Bipolar Junction Transistor (BJT)
Switches
Circuit Symbol & Analogy:
IC
C
C (Collector)
B (Base)
DIO DE
1
IB
Isw
IE
E (Emitter)
Symbol
KCL: IE = IC + IB
B
2
E
Analogy
ECGR 2252
Mehdi Miri
ECE Dept.
UNC Charlotte
ECGR 2252
Mehdi Miri
ECE Dept.
UNC Charlotte
BJT Switches
BJTs have 3 modes of operation
-Cutoff:
IE=IC=IB=0
-Active (amplifier):
IC = βIB; β&VCE large
-Saturation (switch): IC = βIB; β&VCE small
 BJT is an intrinsic current amplifier:
IC = βIB
β is the BJT’s current gain
IC
C (Collector)
IE
E (Emitter)
B (Base)
IB
ECGR 2252
Mehdi Miri
ECE Dept.
UNC Charlotte
BJT Switches
Consider this BJT circuit:
Load Resistance
RB
IB
VBB
IC RC
IE
VCC
0
0
0
KVL in C-E circuit: VCC = RC*IC + VCE
(1)
KVL in B-E circuit: VBB = RB*IB + VBE
(2)
If
RB , IB , IC
& VCE
BJT Switches
BJT power loss:
IC
+
VCE
-
BJT
Ploss ≈ VCE*IC
IC is the required load current.
For switch mode, VCE must be kept as
small as possible.
ECGR 2252
Mehdi Miri
ECE Dept.
UNC Charlotte
BJT Switches
ECGR 2252
Mehdi Miri
ECE Dept.
UNC Charlotte
VCE < 0.5V is acceptable.
Need to inject sufficient IB to bring VCE
down to our target value of 0.2V.
The design process involves only the
determination of RB.
For a BJT’s Base-Emitter junction diode,
VBE ≈ 0.7V.
BJT Control of E.M. Relay
U1
NC
COM
NO
Vac
RC
VCC
B
0
RB
VBB
0
0
Zload
A
ECGR 2252
Mehdi Miri
ECE Dept.
UNC Charlotte
BJT Switch Design Process
From circuit Eq. (1), solve for IC:
VCC = RC*IC + VCE
RB
IB
VBB
IC RC
IE
VCC
0
0
0
From BJT Eq. IC = βIB, solve for IB.
Use a value between 10 and 15 for β.
ECGR 2252
Mehdi Miri
ECE Dept.
UNC Charlotte
BJT Switch Design Process-continued
From circuit Eq. (2), solve for RB and
round as needed:
VBB = RB*IB + VBE
RB
IB
VBB
IC RC
IE
VCC
0
0
0
Check to see if the desired IB can be
sourced by the driving circuit.
ECGR 2252
Mehdi Miri
ECE Dept.
UNC Charlotte
Darlington Switches
ECGR 2252
Mehdi Miri
ECE Dept.
UNC Charlotte
If IB can’t be sourced by driver circuit, we
can use the slower Darlington switch:
RC
RB
IC
VCC
IB
VBB
0
MPSA29
IE
0
0
Darlington’s Eq. is IC = β(β+2)IB, where β
is current gain of a single BJT (prove!).
Darlington Switch Design Process
ECGR 2252
Mehdi Miri
ECE Dept.
UNC Charlotte
Same design process as for BJT except
that design Eqs. are different.
Write KVL for C-E and B-E circuits.
Note that VBE ≈ 0.7+0.7 = 1.4V, and that
target VCE = 0.2+0.7 = 0.9V.
Summarize the design process as in the
BJT case!