Question

2. We will modify example 6.10 from the textbook during recitation #10 (11/14), by taking Re = 0, as shown in figure 2. When we make this modification, we will discover that the transistor will operate in saturation mode using the resistor values in example 6.10. ECE 342 Electronics I Homework # 10 Due Nov. 20, 2019 (Wednesday) a) Calculate the BJT terminal currents Ib, Ic and le, as well as the node voltages VB, Vc and Ve for Rsi = 1 k 2 and RB2 = 500 S2. Which mode is the BJT operating in: forward active, saturation or cut-off? b) Calculate the BJT terminal currents Ib, Ic and le, as well as the node voltages VB, Vc and VE for RB1 = 10 M2 and RB2 = 5 MO2. Which mode is the BJT operating in: forward active, saturation or cut-off? tV cc Roiz RŽlic Tove Re Figure 2. Revised BJT circuit from example 6.10, with RE = 0. Rc = 5 k12. Vcc = 15V.

Answer 1

> Home work #10 04VEHSV ? @ Assume Assume tansistor is active region :- & Ro= 5ks V - RBL ( Veilage divide ckt) von BC os x 15 1 I+05 i RBI=lkoh L RB2=5000h -0. 52 Nu B=5v

Rth= RBill RB2 = 11105 = 11os os = 0.336.2 +0'S 15 Assume Lp81038) ckt is look like! otvec KvL at input side - Vg - IB Rn - UBE=0 0 5-033 10-07 IB 4.3 = 0.33 IB 1 IB = 13.03MA f Ic=ß IB Ve=0 kve at (due to ground) output side :- Vec-Ic RC-VCE=0 L 15-8x13-0345-V6E=0 put value Nee = 15-65.03 $ fit p = 1 VLE S-50'03 Condition V CELO.2V the transistor is going to Saturation region. in this way assumption is wrong. transistor operate in saturation region VCE=0.2 . VE =O (due to ground) VCE = V(-VE =0.2 [vc=02V] ne

5 -0.2 Ie= Vee-ve lo = = 2.96 mA Ic=9.96mA Axuy (f=50) Deze =0.0592 mA IB=0.0592 MA Ang IE = IBt Ic -0.0592+2-96 mal IE = 3.0192 - Ais +Ncc = tisu 'B = RB2 Vec RBTP B2 Кр,-skg XS = 5v - he says - 1015 B-5V = so is OY Ren= 10/15 = 10x Tren= 3:33Me is active region' - ckt can redrawe e centisv Assume transistor + pe=56h *

BE07VB-S 07V, Boso (Assume RUL at input side! VE - Ig Ratu - VBE = 0 5 - 3:33M& IB -0.7=0 4.3 3.33me IB = IB = 1. 29 MA Ic=ß IB =50 P AB=5081-29 M A = 0.0645 mA Ic=0.0645 mA Ang output side for kvL at =15-0.322 VCE = Vec-IcRc 315-0.0695x5 VE=0 [due to VCE = 14.67 VC=1467 V ondition active - the VCE 70.2 - Actine region Assumption is right so transistor is region. All the calculation abone is right. IĘ = Ict IP = 0.06519 MA 12:006579 MA Are