VHDL Adavanced VHDL Arithmetic-Circuits

VHDL Concurrent-Statements

Counters

Flipflops

Logic-Circuits

Logic-Gates

Sequential-Statements

Shift-Registers

Experiment No. 9

Title : Transistor application as a switch.

Pre-lab Requisites :

1. Operation of transistor as a switch.

2. Datasheet specifications of BJT.

Objectives :

1. To operate transistor as a switch.

2. To verify its regions of operations when operated as a switch.

Apparatus :

Sr. No.

Description

Specifications

Qty. Reqd.

1.

DC power supply

0-20 V

02

2.

Signal generator

0-1 MHz

01

3.

DMM

01

4.

Connecting wires

10

Theory :

While transistors have many uses, one of the less known uses is the ability for bipolar transistors to turn things on and off. While there are limitations as to what we can switch on and off, transistor switches offer lower cost and substantial reliability over conventional mechanical relays.

When a bipolar junction transistor (BJT) is used in any circuit, its function is determined by the devices characteristic curves.

For a BJT there are input, output and transfer characteristic curves, the most useful being the output characteristic curves. The output curves dictate the range of collector-emitter voltage Vce for variations in collector current, Ic. For use as an amplifier, biasing is arranged so that the linear part of the output curves (the at most horizontal sections) is used. If the circuit uses the transistor as a switch, then biasing is arranged to operate on regions of the output curve known as saturation and cut-off. See the diagram below.

In ‘cut-off’ region, the operating conditions of the transistor are zero input base current, zero output collector current and maximum (supply rail) collector voltage. In “saturation”, the BJT will be biased so that the maximum amount of base current is applied, resulting in maximum collector current flow and minimum collector emitter voltage. In both the cut-off and saturation, minimum power is dissipated in the transistor.

Knowing the circuit load current and operating voltage, the load line can be constructed. Imagine a transistor designed to switch a 20 mA load, the supply voltage is 5 volts DC. When the transistor is off, the Ic will be zero and Vce will rise to supply voltage (5 V). This is point A on the diagram above. When the transistor is on, Ic will be 20 mA and Vce will be small (close to zero). This is point B in the diagram. This line is called the load line and shows that the transistor can be operated anywhere on this line (by choice of appropriate bias current). However, for use as a switch the device must work in the saturation and cut-off regions of the output curve. The bias circuit should be designed to work with the minimum value of hfe for the given transistor.

Suppose a BJT with a 5 V supply is designed to switch a 5 V, 20 mA lamp on and off. The transistor chosen is from a batch with variations in hfe from 100-500. The switching configuration is for common emitter; the bias circuit is shown below. Find a value for Rb to work with any transistor in the same gain group.

As the transistor chosen may have any hfe between 100 to 500 then the minimum current gain is chosen (100). The collector current is 20 mA, the required base current is therefore:

hfe=Ic/Ib

Ib = Ic/hfe(min)=20/100=0.2 mA

The value of Rb can now be found. As the switching input, Vin is 5 V and the base emitter voltage of the transistor, Vbe is 0.6 V then 4.4 V is developed across Rb. As a base current of 0.2 mA is required, then

Rb = 4.4/0.2=22k

A transistor with a gain higher, equal to or higher than 100 will easily work and light the lamp. The collector emitter voltage of the transistor will be very low (around 0.2 V) the power dissipated in the transistor is also low Ic*Vce=4mW, and almost full power is developed in the load.

You will always know the load current your circuit needs. Therefore use the minimum value of hfe for the transistor in your patch (found from the manufacturer’s data sheet) or catalogue. Calculate the bias current to reach this minimum value. In practice and for heavy load currents and power transistors you may allow 5 times more base current than actually required. This will ensure any transistor within a gain group reaching saturation. The current gain hfe is lower in some power transistors at very high load currents. Therefore it may be wise to calculate the base current Ib and allow the actual value to be two or three times higher. The base emitter voltage Vbe which varies between individual devices should be taken as the highest value. This is generally 0.6 or 0.7 V with small signal transistors, but can be as high as 0.8 Von some power transistors. In saturation a heat sink is rarely required as little power is developed in the transistor. However in a power supply or other circuit where a transistor may be required to control large variations in current and voltage then significant power may be developed. If the power dissipation of the device is exceeded then it will be destroyed. In practice allow for the worst combination of currents and voltages and calculate accordingly.

Procedure :

1. Make the connections as shown in circuit diagram.

2. First of all, connect LED in series with a collector resistor (Rc) between Vcc and collector terminal of BJT.

3. Apply Vcc as +12 V w.r.t. Gnd. Slowly increase the base bias voltage (input voltage to BJT say “Vi”) from 0 to +5 volts till the LED glows with sufficient intensity.

4. Note down Vi. Measure Vbe, Vce, Ic, Ib drop across LED i.e. Vled and note it down. Check in which region the device is operating? Comment.

5. Apply a square wave of 1 Hz frequency, Vi volts peaks as input and see that LED is blinking.

6. Now in place of LED, connect primary coil of electro-magnetic relay in the collector circuit.

7. Again increase Vi till relay operates and then repeat step 4.

Observations :

For LED connected in collector circuit

Vi (Volts)

Vbe (Volts)

Vled (Volts)

Vce (Volts)

Ic (mA)

Ib (µA)

For Relay connected in collector circuit

Vi (Volts)

Vbe (Volts)

Vpri. (Volts)

Vce (Volts)

Ic (mA)

Ib (µA)

Conclusion :----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

FaceBook
Likes
Additonal Information