Question

[P.4] Calculate all node voltage and branch current and small-signal parameters with voltage gain for each of the circuits depicted in (Fig. 4). [20 pts] (Assume 1s = 5x10-16(A), B= 200, VA = 00, and capacitors are very large.) (Fig. 4) Vcc= 2.5 V Vcc= 2.5 V 14 kΩ Reξ 10 kΩ 50 kΩ 1kΩ Lo Vout O Vout C HH 1kΩ 11 kΩ Lai 1 kΩ C1 500Ω C2 2kΩ IHH C2 2 kΩ (b)

Answer 1

# Solution - To calculate all node voltage and branch current and small-signal parameters with voltage gain for each of the circuit depicted. (Assume Ix = 5x10 "amp A, B= 200, VA=co and Capacihers are very large.) Vcc=2.5v Ein sökes lukas • Loka ovo a V. Iku, IK vint HE 2 a G SIKA 35oon 32k1 C2I 2013 C2 II Note > For de analysis, the capaciher six open cinnetét Because Xc = for ($=0). all ac voltage source de short circuit. → For ac analyxia, The Capacitor is short circuit becaus &c=0 for large value of c and frequency, » Alldc vallage source is short cincuet. Scanned with amScanner .

De analysis for bigure a TVC C=25v B = dc lc = Biz IB 5okus Sik kic "Vc JE = vct ÅB & le(+D dig BY VB IVE Loops 32KA Yle apply KVL loop L VCC= iB RB + VBE THERE 25= 18 X 50x10²4 0.7+(B+Dd BX XE Put YBE = 0 7 Valt 2.5=0.7+ 50x103 jg + 402X103 1B 1.8 = 452x103.18 dc = ßle ig= 3.98x156 pic= 0.796 m Amp JE (B+) ja JB= 3.98 M Amp 201x3.99x105 ie= 0.20 X 103 Amp To calculate Nade ME=0.8m Amp voltage Vo = 2.5–50X10°y 3.98xió | VB = 2.3 volt VE = DEXRE = 0.0 X163x 2x103 [VE = 1.6 Volt Ve= Veca Relic = 2.5 - 1X18X0.1986 Ve=1.71 volt = S Scanned with Cam Scanner

at arc VT = 25m Valt Ma= 200 X 25X10 3 Ac analysis for bigure @ vit ii VB rig PrA Elu Sokaş ra= ßre. = BUT ic Vt -- Threshold voltag By KCL law di=latie Vi - VB VB LK 0.719 x103 + VB 6.3K a=6.3K1 Jok Vi = Vol 1+ +1 6.3 VB = Vi volt 1.A |VB=0.25 ve dB = VB - 0,85 ki 6.5K 6.5k Vor 1043 x B le Vor -103 x 2008.0 85ké 6.3X103 Vo Av= 26.98 V small signal signal parameters in 6.3 KM Ava - 26.90 CS Sca ned with Can Scanner

De analysis for figure b. .24 2.50 Rth = 14 Loki ГІЧx11 14+11 Rth=6.16kn НЕ soo Vena 2.5X 11KA & luka+11ka aka C Voltage division Rule Vene sal volt To calculate RenAVth 12.50 loka VBT Vc LN FB 6.16 VE 805k apply KVL in loopt 11 Flig Run + Ver + 2-516 d'I = x6,168103+0.7+2.5X183 if (2004) Loops Eaka VIE 0.4 = 508.06x10lo TiB 0.786 M Amp ica BiB 200X0.786X10 -6 Verdol - Rgx de VB=1.1 - 6.16X18X0.78X106 |VB = 1.ogvolt VE= 2:5X103 LE IVE= 0.395 volt ie= 0,15m Amp JE (Bty le = (200+1 0.78815 Die = 0.150 m Amp V=2.5-10x103 0.15x10 TVC= 1 Volt StarTech CamScanner

For ac Aanlysis of figure b vi ki VB Vo yuz yug ENK 10kA/ loku LUKA Vio 1848 ra= BUT ło 0.5613 ( BD is = 200 x 25x10 0.5X163 VB = Bith + (8 +4 0.5X103 is Ma = 33.39 Uge 33-3x103 ig + 201X0.5X10318 VB Ele 133.8x103 133.8818 By RCL law + VB 133.dk Ij = a +d3 tip Virve VB + VB IK TIK पार VB t I + (ч " VB = Vi 9.16 1 13dd |VB=0.86 Vé Vo=-Brelo Voz-200x10x103, 0.86 vi 138.88103 .-12.39 CamScanner small signal parameter ra= 33.3k Vo Av=-12.39 Vi Vo Cuines wir