Objective:
To calculate the base resistance (RB), in a simple transistor circuit.
Equipment:
- 2 digital voltmeters 6. locktronics 10K potentiometer
- 2 analogue student meters 7. locktronics 1K resistor
- 1 50/100mA shunt 8. locktronics special BFY51 resistor
- locktronics base board 9. links & connection leads
- 1 fairy light 10. 12V power supply
Procedure:
Measure the HFE of the selected transistor and write it on the line below together with the cradle number.
1 (HFE = 96)
Construct the circuit shown below on the locktronic base board. Make sure that the potentiometer and the 1k are placed as shown since this will provide protection to the base-emitter circuit. RB(a) is not provided since its value is determined in response to the calculation.
Calculation:
Calculate the sum of RB(a),(b),(c) for the component configuration you have assembled so that the transistor just saturates and switches on the lamp. Refer to the above diagram and previous notes. Show all-working below.
Ic (sat) = (VCC – VCE) / RL
Ic (sat) = (12 – 0.35) / 170 = 0.069A
IB = IC / HFE = 0.069 / 96 = 7.19 x 10-4A
R = (VCC – VBE) / IB
R = (12 – 0.74) / 7.19 x 10-4 = 15666Ω
RB(a) = 15666 – 1000 – 10000 = 4666Ω
Since RB(a),(b),(c) are known values, the calculation will reflect this before your final choice for RB is made.
Now complete the following tests:
TEST 1: Adjust the potentiometer (RB(c)) to achieve the calculation as measured on an ohmmeter. Apply the power supply and record the readings of the various meters. If VCE (sat) (or a lower value) can’t be achieved, is RB(a) to high a value?
IC (sat) = 68.1mA
IB (sat) = 0.705mA
VCE (sat) = 0.74v
VBE (sat) = 0.24v
TEST 2: State the practical value of HFE using the recorded test figures. State the percentage error between this and the original value of HFE.
HFE = IC / IB
HFE = 68.1 / 0.705 = 96.6
% Error = (Calculated HFE / Measured HFE) x 100%
% Error = (96.6 / 96) x 100%
% Error = 100.625 – 100 = 0.625%
TEST 3: Determine the maximum value that VCE may rise to, before there is a significant fall in the collector current.
VCE = 1.90v
QUESTION 1: When RB(c) was adjusted, at what point would a power rating for the transistor need consideration.
More current flows as the voltage is decreased. Therefore, the lamp will brighten. However, as the current increases, the transistor starts to heat up and could well blow. The power rating needs to be considered before the transistor can heat up.
QUESTION 2: Apart from the meter tolerances, which part of this practical work has the potential to give rise to significant differences between calculated and practical results.
The temperature in the room can increase or decrease and this will have an effect on the transistor’s HFE value. As the circuit is handled, the transistor will have an increase in temperature as a result of the body heat from the handler. Bad connections on the circuit will also effect the results. If the circuit is assembled inaccurately, there will be a greater resistance.
SECTION C (3d & e) - Measuring the Gain Characteristics and the Bandwidth – AC Testing
Once the circuit amplifier had been assembled and all measurements recorded, calculations to add capacitors were done. The calculations were as follows:
Lowest frequency, f = 50Hz
= 1 / T
Period, T = 1 /f
= 1 / 50
= 20 × 10-3
= 20ms
As T = CR
The time constant is usually taken as five times the period of the lowest frequency.
So 5T = CR
5 × 20 × 10-3 = C × 2000
Therefore C = 50μF (a 47μF was used instead)
As R3 = 200Ω
5 × 20 × 10-3 = C × 200
Therefore C = 500μF (a 470μF was used instead)
The following capacitors were used:
A 50μF (a 47μF was used instead),
A 500μF (a 470μF was used instead),
A 10nF capacitor
Once the capacitors had been calculated and assembled, the gain and bandwidth needed to be obtained. A signal generator was set to produce a sine wave. It was connected to channel 1 of the oscilloscope to obtain a reading of 10mV peak-to-peak. The power supply was adjusted to apply 10V to the constructed amplifier. The input lead from the circuit amplifier was also connected to the signal generator. Channel 2 of the oscilloscope was attached to ground and to the 10nF capacitor to measure the output of the amplifier.
Input Voltage: 10mV
Output Voltage: 1.2V
The amplifier gain is equal to the ratio of the input to the output voltages.
e.g. gain = output / input
= 1.2/10 × 10-3
= 120
The bandwidth for current or voltage amplifiers is defined as the range of frequencies at which the gain does not drop down further than 70.7%. The cut-off frequencies are the points at which fulfil the requirements for the amplifier gain. Below is a typical frequency response curve:
The bandwidth is calculated using the formula:
Bandwidth = Upper cut-off frequency – Lower cut-off frequency
Upper cut-off frequency: 21kHz
Lower cut-off frequency: 29Hz
Bandwidth = 21000 – 29
= 20971Hz or 20.97kHz
The bandwidth of the circuit amplifier is 20.97kHz
SECTION B (2a) - Specific function of two different types of transistor within a genuine circuit
- Metal Oxide Semiconductor Field Effect Transistor (MOSFET)
Below is a diagram of a typical MOSFET common-source amplifier. Low power MOSFETs are useful for radio frequency circuits. This is because they have a rapid response time. Amplifiers based on power MOSFETs are useful for power-control circuits. They have better thermal stability than bipolar transistors (less subject to thermal runaway).
- Unijunction Transistor (UJT)
This is one of the simplest oscillator circuits based on a unijunction transistor. In the above circuit, the potential difference across the capacitor gradually builds up and is released almost instantly. This type of oscillator is known as a ‘relaxation oscillator’.
