ECE20A, Lab Quiz, Fall 2001
NAME:
Specs: All resistors are 0.25 W and 5%. All semiconductors are made of silicon with the
exception of the LEDs which have a forward bias voltage of 1.7 V and need about 10 mA
of current to be brightly lit. All transistors have β = 200. The zener diode has a zener
voltage of 5.7 V.
Problem 1. Consider the power supply circuit below. Vin is supplied by a 9 V commercial
battery. Depending on the battery life time and the current drawn from it, Vin ranges from
7.5 to 9.5 V. The current IL is well within the range for this circuit to operate as a regulator.
a) What is VL b) Show that the BJT is always in the active-linear region. (10pts)
v
+
CE
−
IL
−
V
in
+
−
+
−
1 kΩ
+
vBE
VL
v
D
+
−
1) The zener diode is in the zener region because it is reversed biased by Vin and Vin >
5.7 V. Thus, VD = −5.7 V.
2) BJT is not in cut-off because IE = IL > 0. Thus VBE = 0.7 V.
a) From KVL:
VD + VBE + VL = 0
→
VL = −VD − VBE = 5.7 − 0.7 = 5 V.
b) By KVL:
−Vin + VCE + VL = 0 → VCE = Vin − VL = Vin − 5
But, 7.5 < Vin < 9.5 → 7.5 − VL < Vin − VL < 9.5 − VL
Substituting for VL = 5 V, we get 2.5 < VCE < 4.5 V
Since, VCE > Vγ = 0.7V, BJT has to be in active-linear state.
ECE20A, Lab Quiz
Page 2
Problem 2. Consider the circuit below. a) What is the current in LED when Vin = 0, b)
What is the current in LED when Vin = 5, c) For what range of Vin LED will be ON, d)
Modify the circuit such that LED will switch ON (or OFF) at Vin = 2.5 V (20pts).
From the circuit, we note that the current in LED is IC
and the voltage across resistor R2 is VBE . Thus:
Ohm’s Law: I1 = (Vin − VBE )/10k
Ohm’s Law: I2 = VBE /10k
KCL: I1 = I2 + IB
+5 V
330
IC
R =10 k
1
IB
+
a) Vin = 0.
+
I
VBE
1
+ V
Assume that BJT is in cut-off: IB = 0, VBE < 0.7 V.
I
R
=10
k
2
2
− in
Using IB = 0 in KCL above, we have I1 = I2 and
−
substituting for I1 and I2 from Ohm’s Law above we get:
(Vin − VBE )/10k = VBE /10k → VBE = 0.5Vin = 0.
Since VBE = 0 < 0.7 V, BJT is in cut-off, IB = 0 and IC = 0 and LED is OFF.
+
VCE
−
−
b) Vin = 5. Assume that BJT is NOT in cut-off (either active-linear or saturation):
VBE = 0.7 V, IB > 0. Then:
I1 = (Vin − VBE )/10k = 4.3/10k = 0.43 mA
I2 = VBE /10k = 0.7/10k = 0.07 mA
IB = I1 − I2 = 0.36 mA.
Since IB > 0, our assumption of BJT NOT in cut-off is justified.
To find, IC , we assume that BJT is in saturation: VCE = 0.2 V, IC /IB < β
Writing KVL in CE circuit gives:
5 = 330IC + VLED + VCE → 330IC = 5 − 1.7 − 0.2 = 3.1 → IC = 9.4 mA
Since IC /IB = 9.4/0.36 = 26 < β = 200 our assumption of BJT in saturation is justified.
Thus, LED current is IC = 9.4 mA and LED is ON.
c) LED will turn on at a certain “threshold” Vin such that BE junction is forward biased,
VBE = 0.7 V and IB > 0, but IB can be very small (the threshold is when VBE = 0.7 V
and IB ≈ 0). From KCL and Ohm’s Law equations above, we have:
I2 = VBE /10k = 0.7/10k = 0.07 mA → I1 = I2 + IB ≈ 0.07 mA
I1 = (Vin − VBE )/10k = 0.07 mA → Vin = 0.7 + 0.7 = 1.4 V.
So, LED is ON for 1.4 ≤ VIN ≤ 5 V.
d) At the “threshold” value of Vin that switches the LED ON (or OFF), we have VBE =
0.7 V and IB ≈ 0. Then, from KCL above I1 = I2 and Ohms’ Law across R1 & R2 gives:
(Vin − VBE )/R1 = VBE /R2 → Vin = VBE [1 + R1 /R2 ]
For Vin = 2.5 V and VBE = 0.7 V, we get: R1 /R2 = 2.57. Thus, we should modify the
above circuit by either:
1) Keep R1 = 10 kΩ and change R2 = 10/2.57 = 3.9 kΩ
2) Keep R2 = 10 kΩ and change R1 = 10 × 2.57 ≈ 26 kΩ.