EH. - I :0
Wﬂw M4 at L' h" arr-Cf. In"
'I'CI'I' A f .- Hg 113.1; .- Iaﬁf, , .5 “ff-j
'14: +51 Kw! 1. mi Irr' w i.
I” ~ m _.
C“ [in A)_00“‘ _ﬂf'
H F' (a 'l :1 5,1 F
Vt: . My
m Tﬁ‘ Fir a ,4
h‘ 5'0 _.,J 2 1 1'
VS = 120 /(10/1) = 12V
12/.707 = 17Vdc
Subtract voltage drop across diodes
17 1.7 = 15.6Vdc
15.6Vdc across the capacitor
Current through 390 ohms resistor =
15.6/390 = 40mA
Current through RS =
Vz = 6V
Rs = 270ohms
6/270 = 22mA
Capacitor supplies the sum
IB = 10V-.7v/ 330 kOhms = 28.2uA
VCE = 7.76V
V = 10V = 7.76/ 470ohms = 16.5mA
Bdc = 16.5mA/ 28.2 uA = 579 Gain
I found that the current gain that was measured to be significantly higher then the 100- 300 found
in the the data sheet. If I used the current
VE = 5v - .7v = 4.3v
IE = 4.3/1.5kohms = 2.9mA
ICRC = (2.9mA)(4kohms) = 3.8V
VCE = 4.3v 3.4V = .9V
There is somewhat of a discrepancy between the measured values and the theoretical values.
This is somewhat due to the difference between potential transist
3-20 There would be zero Forward volts. It would most likely short the diode from excessive
current, since you would be going from 100k ohms to zero ohms.
3-21 The resistor or the diode shorted
3-22 The diode could be open, a wire between the components a