SCR / Thyristor
• Circuit Symbol and Terminal Identification
ANODE
GATE
SCR
2N3668
CATHODE
SCR / Thyristor
• Anode and Cathode
terminals as
conventional pn
junction diode
• Gate terminal for a
controlling input
signal
ANODE
GATE
SCR
2N3668
CATHODE
SCR/ Thyristor
• An SCR (Thyristor) is a “controlled”
rectifier (diode)
• Control the conduction under forward bias
by applying a current into the Gate
terminal
• Under reverse bias, looks like
conventional pn junction diode
SCR / Thyristor
Anode
• 4-layer (pnpn) device
• Anode, Cathode as
for a conventional pn
junction diode
• Cathode Gate
brought out for
controlling input
P
N
Gate
P
N
Cathode
ANODE
Equivalent Circuit
ANODE
P
Q1
N
N
BJT_PNP_VIRTUAL
GATE
P
Q2
P
GATE
BJT_NPN_VIRTUAL
N
CATHODE
CATHODE
Apply Biasing
Variable
50V
With the Gate terminal
OPEN, both transistors are
OFF. As the applied
voltage increases, there will
be a “breakdown” that
causes both transistors to
conduct (saturate) making GATE (G)
IF > 0 and VAK = 0.
IF
ANODE (A)
Q1
IC2=IB1
BJT_PNP_VIRTUAL
IC1 = IB2
Q2
BJT_NPN_VIRTUAL
VBreakdown = VBR(F)
IF
CATHODE (K)
Volt-Ampere Characteristic
IF
Holding Current IH
VBR(F)
VAK
Breakdown Voltage
Apply a Gate Current
For 0 < VAK < VBR(F),
Variable
50V
Turn Q2 ON by applying a
current into the Gate
IF
ANODE (A)
Q1
This causes Q1 to turn ON, and
eventually both transistors
SATURATE
VAK = VCEsat + VBEsat
IC2 = IB1
BJT_PNP_VIRTUAL
GATE (G)
IB2
Q2
BJT_NPN_VIRTUAL
If the Gate pulse is removed,
Q1 and Q2 still stay ON!
VG
IF
CATHODE (K)
How do you turn it OFF?
• Cause the forward current to fall below the
value if the “holding” current, IH
• Reverse bias the device
SCR Application – Power Control
XSC1
G
A
B
T
R
25kOhm
Key = a
Vs
170V
120.21V_rms
60Hz
0Deg
60%
When the voltage across
the capacitor reaches the
“trigger-point” voltage of
the device, the SCR turns
ON, current flows in the
Load for the remainder of
the positive half-cycle.
D1
2N1776
Rload
15ohm
C
0.01uF
Current flow stops when
the applied voltage goes
negative.
Input / Output Voltages
Look at the LOAD Current
Conduction time  Conduction Angle = 180 - 
“Firing” time  Firing Angle ()
Average Load Current
i
1 V

sin td (t )

2 R
V

(1  cos  )
2 R

p
L ,A V E

LOA D
i
p
L ,A V E
L
   t an (R C )
1