For the 2N3904 transistor, the data sheet says that (3=1.4*(T degree C) + 195. S

Question

For the 2N3904 transistor, the data sheet says that (3=1.4*(T degree C) + 195. So at room temperature T=25degreeC and (3=230. It also says that for each Watt of power dissipated in the transistor, the temperature will rise by 200:C above room temperature. Find Ic, and Vce at room temperature. Determine the power dissipated in the transistor and how much the temperature will rise. Find the new (3 Continue this for a couple of iterations until the temperature stops increasing. What is the final temperature of the transistor? Assume Vbe remains at 0.7 V.

Explanation / Answer

beta =1.4*T + 195

V(BE) =0.7 V, I(RB2) = 0.7/680 =1.029 mA

I(RB1) =(20-0.7)/18 k= 1.072 mA

IB =I(RB1) -I(RB2) =42.81 microA

1)

T =25 C, beta =230

IC = beta*IB =9.84 mA

V(CE) =20-IC*RC =20-9.84 m*470 =15.37 V

disipated power is P =V(CE)*IC =151.26 mW

Delta(T) =200*P =200*0.15126 = 30.25 C

T2 = T1 +Delta(T) =25+ 30.25 = 55.25 C

2)

T2 = 55.25 C, beta =272.35

IC =beta*IB =272.35 *42.81 micro = 11.66 mA

V(CE) =20 -470*11.66 m = 14.52 V

P =V(CE)*IC =14.52*11.66 m = 169.3 mW

T3 = T1+ delta(T) = 25 +200*0.169.3 =58.86 C

3)

T3 =58.86 C   beta =277.4

IC =277.4*42.81 micro =11.86 mA

V(CE) =20-11.86 m*470 =14.43 V

P =V(CE)*IC =14.43*11.86 m = 171 mW

T4 =25 +200*0.171 =59.2 C

Final temperature is somwhere a bit above 59.2 degree C

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