Question

6. The following LED (Light Emitting Diode) circuit has NMOS transistors with a Vto = 1V, and the assigned gate-source voltages (Vgs). Determine the current flowing through each LED (D1, D2, and D3). Next, determine the power dissipated in each resistor. Note: The diodes have a forward voltage (V+) drop of D1 = 1.9V, D2 = 2.2V, and D3 = 2.5V, and have a reverse voltage (V.) of D1 = 12V, D2 = 10V, and D3 = 4V. (15 points) 12 VDC 4251 3 R2 970 92 3 R3 1.3 kN Vgs = 3.3V Vgs = OV Vgs = 5V ID= PRIE ID2= PR2= ID= PR3=

Answer 1

₂ {gfar R₂ & 1.3 kr Ruasr VID $123 Vasov 3V/S to assume ov given & Vto = threshold Voltage of NMOS = 1 V. Von = 1.9V; Vrem= 9.2 V; Vf9;=2.5 V Vo= also given in the quation Dear student first under stand MOSFET Concept - Nanosi 117 NMOS 1. NMOS ON } Vers <Vto > cut off rigion Nmos off I os=0 Vas = 3.3V Vte = 1 V { Vus > Vio Vgs > Vt. ON condition ! Iss #0 NMOS 2: Voa=ov Vto= 1 V 1 We can canculate Ins s<V to off I se=0 NMOS 3: Ves = 5V, Vte = 1V - Vgs - Vt: Vgs > Vto ON condition

V og = Vgs - Vt Vos, = 3-3-1 - 2.3V Voso = 0-1 = -1 off Vos: = 5-1 = 4 V 12-1.9 -2.3 uige LED -1.9-23 = 425 7.8 -0.01835 A Voltage drop Is = 425 = 18.35 MA I),= OA became NMOS 2 off condition | 9.5=LED T 12-2.5-4 - 5.5 - 4.23 MA | Voltage drop 1.3kr 1.3k2 Pr. = I 2 R, = (18.35 x 10 x 10 x 425 = 143107 x 10-6= 0.1431 Watt Pre = I S R a = 0 Watt PEx = I3, R3 = (4.23) *x 10310X1-3 x 103 = 23.26 X10?– 23.26 mW =0.0232 Watt ID 1 = 18.35mA PR = 0.1431 Watt Isa=0 Pre=0 I vz = 4.23 MA PRA = 23.76 mW = 0.02326 Watt please comment query 3 Any Please like Hit
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