Question

Fig. shows a transistor with proper voltages established across its base, collector are emitter for proper function. Assume that there is a voltage drop V_BE across the base-emitter junction of 0.6 V and collector current I_C is equal to collector current IE. Calculate (a) V₁ (b) V₂ and V_B (c) V₄ and V_E (d) I_E and I_C (e) V₃ (f) V_C (g) V_CE. All resistances are given in kilo-ohm

Answer 1

(a) The 250 k and 50 k resistors form a voltage-divider bias network across 20V supply. 

∴ V₁ = 20 × 250/300 = 16.7V 

(b) V₂ = 20 − 16.7 = 3.3V 

The voltage of point B with respect to ground is V₂ = 3.3 V 

(c) V_E = V₂ − V_BE = 3.3 − 0.6 = 2.7 V. Also V₄ = 2.7 V 

(d) I_E = V₄/2 = 2.7V/2 k = 1.35mA. It also equals I_C. 

(e) V₃ = drop across collector resistor = 1.35mA × 8k = 10.8V 

(f) Potential of point C is V_C = 20 − 10.8 = 9.2V 

(g) V_CE = V_C − V_E = 9.2 − 2.7 = 6.5V