Programmable Unijunction Transistor A programmable unijunction transistor (PUT for short) is also a 4-layer p-n-p-n device with an anode gate G as shown in Figure-1. The anode A and the gate G form a p-n junction that controls the operation of the device. The word programmable in the name simply highlights the fact that the gate voltage is externally controlled.
Anode (A) A IA
p J1 Gate (G)
n
I BP
p-n-p
J2
G
I CP
I CN
p J3 n-p-n
n I BN
IK
Cathode (K) K (a)
(b)
Figure-1: (a) Programmable unijunction transistor, and (b) Its equivalent circuit When the cathode terminal is taken as a reference, the gate voltage is positive with respect to the cathode. The device will remain in the off state as long as the gate voltage is positive with respect to the anode. The device will switch from its off state to on state only when the anode voltage is one diode voltage drop higher than the gate. The voltage current characteristic and the circuit symbol of a PUT are given in Figure-2.
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It is clear that the voltage-current characteristic of a put is very similar to that of a unijunction transistor. For this reason, the unijunction transistor terminology is used to describe its parameters.
IA
+ Negative resistance region VAK IV IP
_ VV
VP
VAK
A + VD _ + G VG _ K (b)
(a)
Figure-2: (a) Voltage-current characteristic of a PUT, and (b) Its circuit symbol Advantages of a PUT over a UJT: (1)
The peak (switching) voltage can be easily varied by adjusting the gate voltage.
(2)
The PUT can operate at low voltages (as low as 3 V). The operation at the low voltage makes it compatible with other digital circuits.
(3)
It has low peak-point current. As the peak-point current decreases, the maximum value of the resistance in the anode circuit increases. For a timing circuit, the time constant can be very large. Therefore, a PUT is very suitable for a long delay.
All the equations we have developed for the UJT are valid for the PUT. When the PUT turns on, the gate voltage VG drops to about 0.5 V or so. The diode voltage drop VD can also be taken as 0.5 V. Therefore, the anode-to-cathode voltage
VAK is about 1 V when the device is on.
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The electrical characteristics of a 2N6027 PUT are given below: Minimum
Typical
Maximum
( VG = 10 V, R G = 1 MΩ )
---------
1.25
2
( VG = 10 V, R G = 10 kΩ )
---------
4
5
( VG = 10 V, R G = 1 MΩ )
---------
18
50
( VG = 10 V, R G = 10 kΩ )
70
150
-----
( VG = 10 V, R G = 200 Ω )
1.5
------
------
-------
0.8
1.5
Peak current, I P µA
Valley Current, I V , µA
Forward Voltage, VF , V
(I F = 50 mA, max )
•
•
•
t=0
R2
R VS
+ VD −
A•
+ _
+ v( t )
G•
C
+ v o (t ) _
r _ •
VG R1
•
Figure-3: Typical application of a PUT
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Typical Application: The circuit given in Figure-3 can be used either for a time-delay or an oscillator circuit. When the PUT is off, the gate voltage is
VG = VS
R1 R1 + R 2
The peak-point voltage at the anode is VP = VG + VD The capacitor begins to charge toward the applied voltage VS . As soon as the voltage across C reaches VP the PUT switches on if R is less than or equal to
R MAX . The maximum value of R is given as R MAX =
VS − VP IP
The capacitor now discharges through the small resistor r and its voltage falls to the valley voltage VV in no time. The PUT will continue to conduct as long as R is less than R MIN where
R MIN =
VS − VV IV
For the PUT to turn off, R has to greater than or at least equal to R MIN . Once the PUT turns off, the capacitor charging and discharging cycle begins again. When the PUT is used as an oscillator, the minimum voltage across the capacitor is VV . For a time delay circuit, the minimum voltage across C can be assumed as zero. The time taken by the capacitor to charge to VP is the charging time and is given as
V − VV TCH = RC ln S VS − VP
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Example: Analyze the operation of a 2N6027 PUT of Figure-3 when
R 1 = 15 kΩ, R 2 = 10 kΩ , R = 500 kΩ, C = 0.2 µF, r = 100 Ω, and VS = 10 V. Solution: 15 VG = 10 =6 V 10 + 15 VP = VG + VD = 6 + 0.5 = 6.5 V RG =
10 × 15 = 6 kΩ 10 + 15
For a 2N6027 PUT, typical value of the peak-point current is 4 µ A.
R MAX =
VS − VP 10 − 6.5 = = 875 k Ω IP 4 × 10 −6
Assuming VV = 0.8 V, and I V = 150 µ A, the minimum resistance is
R MIN =
VS − VV 10 − 0.8 = = 61.3 k Ω IV 150 × 10 − 6
Since R is greater than R MIN and less than R MAX , the circuit must oscillate. The time-period is nearly equal to the charging time. That is,
V − VV TCH = RC ln S VS − VP
10 − 0.8 = (500 × 10 3 )(0.2 × 10 − 6 ) ln = 96.64 ms 10 − 6.5
Finally, the frequency of oscillations is f=
1 1 ≅ = 10.35 Hz T TCH
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Example: Design a time delay circuit with a delay of 30 seconds. Solution: Since we have to select all the components, we make the following selections for the circuit in Figure-3: VS = 10 V, VG = 10
R 2 = 27 k Ω , R 1 = 47 k Ω , and r = 47 Ω
47 = 6.35 V 27 + 47
VP = 6.35 + 0.5 = 6.85 V RG =
27 × 47 = 17.15 k Ω 27 + 47
For a 2N6027 PUT, typical valley-point values are I V = 18 µ A
R MIN =
and
VV = 0.8 V
VS − VV 10 − 0.8 = = 511 k Ω IV 18 × 10 − 6
For a time-delay circuit, R must be less than R MIN . So, let us select R = 390 k Ω . By setting TCH = 30 s in the charging time equation, we get 10 − 0.8 30 = (390 × 10 3 ) C ln 10 − 6.85
or
C = 71.76 µ F
Let us select a standard value for C as 100 µ F. We can now recalculate R as 10 − 0.8 30 = R (100 × 10 − 6 ) ln 10 − 6.85
or
R = 280 k Ω
We can use 270-k Ω resistor in series with a 10-k Ω potentiometer for R . The potentiometer will help us adjust the value of R to obtain the desired delay. Since the capacitor charges to 6.85 V, the peak value of the output pulse will be nearly equal to 5.85 V allowing for a 1-volt drop across the device. The resistance r controls the duration of the pulse.
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Problems: PUT
1.
A student built the circuit of Figure P1 using a programmable unijunction transistor and found it oscillating. Determine its charging time. If the discharging time is ignored, what is its frequency of oscillation? Sketch the voltages v C ( t ) and v K ( t ) .
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