Question

THE STEPS TO DO SO:

Design a BJT amplifier based on the specifications provided in the table below. Your design should be insensitive to β variations, and both the input and the output should be AC coupled as in Fig. 1. Supply Voltage, Vcc Load Resistance, RL Transistor's Current Gain, β Relative Variation of lc for VBE-0.7 ± 0.1 V 0-to-Peak Output Swing, Vo Voltage Gain, A Input Resistance, R THD for 5kHz IV (0-to-peak) Sine Wave Output Voltage, V。S5% 5V 100Ω 100 10% -1V 20 RB1 RB1 RH RVRERERa Figure 1: Two-stage BJT amplifier Assuming that one gain stage followed by a buffer is sufficient for the design requirements, the circuits in Fig. 1 can be used as a starting point. Typical specifications include, but not limited to: 0-to-peak output swing: Vo Gain: Av Av1Av2 (where Avı is the amplifier gain and Av2 is the buffer gain) . Input resistance: R; Load resistance: RL . Linearity: vbel《VT ·3-insensitivity: IRBI 》 'B1' lcl 》 ,B2 . VBE-insensitivity: VRE 》 0.1V Using one of the circuits in Fig. 1, the design procedure can be given as follows: Assume Ay2 1 and Ri2 》 Rc, which is necessary to make the overall gain insensitive to β, as Rī21s directly dependent on β Choose VRE 0.1V, which requires VRE21V. Smaller values of VRE cause the DC biasing to be sensitive to variations of VBE, whereas larger values decrease the available output swing Since the buffer gain Av2 is assumed to be close to unity, the AC signal magnitudes at the output, at the emitter of Q2, and at the collector of Q will be similar, where the main difference among these voltages will be their DC levels. To avoid voltage clipping, VRc should be chosen such that Vecーら-VeE-WE.satl Vic2ら To maximize the available gain and linearity, VRc should be maximized, however the value of VCE sat can be chosen slightly larger (0.3 to 0.5V) than the typical value of 0.2V to account for variations in the circuit, as well as the small AC signal content at VRE The output voltage is a sine wave with the amplitude Vo, therefore the maximum value of in Fig. 1 is Vo/RL. When li reaches its maximum level, the voltage drop on RH reaches its minimum. Since lE2 is always a positive current (leaving the emitter for NPN, or entering the emitter for PNP), the current on RH must always be larger than I to avoid current clipping. Therefore, RH should be chosen such that RH 3 Once Ry is chosen, Ica can be determined from VRC 0.7 The maximum value of Ici is determined by the input resistance requirement, whereas the minimum value of that where N 10. Calculate the resistor values: VRC REIC1 (Vcc -VRE 0.7) Nlcı C1 NIcı RH R re2+ (RH l RL) and Ra = (β + 1)(n2+ (RH ll RL)) to verify the initial assumptions.

Answer 1

SOLUTION:

Given that
Solutton Given thob Supply voltagusvee Sv lond Resistant. RL 100」- Tsan s/stows cuvrort Bain B-2 loo THD of the 5kH2 、

Vce RH RB Commom) Colle 1twf ollutuy Common Emiv Lot us considuv

の cat culae oc analysic RBitRB lo+2o 81 RB 2oo 3 0 RTH 6-66 k

CRC S91) 519

100 (475)

st sta RE 12-67 The volase 9ain cE S RET 12.69 Avi 1 39

Source Pyodu (Y. 20mv yms voltag collecuv o c 3.03 t-4/4 (0.9 13 303-1.4,4 (07135)

Volfa% Swinq - vella swing 2 o v Soin vffVolfa