BJT Amplifier

[eulyn]

Can anyone give me some guidance? Please show me some formula, thanks.

The transistor of Fig.1 has β=99 and a base current of 35μA.

(a) Name the circuit configuration
- COMMON EMITTER CONFIGURATION

(b) Find

(i) IEQ
(ii) Current gain
(iii) Voltage gain

Assuming leakage current is negligible. State also additional assumptions where necessary.

 

[MrAl]

Hi,

Can you post your circuit in a more common format like gif or jpg.

 

[ericgibbs]

Hi,

Can you post your circuit in a more common format like gif or jpg.

hi Al,
Here you go.

E

 

[MrAl]

Eric:
Thanks for posting the pic in a better format.

Can anyone give me some guidance? Please show me some formula, thanks.

The transistor of Fig.1 has β=99 and a base current of 35μA.

(a) Name the circuit configuration
- COMMON EMITTER CONFIGURATION

(b) Find

(i) IEQ
(ii) Current gain
(iii) Voltage gain

Assuming leakage current is negligible. State also additional assumptions where necessary.

Hi,

First, your circuit shows a common collector circuit (also called a voltage follower) not a common emitter.
The current gain is the ratio of the output current to the input current Ic/Ib which is close to Ie/Ib.
The voltage gain is considered to be 1 or slightly less than 1 because of the configuration.
The quiescent emitter current is Ib*(1+Beta) where Beta is the DC current gain of the transistor.

Since there is an input voltage Vbb, we can also calculate the base current Ib.
Emitter current is Ib*(1+Beta) and base voltage is Vd+(Ib*(1+Beta)*Re), so base current is:
Ib=(Vbb-Vd-Ib*(1+Beta)*Re)/Rb

and solving explicitly for Ib we have:
Ib=(Vbb-Vd)/((Beta+1)*Re+Rb)

where Vd is the base emitter diode drop.

So with Rb=1k, Re=100, Vbb=5, Vd=0.7 and Beta=50 we get about 705ua base current provided that Vcc is high enough to support the voltage across the emitter resistor plus maybe another volt for the transistor collector to emitter. This means we see about 36ma through the emitter resistor, and that means about 3.6v across that resistor.

Checking, we see that the sum of the voltage drops around the emitter base loop equals 5 volts which is the assumed voltage for Vbb.

 

[eulyn]

Sorry, i still don't understand. Given beta is 99 and base current of 35uA, why we still need to calculate the base current? Why your Beta sudden chage to 50? Can you explain more detail? Thanks.

 

[MrAl]

Hi,

That was an example that was meant to show you how to do these problems. But if your base current is given, then you dont have to calculate that.

With a base current of 35ua and a Beta of 99, that means the collector current is 99*35ua and so the emitter current is (99+1)*35ua. The voltage between the emitter and ground then is that current times the emitter resistor Re. You can then calculate Vbb if you like.

Does that help?

 

[eulyn]

Noted, thanks for your explanation.

 

[eulyn]

May I get some advice for this lab? I'm really confuse the theory and it's calculation.

Procedures:

1. Construct the circuit shown in figure 11 on a breadboard carefully.
2. Before connect the function generator to the circuit, perform DC analysis for the entire circuit and record your answers in the Table 1 below.
3. Next, connect the function generator to the circuit.
4. Set the function generator to output a sine wave at 10 kHz and amplitude equal to 60 mVpp. Use a scope to ensure the settings are correct.
5. Turn on the output of the function generator. Measure and record all the AC values in Table 2 below using the scope.
6. Using the measured values of VB(Q1) and VC(Q1), calculate AV(Q1). Insert this value in the measured section of Table 2.
7. Using the measured values of VB(Q2) and VC(Q2), calculate AV(Q2). Insert this value in the measured section of Table 2.
8. Using the measured values of VB(Q1) and VC(Q2), calculate AVT. Insert this value in the measured section of Table 2.



Figure 11 A 2-satge common-emitter amplifier circuit


Questions:
1. Perform DC analysis on circuit in this lab by filling up table 1 below. Be sure to show all your steps and calculations.




Table 1 DC analysis values


2. Perform AC analysis on the circuit in this lab by filling up table 2 below. Be sure to show all your steps and calculations.


Table 2 AC analysis values


3. Determine the values of Zbase for Q2, Zin for Q2 and rc for stage 1. Calculate Av1.


4. Determine the values of rc for stage 2. Calculate AV2.


5. Calculate AVT.


6. Calculate the percentage error by comparing the values of AVT measured vs AVT calculated. Explain discrepancy in the results between measured and calculated.

 

[eulyn]

**broken link removed**

 

[ericgibbs]

May I get some advice for this lab? I'm really confuse the theory and it's calculation

hi,
Have you built the circuit on the breadboard.? If yes, have you measured the voltages as required in Step 3 thru 8.?

Please post your calculated values for the DC and AC analysis, so that we can see what you have done.??

E.

 

[eulyn]

This is the result, now i meet problem at question 4,5, and 6

 

[ericgibbs]

View attachment 73019

This is the result, now i meet problem at question 4,5, and 6

hi,
Your 1st table results are close enough.

On Table 2, Recheck the gain of Q2... remember the input to Q2 is the output of Q1, not 60mVppk.

E.

EDIT: dont forget R11 is in parallel with Q2 collector resistor.

4. Determine the values of rc for stage 2. Calculate AV2

Do you mean re' or the total Resistance load on Q2 collector..???

 

[NANIGESWARAN]

View attachment 73019

This is the result, now i meet problem at question 4,5, and 6

Hi eulyn,
Can u explain how u get answer for DC table 1 and AC table 2.
The circuit is confusing me cause it have 2 sources.

thank you.