Cascode amplifier. Task 1. Ucc

V1 R1

RC Out

UB Q1 Q1 C1 Cin

UCE Q1 Q2

R2 UB Q2 Q2

+

UE Q2 In R3 RE2

CE

Calculation of the DC values. URE is chosen to be 2V U 2 RE = RE ⇒ = 1,66KΩ ≈ 1,5KΩ E12 IC 1,2m U RE = RE * IC ⇒ 1,5K * 1,2m = 1,8V The transistors is chosen to be BC547A the hFE is 463 this value is from Tina. U Basic of Q2 = U E + U BE = 1,8 + 0,7 = 2,5V I B Q2 =

I RE 1,2m ⇒ = 2.59µA h FE 463

I B Q2 = rufly 10% of I R3 I B Q2 * 100 2.59µ * 100 ⇒ = 25,9µA 10 10 U 2,5 R3 = R3 = = 96,5KΩ ≈ 100KΩ E12 I R3 25,9µ I R3 =

I R3 =

U R3 2,5 = = 25µA R3 100K

UC Q1 chosen to 10V Ucc − U C Q1 15 − 10 RC = = = 4,2KΩ ≈ 4,7KΩ E12 IC 1,2m

U C Q1 = Ucc − (I C * RC ) ⇒ 15 − (1,2m * 4,7K ) = 9,36V

I R2 = I R3 + I B Q2 ⇒ 25µ + 2,59µ = 27,59µA U CE = Ucc − (U RC + U RE ) ⇒ 15 − ((15 − 9,36) + 1,8) = 7,56V U CE 7,56 ⇒ = 3,78V 2 2 U B Q1 = (U C Q1 − U CE Q1 ) + U BE Q1 ⇒ (9,36 − 3,78) + 0,7 = 6,28V U CE Q1 and Q2 =

R2 =

(U

− U B Q2 ) (6,28 − 2,5) ⇒ = 137KΩ ≈ 139KΩ E12 I R2 27,59µ

B Q1

U B Q1 = U B Q2 + (I R2 * R2 ) ⇒ 2,5 + (27,59µ * 139K ) = 6,33V I B Q1 = rufly 10% of I R2 I R2 *10 27,59µ *10 ⇒ = 2,759µA 100 100 I R1 = I R2 + I B Q1 ⇒ 27,59µ + 2,759µ = 30,35µA I B Q1 =

R1 =

(Ucc − U ) ⇒ (15 − 6,33) = 285KΩ ≈ 270KΩ B Q1

I R1

30,35µ

E12

Simulation of the DC values in Tina shows that the DC values all most is correct. Ucc 15V

V1 15V R1 270k

RC 4.7k Out 10.89V

UB Q1 5.74V Q1 BC547A C1 1uF Cin 1uF

R2 139k

UCE Q1 Q2 5.1V

UB Q2 1.97V

+

Q2 BC547A UE Q2 1.33V In R3 100k RE2 1.5k

CE 1uF

Calculation of the AC values. The gain is 30 and the Uout is ±600mV from that the Uin can be calculated Uout AU = Uin AU = 30gg Uout = ±600mV Uout ± 600m Uin = ⇒ = ± 20mV AU 30 To get a bigger gain in the AC mode the RE1 is added.

V1 15 R1 270k

RC 4.7k Out

T1 BC547A C1 1u

R2 139k

C2 1u +

T2 BC547A In R3 100k

RE1 130

RE2 1.3k

RE1. re ≈

1 1 ⇒ = 20,83Ω 40 * I E 40 * 1,2m

RC RC RC ⇒ = re + RE1 ⇒ − re = RE1 AU re + RE1 AU RC 4,7 K RE1 = − re ⇒ − 20,83 = 135,8Ω ≈ 130Ω E 24 AU 30 RE = RE1 + RE 2 ⇒ RE 2 = RE − RE1 AU =

RE 2 = RE − RE1 ⇒ 1,5 K − 130 = 1,37 K ≈ 1,3K E 24

C4 10u

The AC gain for the cascode amplifier. The input and the output show a gain with RE1 at 130ΩE24 and RE2 at 1.3KΩE24: 580,23mVpeak Uout ⇒ = 29,01gg AU = Uin 20mVpeak 20.00m

T

In

-20.00m 580.23m

Out

-580.23m 0.00

500.00u Time (s)

1.00m

The input and the output shows a gain of with RE1 at 135,8Ω and at RE2 = 1,5K − 135,8 = 1,36KΩ : 557,19mVpeak Uout AU = ⇒ = 27,86gg Uin 20mVpeak T

20.00m

In

-20.00m 557.19m

Out

-557.19m 0.00

500.00u Time (s)

1.00m

To get the exactly output of the cascade amplifier the optimiser tool in Tina can be used to get the exactly value of RE1 and RE2, RE1 is 123,83Ω and RE2 is 1,38KΩ T

20.00m

In

-20.00m 599.99m

Out

-599.99m 0.00

AU =

599,99mVpeak Uout ⇒ = 29,99gg Uin 20mVpeak

500.00u Time (s)

