# Dissipation of 2N3904

#### The 2N3904 Transistor

The current amplification $\beta$ value of the 2N3904 transistor is given by the relation $\beta=1.4T(C)+195$. For a temperature of T=25 C the value of $\beta=230$ from the previous equation. The temperature increase for each W dissipated is T=2.00 C. Please find (at T=25 C) the values of  Vce and Ic, the increase of temperature and the power dissipated. Do three iterations. Assuming Vbe remains 0.7 V what is the transistor temperature after these three iterations?

Find how to compute the power disipation of a transistor here

#### The computation

$\beta =1.4*T + 195$

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

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

$IB =I(RB1) -I(RB2) =42.81 \mu A$

1) In the first iteration we begin with initial temperature and initial current amplification:

$T =25 C$, $\beta =230$

$IC = \beta*IB =9.84 mA$

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

Thus we find that the dissipated power of the 2n3904 transistor 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) We begin the second itaration with the values that resulted from the previous iteration:

$T2 = 55.25 C$, $\beta =272.35$

$IC =\beta*IB =272.35 *42.81 \mu A= 11.66 mA$

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

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

$T3 = T1+ \Delta(T) = 25 +200*0.169.3 =58.86 C$

Hence, after the second iterationthe temperature increases to $58.86 C$ and the disipation to $169.3 mW$

3)  The third iteration increases further the temperature and the disipation of this device:

$T3 =58.86 C$   $\beta =277.4$

$IC =277.4*42.81 \mu =11.86 mA$

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

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

$T4 =25 +200*0.171 =59.2 C$

$T4 =25 +200*0.171 =59.2 C$

We observe that the although the temperature and disipation have increased after this third iteration, this increase is 3-5 times lower than the previous one (found in the second iteration). Therefore a fourth iteration will be sufficient to deterime with good precition the final temperature and disipation.

Hence, the final temperature is some where higher than 59.2 degree C