Electronics Revision Worksheet # 2 Solutions
Solution 1
Solution 5
Note you needed this graph
Solution 2
a) Y and Z are non ohmic, the graphs
are not straight (linear)
b) 80/510-3 = 16k
c) Use the graph to see that when
device Z has 2mA it has 50V.
Therefore the device X must have
50V over it as the two devices are in
parallel. If device X has 50V over it,
use the graph and you see that the
current must be 3.1~3.2mA. Use
V=IR and you would find that the
current is 3.13mA. (from question 2 R
= 16k)
a)
(i) The switch on voltage is the
potential difference that the diode requires in
order to work.
(ii) The switch on voltage for this
diode is 0.7V
(iii) Silicon
(iv) If the diode is reverse biased the
Vy > Vx
(v) If the diode is forward biased the
conventional current flows from X to Y
b) There are two diodes, so each will take
0.7V. So the two diodes take 1.4V. Leaving
0.1V for the resistor. The current would be
0.1/60 = 0.001666A. Use this current and
multiply by the voltage 1.5V and you will
know the power. 2.5mW.
Solution 3
a) Answer is E. The new current would
be much less then half the old current.
Note the graph for a diode is not
linear.
b) Answer is C, the new current would
be similar to the old current, which
would almost be 0A.
Solution 4
a)
(i) The diode is forward biased during
the BC period.
(ii) During the time interval A the diode
is reversed biased.
b) A sine graph
Solution 6
Solution 7
a)This is the classic de-coupler, where the D.C.
voltage goes over the capacitor and the A.C.
voltage goes over the resistor
b) Vout and Vin are parallel so Vout is Vin
Solution 8