Question

Question 3. Unregulated supply Rz IL Vin IR Ib (a) The circuit on the right shows a series regulator connected to the output of an unregulated power supply. The transistor has B =50, and a 6 volt Zener diode is used. When the load current, Il, is 1 amp the de input voltage from the unregulated supply, Vin, is 11 volt, VBE = 1 volt and the Zener diode current, Iz, is 20 mA. For these conditions, calculate Iz (i) the transistor base current, Io, and the power dissipated in the transistor, (ii) the value of Rz required, and the current Ir passing through it, (iii) the power dissipated in the Zener diode. +12.7V R (b) The diagram at right shows a common-emitter transistor amplifier operating from a DC power supply of 12.7 volt. The transistor has B = 150 and VCE = 0.3 V when the transistor is saturated. With no signal applied, the de voltage between the collector and ground is measured as 7.7 volts. 5k2 C) Calculate the transistor's base, emitter and collector currents. V VI C 3.3k_2 (ii) What is Vce, the dc voltage between the collector and emitter of the transistor? (iii) What is the peak value of the largest ac signal at the collector of the transistor that the amplifier can handle without distortion? Briefly describe what form this distortion will take. (Hint: Remember that Ce bypasses all signal currents so that they don't go through the emitter resistor). (iv) Briefly describe the purposes (one sentence each) of the capacitors C1 and C2. (c) Draw the circuit of a summing amplifier which uses a feedback resistor of 12k12 and has an output voltage of Vout = -(6Va + 2V6). Make sure that you label the values of the other resistors i.e. what they are in kΩ. KR (d) The circuit on the right shows an operational amplifier circuit. All voltages in the circuit diagram are measured with respect to the ground. Assuming the operational amplifier is not in saturation and using R the properties of an opamp: R (i) Find the voltage at the non-inverting input of the opamp. VB 1 V2 KR (ii) Find the voltage at the inverting input of the opamp using Millman's theorem. (iii) Find an expression for V2 in terms of VB and VA.

Answer 1

pown dissipation in transistor. 0 Given 3 VBE = IV I = lamp Iz=20mA Vin- IV ß = 50 Uz = 6V Ve IL VE + vin Roz tlu lov od IR Ib Iz GV 온 IL = TESTA Ib= IE -15 HB Ib=19.607mA (Pd) = Pa = UCE IC VCE = Verve = Vin - VL VL = 2 LOUP 0 -6 +1UL = 0 VL = SV vin in-vcluce = 11-5 = 60 CS Scanned with CamScanner

2 Ic = IE-IB 19.60TMA 1000 MA Tc Ica 980.39 mA Pd = UCE IC T Pdq= 6x980.39m w P<= 5.882 watts (ii) Rz = ? IR vin-vz RZ IR = Ibt Iz 19.607+20 MA IR = 39; 607MA 11-6 39.607MA= RZ Rz 5 IL 396607 Rz = 126.242 TID CPd) z = P.d in zenor 모 diode . (PdDz = Uz Iz = 6V*20 mA (Pd) 120 mW Z CS Scanned with CamScanner

b VIESO.3, Vco= 7.70 12.7 HY 5k RI ot + 0.3 7.7V R2 UTE 3.3k VEO 70 7-0.3=1.4u MA IE= 7.4 3.3 TE = P242mA Ic = 12.7- 7.7 5 Тc= IMA IB=IE-Ic IB = 2.242-1 ТВ - 10242 MA CS Scanned with CamScanner