amplifiertransistors
I need your help
Given:
Find :
a) Rin (TOTAL)
b) Rin (BASE)
c) Av (voltage gain)
I am uncertain of the process.
I have Rin (Base) = Rs + R1||R2 or is it the Rin (total)? If not, where is Rin (total)?
Then I compute for Ib = Vs/Rin(Base)
Substitute Bac = Ic/Ib values to compute for Ic
Then formula for re'=25mV/Ie
Then c) Av = Rc/re'
There are many places we can begin looking at this.
We know that the collector current is 2 mA, and that \$V_{CE}\$ is 5V. Since the emitter is at ground, the collector must be at 5V, and so the voltage across \$R_C\$ is therefore 10V: the difference between 5V and 15V. So \$R_C\$ must be \$10V/.002A = 5K\Omega\$.
Next, \$R_2\$ and \$R_1\$ which span a voltage from \$-15V\$ to \$+5V\$ must form a voltage divider such that the top of \$R_1\$ is at 0.7V (\$V_{BE}\$). Hint: the voltage divider spans a range of 20V, and the 0.7V transistor base voltage is 15.7V above the bottom of the voltage divider.
This approach assumes that we can ignore the base current because it is small. Often when analyzing transistor circuits we can do that, but not in this case because \$R_1\$ is such a high resistor. The voltage divider is not "stiff" at all with regard to the resistance in the base circuit of the transistor (which has no emitter resistor at all).
A more exact answer requires that we account for the base current. The transistor is carrying only 2mA of current, and so is nowhere near hard saturation, and so the base current is only 0.02 mA, or 20 micro-Amperes (2mA divided by \$\beta\$).
Determine how much current is flowing through \$R_1\$ from its resistance, and 15.7 voltage. The current flowing through \$R_2\$ is the sum of the transistor base current (.02 mA) and the current through \$R_1\$. Knowing the current through \$R_2\$ and the voltage across it, we can calculate its resistance.
There are several gains associated with voltage amplifiers. Consider the following model
In this model, the gain \$A_{VO} \$ is the open circuit voltage gain of the amplifier which, in your circuit, is given by \$R_C/r_e\$
But note that the output resistance of the amplifier (which is about \$R_C\$ in your circuit) forms a voltage divider with the load \$R_L\$.
So, the loaded voltage gain is:
\$A_{V} = A_{VO} \dfrac{R_L}{R_{out}+ R_L}\$
But, note that the input voltage \$V\$ is less than the source voltage due to the voltage division between the source resistance and the input resistance of your amplifier. Thus, the loaded gain with respect to the source is:
\$A_{VS} = A_V \dfrac{R_{in}}{R_S + R_{in}} = A_{VO} \dfrac{R_L}{R_{out}+ R_L} \dfrac{R_{in}}{R_S + R_{in}}\$
So, you cannot expect to measure anything close to the open circuit gain \$R_C/r_e \$
Best Answer
In order to compute the given quantities, You must be aware of small signal equivalent circuit of various BJT configurations. A little more experienced users can directly determine these quantities from the given circuit diagram itself.
Here is a commonly used circuit :
It is called as Voltage Divider bias and has an equivalent small signal of
First replace all the Dc supplies by Gnd. They have no role to play in ac analysis except for biasing the BJT in proper region.
Next see that R1 and R2 are both connected with one end tied to base and other end to Gnd. So they appear in parallel.
The small signal resistance called re is connected to the emitter. Its value is Ie / Vt . However, it is 'reflected' to the input side by a factor of (B + 1). The Emitter resistance RE is also reflected similarly. However, here it is bypassed by the capacitor which acts as a short circuit for small ac signals, and hence does not appear in the picture.
The resistance ro is there to model the Early Effect and has large value ( in 10's of K ohm)
As a rule of thumb, it is always best to begin by finding the Emitter current, which depends on your particular biasing scheme. Here however, you are already given its value as 3.8 mA.
Next determine the value of re as Ie / Vt and optionally, transconductance gm, which here equals (1 / re ).
So the base resistance is parallel combo of R1 and R2. Call it Zb (or anything).
The input resistance is ( Zb || B.re ) + Source Resistance Rs.
The Output resistance is ro || Rc . Call it Zo
The voltage gain is (gm.Zo)