1.00m

When the load resistor is added on the output, a buffer is needed (Q3) because a load resistor of 100Ω will destroy the gain in the cascode amplifier. R4 is chosen to be 39KΩE12 but when the buffer is added the value of RC is changed, RC and the resister value of Q3 is in parallel, and the buffer circuit Q3 is attenuating the final signal of the cascade amplifier and there for the total value of RC is getting smaller and therefore the value of RE1 has to be change. But there is no good way to calculate the value of RE1 with the buffer transistor added to the circuit, so the optimizing tool in Tina is used to get the value of RE1. V1 15V

RC 4.7k

Out 1

R1 270k Q3 BC547A C3 1uF

Out 2

Q1 BC547A C1 1uF

R2 139k

R4 39k

RL 100

Cin 1uF +

Q2 BC547A In R3 100k

RE1 33.92

RE2 1.46k

CE 10uF

20.00m

T

In

-20.00m 1.21 Out 1

-1.21 600.03m Out 2

-600.03m 0.00

500.00u Time (s)

If the circuit is to be build the resistor values is showed below. RE1 = 33,92Ω ≈ 33Ω E12 RE2 = 1,46KΩ ≈ 1,47KΩ E48

1.00m

Task 2. Now the RC resistor is to be changes with a LC circuit so the cascode amplifier be come a Tuned Amplifier with a centre frequency at 470KHz and a bandwidth on 20KHz. LC circuit.

L Rp

C

Rs

The resistor Rs in series with the coil is there because the coil is are short in DC and the capacitor is an infinity resistor. The value of Rs shall be somewhere below 500Ω. Rp is the total resistor of the LC circuit and is not a visible resistor but is used in the calculations of the LC circuit. f C = 470KHz BW = 20KHz Q=

ωL f = C Rs BW

Rp = Q 2 * Rs

Rs =

ωL Q

L is chosen to be 1mH 470 K f = 23,5 Q= C ⇒ 20 K BW 2 * π * fC * L 2 * π * 470 K *1m ωL ⇒ ⇒ = 125,66Ω ≈ 120Ω E12 Rs = 23,5 Q Q Rp = Q 2 * Rs ⇒ 23,52 * 125,66 = 69,39 KΩ 23,5 Q ⇒ = 114,67 pF C= 2 * π * f C * Rp 2 * π * 470 K * 69,39 K Rp = Q 2 * Rs ⇒ 23,52 * 120 = 66,27 KΩ 23,5 Q C= ⇒ = 120 pFE12 2 * π * f C * Rp 2 * π * 470 K * 66,27 K

Test of the LC circuit with the calculated values. R2 1M

out

+

VG1 L 1mH C 114.67pF Rs 125.66

T

-20.00 470KHz / -23,90dB

Gain (dB)

460KHz / -26.9dB

480KHz / -26,9dB

-49.01

-78.02 10.00

10.00k Frequency (Hz)

10.00M

Test of the LC circuit with the values from the E-series. R=120Ω C=120pF T

-20.00 459KHz / -23,79dB

Gain (dB)

449,3KHz / -26,79dB

469,7KHz / -26,73dB

-49.21

-78.42 10.00

10.00k Frequency (Hz)

The fC is 11KHz from it’s place with the values from the E-series.

10.00M

The final circuit with the LC circuit as a replacement for the RC resistor. V1 15V L1 1mH R1 270k

C5 114.67pF

R7 125.66

out 1 T3 BC547A C3 1uF

Out

T1 BC547A C1 1uF

R2 139k

R4 39k

R5 100

C2 1uF +

T2 BC547A In R3 100k

R6 33.92

RE 1.46k

T

43.03

C4 10uF

fc 445KHz / 43dB

f1 408KHz / 40dB

Out

-96.46 47.78

f2 486KHz / 40dB

fc 445KHz / 47,77dB f1 408KHz / 44,79dB

f2 485.8KHz / 44,74dB

out 1

-32.41 1.00

3.16k Frequency (Hz)

Out. BW = f 2 − f 1 ⇒ 486 K − 408 K = 78KHz 445K f = 5,7 Q= C ⇒ 78 K Bw

10.00M

To get the correctly fC at 470KHz the optimizing tool in Tina bee used. V1 15V L1 1mH R1 270k

C5 99.8pF

R7 125.66

out 1 T3 BC547A C3 1uF

Out

T1 BC547A C1 1uF

R2 139k

R4 39k

R5 100

C2 1uF +

T2 BC547A In R3 100k

R6 33.92

RE 1.46k

C4 10uF

C5=99,8pF T

43.28 fc 470KHz / 43,24dB

f1 432KHz / 40,23dB

Out

-96.46 48.03

f2 518KHz / 40,26dB

fc 470KHz / 48dB f1 441,3KHz / 46dB

f2 507KHz / 46dB

out 1

-32.41 1.00

3.16k Frequency (Hz)

Out. BW = f 2 − f 1 ⇒ 518K − 432 K = 86 KHz 470 K f Q= C ⇒ = 5,47 86 K BW

10.00M