(a) The current gain factor (b)= DIc/DIb = slope of curve above.
This calculates as (21.5-3.6 mA)/(120-20 uA) = 180 to nearest integer.
(b) The circuit to allow these measurements is the same as Figure 24.6 in the text.
My thanksto Peter Finch from St Joseph's College, Gregory Terrace, for these (pdf) solutions.
The rest are from Glenn Rossiter:
Q21 :-The data plotted as an Excel chart is :-
(a) The current gain factor (b)= DIc/DIb = slope of curve above.
This calculates as (21.5-3.6 mA)/(120-20 uA) = 180 to nearest integer.
(b) The circuit to allow these measurements is the same as Figure 24.6 in the text.
Given the bulb is 0.3W at 3.0 volts, then power dissipated P = V.I or current I = 0.3/3.0 = 0.1 A.
(a) If the transistor is ON collector current I = 100 mA.
also VRC = 6 - 3.0 = 3.0 V.
(b) Collector resistor RC = VRc/Ic = 30 W.
(c) If b = 100, then IB = IC/b = 1.0 mA.
Hence RB = (6 - 0.7)/1.0 mA = 5.2 kW as nearest value.
Given Rc = 10 kW, Rb = 100 kW, Vcc = 10V and b = 120.
(a) By Kirchhoff's voltage law Vcc = VBE + IB.RB + IC.RC.
Hence by rearrangement the required equation is produced.
(b) By substitution IB = (10 - 0.7 - 120.IB.10k)/100k.
which yields IB = 7.15 mA.
As current gain b = IC/IB, then IC = 0.86 mA.
Now VCE = VBE + IB.RB = 0.7 + 7.15 mA x 100k = 1.4 V.
(a) The input pots vary or control the input voltage being applied to the amplifier.
(b) Vout = RF.(V2 + V1)/R.
(c) Amplifier gain is unity if RF = R in the circuit.
(d) Vout = V2 + V1 + VDC if an extra DC voltage exists.
(a) The 555 timer IC is being used in its astable or free running multivibrator mode with VR1 controlling the output frequency, probably over 20-20KHz range. This signal would be a square wave.
Component VR2 controls the voltage at the output probes, signal output probe voltage.
In an audio lab, this circuit could be used to provide an oscillating audio test signal for injection into circuits under test ie as an input signal oscillator.
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