# Objective. To design and simulate a cascode amplifier circuit using bipolar transistors.

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1 ascode Amplifier Design. Objective. o design and simulate a cascode amplifier circuit using bipolar transistors. Assignment description he cascode amplifier utilises the advantage of the common-emitter and common-base circuits. his type of amplifier has been used in high-frequency applications. 1. Design a cascode amplifier stage using bipolar transistors to drive a load resistor of 100Ω. he amplifier output is to be time varying signal of ±600mV. he overall performance of the amplifier is specified as: Operating current for cascode stage collectors = 1,mA Overall gain = -30 D power supply 15V Showing clearly any design considerations, design and simulate the amplifier circuit.. An extension to the design is that the amplifier should be a uned Amplifier with a centre frequency of 470KHz and a Bandwidth of 0KHz. Modify the original circuit to meet this requirement and simulate the modified circuit.

2 ascode amplifier. ask 1. cc V1 R1 R Q1 1 in R B Q R3 Q RE E Q1 Q E Q E alculation of the D values. RE is chosen to be V RE RE = = 1,66KΩ 1,5KΩ 1,m = RE * 1,5K *1,m = 1,8V RE he transistors is chosen to be B547A the h FE is 463 this value is from ina. Basic of Q = E BE = 1,8 0,7 =,5V RE 1,m B Q = =.59µA h FE 463 B Q = rufly 10% of R3 B Q *100.59µ*100 R3 = = 5,9µA R3,5 R3 = = = 96,5KΩ 100KΩE1 R3 5,9µ R3,5 R3 = = = 5µA R3 100K E1

3 Q1 chosen to 10V cc R = R R1 Q1 E = E Q1and Q R = R1 = R3 ( * R) 15 ( 1,m * 4,7K) ( ) 15 (( 15 9,36) 1,8) = ( Q1 E Q1) BE Q1 ( 9,36 3,78) ( ) ( 6,8,5) R = = cc = cc = B Q B Q R ( * R),5 ( 7,59µ*139K) ( cc ) ( 15 6,33) R1 Q1 = R E B Q R rufly 10% of = = 4,KΩ 4,7KΩ 1,m 5µ,59µ = 7,59µA R RE 7,56 = 3,78V 7,59µ R *10 7,59µ*10 = =,759µA = 7,59µ,759µ = 30,35µA 30,35µ E1 = 9,36V = 137KΩ 139KΩ = 85KΩ 70KΩ = 7,56V 0,7 = 6,8V E1 E1 = 6,33V Simulation of the D values in ina shows that the D values all most is correct. cc 15V V1 15V R1 70k R 4.7k 10.89V 5.74V Q1 B547A 1 1uF in 1uF R 139k B Q 1.97V R3 100k E Q1 Q 5.1V Q B547A E Q 1.33V RE 1.5k E 1uF

4 alculation of the A values. he gain is 30 and the out is ±600mV from that the in can be calculated out A = in A = 30gg out = ± 600mV out ± 600m in = = ± 0mV A 30 o get a bigger gain in the A mode the RE1 is added. V1 15 R1 70k R 4.7k 1 B547A 1 1u R 139k 1u B547A R3 100k RE1 130 RE 1.3k 4 10u RE1. 1 re 40* E 1 = 0,83Ω 40 *1,m R R R A = = re RE1 re = RE1 re RE1 A A R 4,7K RE1 = re 0,83 = 135,8Ω 130ΩE 4 A 30 RE = RE1 RE RE = RE RE1 RE = RE RE1 1,5 K 130 = 1,37K 1,3 K E 4

