# cascade amplifier with common emitter amp and common collector amp

Discussion in 'Electronics Homework Help' started by majdi, Mar 12, 2014.

1. ### majdi

33
0
Jul 10, 2010 I need to find IE1, IE2, re1, re2 and overal voltage gain.

First i should open all capacitor first right?. it should be like this: How possible i can calculate since there dont have enough value in this circuit? like Ic or Ib.

2. ### Harald KappModeratorModerator

11,522
2,654
Nov 17, 2011
Start by noting that for each transistor IE=(1+beta)*Ib
Set up the equations applying Kirchhoff's laws and solve for the operating point.

3. ### majdi

33
0
Jul 10, 2010
Here my final draw: IC1 = IB2

IC1 = VR1/R1

VR1 = Vcc - (Vceq1+VRE)

----------------------------------------------
or

Vceq2 = Vcc - (VR4 + VR5)

I should find IB2 first..let me think........

4. ### Harald KappModeratorModerator

11,522
2,654
Nov 17, 2011
IC1 <> IB1
IC1 flows through the collector of Q1, not into the base of Q2.
IC1+IB2 = VR1/R1
...

5. ### majdi

33
0
Jul 10, 2010
IE1 = (VCC – VCE) / (R1 + R2 + R3)
= 12v – 0v / 20k + 2k + 8k
= 12v / 30k
= 0.4mA

Let assume VCE Q1 = 6.0v

IE1 = (VCC – VCE) / (R1 + R2 + R3)
= 12v – 6v / 20k + 2k + 8k
= 6v / 30k
= 0.2mA

VE Q1 = IE1 x (R2+R3)
= 0.2mA x 10k
= 2v

VB Q1 = VE1 + VBE1
= 2v + 0.7v
= 2.7v

VB Q2 = VCC – (R1 x IE1)
= 12v – 4v
= 8v

VE Q2 = VB Q2 – VBE Q2
= 8v – 0.7v
=7.3v

VR5 = ½ x VE Q2
= 3.65v

*VR5 should be near to VB Q1 = 2.7v, this show IE1 is too low, we need to increase it. Try IE1 = 0.3mA

VE Q1 = IE1 x (R2+R3)
= 0.3mA x 10k
= 3v

VB Q1 = VE1 + VBE1
= 3v + 0.7v
= 3.7v

VB Q2 = VCC – (R1 x IE1)
= 12v – 6v
= 6v

VE Q2 = VB Q2 – VBE Q2
= 6v – 0.7v
=5.3v

VR5 = ½ x VE Q2
= 2.65v

*VR5 should be near to VB Q1 = 2.7v, this show equal to VB Q1 to low. Try with IE1 = 0.25mA

VE Q1 = IE1 x (R2+R3)
= 0.25mA x 10k
= 2.5v

VB Q1 = VE1 + VBE1
= 2.5v + 0.7v
= 3.2v

VB Q2 = VCC – (R1 x IE1)
= 12v – 5v
= 7v

VE Q2 = VB Q2 – VBE Q2
= 7v – 0.7v
=6.3v

VR5 = ½ x VE Q2
= 3.15v

*This result is close for VB Q1.

Let IE1 = 0.24mA

VE Q1 = IE1 x (R2+R3)
= 0.24mA x 10k
= 2.4v

VB Q1 = VE Q1 + VBE Q1
= 2.4v + 0.7v
= 3.1v

VC Q1 = VCC – (IE1 x R1)
= 12v – 4.8v
= 7.2v

VB Q2 = VC Q1
= 7.2v

VE Q3 = VB Q2 – VBE Q2
= 7.2v – 0.7v
= 6.5v

IE2 = VE Q2 / (R4 + R5)
= 6.3v / 2k
= 3.15mA

So: IE1 = 0.25mA and IE2 = 3.15mA.

re1 = VT / IE1
= 26mV / 0.25mA
= 104 Ohm.

re2 = VT / IE2
= 26mV / 3.15 mA
= 8.25 Ohm.

IB1 = IE1 / Beta
= 0.25mA / 100
= 0.0025mA

IC1 = 1+Beta x IB1
= 101 x 0.0025
= 0.253mA

Av1 = IC1xRC / IE1 x re1
= 0.253mA x 20k / 0.25mA x 104 Ohm
= 5.06v / 26mV
= 0.19 v

Av2 = 12v / IE2 x re2
= 12v / 3.15mA x 8.25 Ohm
= 12v / 25.99mV
= 0.462 v

6. ### Harald KappModeratorModerator

11,522
2,654
Nov 17, 2011
Why should this be so?
What about IB2 which also flows through R1?

What makes you think so? Why not 5V or 7V?

Some equations to work with:
IE1=(1+beta)*IB1
IE2=(1+beta)*IB2
voltage across R5: (IE2-IB1)*R5=IB1*R6+VBEQ1+IE1*(R2+R3)
...
I hope that helps.  