Page 290 - Mechatronics with Experiments

JWST499-Cetinkunt
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the maximum voltage drop across the base and emitter is V BE = 0.7 V, and the minimum
voltage drop across the collector and emitter is V CE = 0.2V.
1. Case 1: V CC1 = 0.0 V. Then i = 0, hence, i = 0. Therefore, the voltage measured
C
B
between the collector and the emitter (let us call it the output voltage, V out )is
V out = V CC2 = 25 VDC. We can call this the OFF state of the transistor.
2. Case 2: V CC1 = 30.7 V (more precisely, the input voltage at the gate is large enough
to generate a large base current that will saturate the transistor output circuit). Then
i = (30.7 − 0.7)∕100 K = 0.3 mA, hence, i = i = 30 mA. Therefore, the voltage
B C B
measured between the collector and the emitter (let us call it the output voltage, V )
out
is V = V − R ⋅ i = 25 − 1000 ⋅ 30 mA =−5 V. Since the output voltage at
out CC2 2 C
the collector cannot be less than the emitter voltage plus the 0.2 V minimum voltage
drop across the collector and emitter, it must be, V = 0.2 V. Hence the actual
out
collector current saturates at i = (V − V )∕R = (25 − 0.2)∕1000 = 24.8mA
C CC2 CE 2
and the output voltage at the collector is V out = 0.2 V. We can call this the ON state
of the transistor.
3. Case 3: When the input voltage is above cutoff voltage (V BE = 0.7 V) but below the
saturation voltage, the transistor operates as a proportional voltage amplifier. The
saturation voltage can be calculated as follows. The minimum voltage drop across C
and E when the transistor is saturated is
V CE = 0.2V; when saturated (5.148)
At this point, the current is
V CC2 − V CE 25 − 0.2
i = = = 24.8 mA (5.149)
C
R 1000
2
The base current must be
1 1
i = i = i (5.150)
B C C
100
V in,sat − V BE
= (5.151)
R 1
V = 0.248 mA ⋅ 100 KΩ− 0.7 = 24.1 V (5.152)
in,sat
Base voltage value larger than that operates the transistor in the saturation region.
V < V ; transistor is in fully OFF state (5.153)
in BE
V < V < V ; linear region (5.154)
BE in in,sat
V > V ; saturated region (fully ON state) (5.155)
in in,sat
Case 1 and case 2 are shown as a mechanical switch analogy in Figure 5.18. Figure 5.18a
shows the so-called sinking connection and Figure 5.18b shows the so-called sourcing
connection of the transistor to the load. The names sinking and sourcing are given from the
point of view of the transistor in that it either sinks the current from the load or sources
current to the load.

Example A transistor circuit shown in Figure 5.19 can be used to switch the power
on a load which is controlled by a mechanical switch. Let’s assume that V BE = 0.7 V and
V CE = 0.2 V. If the load is of inductive type, a suppression diode should be used in parallel
with the load. When the transistor goes from the ON to OFF stage very fast, large voltage
transients develop due to inductance. These large voltage transients can be easily large
enough to damage the transistor if it is larger than the forward break-down voltage of the