TRANSISTOR SWITCHES
Transistor Characteristics (dc)
• Operational characteristics for npn
→ collector must be more positive than emitter
→ BE & BC circuits behave like diodes
→ BE diode conducts,
BC diode is reverse biased
• Max values of IC, IB, VCE and power dissipation must not be exceeded (given in manufacturers
dta sheeets)
• Usual circuit configuration is COMMON-EMITTER (CE)
+ VS
→ emitter appears in both input & output circuits
RL
• Collector current
IC = I0[exp(qVBE/kBT)-1]
• β = IC/IB CURRENT GAIN
IC
RB
IB
(Also denoted by hFE)
∴IC = βIB
Vo
• But IE = IC + IB
V BE
I
+V
= IB + βIB = (β+1)IB
i
E
0V
≈ βIB (since β>>1)
∴IE ≈ IC
• pnp characteristics are same but with opposite polarity
• Current will not flow unless Vi > VBE = 0.7 V
Output Characteristics & Load Line
• Static OUTPUT CHARACTERISTICS define steady state conditions
→ family of curves of IC vs VCE for increasing IB
• Graphical design procedure gives
→ performance under non-linear conditions
→ start & finish points in analytical procedure
IC
• For Vi = 0, IB = 0
(mA)
→ no current flows through RL thus IC = 0
10 mA
→ so VS = VCE(max)
• For Vi >> VBE, (say Vi = Vs), IB is high and IC is at
maximum
→ nearly all volts dropped across RS
∴VCE = 0 & IC(max) = VS/RL
⇒ Transistor is said to be SATURATED
• A LOAD LINE drawn from IC(max) to VCE(max)
Saturation Region
LOAD LINE
50 µ A
40 µ A
Q-point
30 µ A
IB
25 µ A
20 µ A
5 mA
10 µ A
0 µA
0
• Slope of load line is -(1/RL)
10 V
5V
Cut-off region
Transistor switches
• Can use load line to show how transistor behaves as a switch
• When no base current flows, no collector current flows
∴transistor is said to be OFF ⇒ analogous to switch being open
• When max base current flows, max collector current flows
→ transistor is ON, output saturated ⇒ analogous to switch closed
EXAMPLE: TRANSISTOR SWITCH
Draw a load line for circuit shown; show how circuit operates as a switch. Assume β in range 110800, load resistor is 1 kΩ, VS = 10 V.
Solution
V CE (V)
• When applied base voltage is zero (ie Vi = 0)
→ i.e. IB = 0 and IC = 0
→ ∴VCE ~ VS = 10 V since no volts drop across resistor
→ transistor is OFF
• When Vi = VS applied (ie Vi >> VBE)
→ IB can flow ⇒ IC can flow
10 V
+ VS
RL
1 kΩ
→ IC(max) = VS/RL = 10 mA
→ ∴VCE ~ 0
→ transistor is ON
• Load line drawn from 10 mA (on IC scale) to 10 V (on
VCE scale)
• Value of RB chosen to ensure that transistor saturates
RB
IC
IB
Vo
+ Vi
→ max collector current obtained when β is at
minimum
→ thus a min base current needed to ensure this
→ IB(min) = IC/β(min) = 10 mA/110 = 91 µA
• Current flows from Vi through RB to base
→ RB controls current to base
→ RB = (VS-VBE)/IB = (10 V-0.7V)/91 µA = 103 kΩ
→ Take 100 kΩ as next lowest SV
⇒ ensures that IB is at least 91 µA
→ Recalculate new base current
• When input voltage is low (0 V), output voltage is high (= VS)
• When input voltage is high (= VS), output voltage is low (0 V)
V BE
IE
0V
V
i
10 V
t
IC
10 mA
t
V
o
10 V
t