remove

CodePudding user response:

refer to the second floor stone_qaq response:

Quote: refer to 1st floor core electrical world response:

Ic=beta Ib, beta is generally much

But this discussion is I1 shunt under static condition,,
Want to know is

Ic=beta Ib, this no problem,
Calculated according to I1=I2,
I1, I2==133 ua, Ub=1.3 V, Ue=0.6 V, the Ie=0.6/1=0.6 k mA
And static conducting Ic=[Vcc - Ue]/Rc=7.2 mA,
So it is not in conformity with the Ie=(1 + beta) Ib,
And suppose beta for 100 times, back out of the Ib uA level,
So, I want to know why I1=I2 + Ib can approximate the Ib remove

Static conduction, not in a fully conducting state, should calculate the Ic and Uce pressure drop, this section pressure drop size is changing with the Ib size

CodePudding user response:

The Ib=I2 - I3 (I2 Rb current, I3 rb2’s much-publicised branch current, Rb base current)

With Vb=(beta + 1) * (I2 - I3) * Re + Vbe varies=rb2’s much-publicised * I3

(beta + 1) Re I2 - (beta + 1) Re I3 + Vbe varies=rb2’s much-publicised * I3

I2=I3 * [(beta + 1) Re + rb2’s much-publicised]/[Re] (beta + 1) - Vbe varies//(beta + 1) Re

According to you the parameters of the circuit, and assuming beta=100:
I2 is 15 v/110 k=0.13 mA
Vbe varies/[Re] (beta + 1)=6.9 uA
+ 1) and (beta Re> Rb2’s much-publicised so [(beta + 1) Re + rb2’s much-publicised] is approximately equal to + 1 (beta) Re (numerical brought in to verify yourself)
So I2=I3 ~