5 he A gain for the cascode amplifier. he input and the output show a gain with RE1 at 130Ω E4 and RE at 1.3KΩ E4 : out 580,3mV A = peak = 9,01gg in 0mV peak 0.00m -0.00m 580.3m m u 1.00m ime (s) he input and the output shows a gain of with RE1 at 135,8Ω and at RE = 1,5K 135,8 = 1,36KΩ : out 557,19mV A = peak = 7,86gg in 0mV peak 0.00m -0.00m m m u 1.00m ime (s) o get the exactly output of the cascade amplifier the optimiser tool in ina can be used to get the exactly value of RE1 and RE, RE1 is 13,83Ω and RE is 1,38KΩ 0.00m -0.00m m m out 599,99mV A = in 0mV peak u 1.00m ime (s) peak = 9,99gg

6 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 39KΩ E1 but when the buffer is added the value of R is changed, R 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 R 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 ina is used to get the value of RE1. V1 15V R 4.7k 1 R1 70k Q3 B547A Q1 B547A 3 1uF 1 1uF R 139k R4 39k RL 100 in 1uF Q B547A R3 100k RE RE 1.46k E 10uF 0.00m -0.00m m m u 1.00m ime (s) f the circuit is to be build the resistor values is showed below. RE1 = 33,9Ω 33Ω E1 RE = 1,46KΩ 1,47KΩ E48

7 ask. Now the R resistor is to be changes with a L circuit so the cascode amplifier be come a uned Amplifier with a centre frequency at 470KHz and a bandwidth on 0KHz. L circuit. L Rs Rp he resistor Rs in series with the coil is there because the coil is are short in D and the capacitor is an infinity resistor. he value of Rs shall be somewhere below 500Ω. Rp is the total resistor of the L circuit and is not a visible resistor but is used in the calculations of the L circuit. f = 470KHz BW = 0KHz ωl f Q = = Rs BW ωl Rp = Q * Rs Rs = Q L is chosen to be 1mH f 470K Q = = 3,5 BW 0K ωl * π * f * L * π * 470K *1m Rs = = 15,66Ω 10ΩE1 Q Q 3,5 Rp = Q Q = * π * f Rp = Q Q = * π * f * Rs 3,5 *15,66 = 69,39KΩ 3,5 * Rp * π * 470K *69,39K * Rs 3,5 *10 = 66,7KΩ 3,5 * Rp * π * 470K *66,7K = 114,67 pf = 10 pf E1

8 est of the L circuit with the calculated values. R 1M out VG1 L 1mH pF Rs KHz / -3,90dB 460KHz / -6.9dB 480KHz / -6,9dB Gain (db) k 10.00M Frequency (Hz) est of the L circuit with the values from the E-series. R=10Ω =10pF KHz / -3,79dB 449,3KHz / -6,79dB 469,7KHz / -6,73dB Gain (db) k 10.00M Frequency (Hz) he f is 11KHz from it s place with the values from the E-series.

9 he final circuit with the L circuit as a replacement for the R resistor. V1 15V R1 70k L1 1mH pF R B547A out 1 3 B547A 3 1uF 1 1uF R 139k R4 39k R uF B547A R3 100k R RE 1.46k 4 10uF fc 445KHz / 43dB f1 408KHz / 40dB f 486KHz / 40dB fc 445KHz / 47,77dB out 1 f1 408KHz / 44,79dB f 485.8KHz / 44,74dB k 10.00M Frequency (Hz) BW = f f 445K Q = Bw 78K f 1 486K 408K = 78KHz = 5,7

10 o get the correctly f at 470KHz the optimizing tool in ina bee used. V1 15V R1 70k L1 1mH pF R B547A out 1 3 B547A 3 1uF 1 1uF R 139k R4 39k R uF B547A R3 100k R RE 1.46k 4 10uF 5=99,8pF 43.8 fc 470KHz / 43,4dB f1 43KHz / 40,3dB f 518KHz / 40,6dB fc 470KHz / 48dB out 1 f1 441,3KHz / 46dB f 507KHz / 46dB k 10.00M Frequency (Hz) BW = f f Q = BW f 1 518K 43K = 86KHz 470K 86K = 5,47

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