load line and determine the operating point. 3 (i) shows that a silicon transistor with β = 100 is biased by base resistor method. (ii) When R B is made equal to 50 kΩ, then it is easy to see that base current is doubled i.e. (i) Referring to Fig.2 (ii) and applying Kirchhoff ’s voltage law to the circuit ABEN, we get, It may be noted that negative terminals of the power supplies are grounded to get a complete path of current. Here, we need show only the supply voltages, + 2V and +9V. The same circuit is shown in a simplified way in Fig. 2 (i), biasing is provided by a battery V BB (= 2V) in the base circuit which is separate from the battery V CC (= 9V) used in the output circuit. (ii) If R B in this circuit is changed to 50 kΩ, find the new operating point. (i) Determine the collector current I C and collector-emitter voltage V CE . 2 (i) shows biasing with base resistor method. ∴ Base voltage (signal voltage) = Collector current / (5 mA/V )= 3 mA /( 5 mA/V) = 600 mV Q3. Fig. Now Collector current / Base voltage (signal voltage) = 5 mA/V allowed collector current, i C =12 V /R C = 12 V/ 4 KΩ = 3 mA allowed voltage across R C = 13 − 1 = 12 V What is the maximum input signal if β = 100 ? Given V knee = 1V and a change of 1V in V BE causes a change of 5mA in collector current. And the emitter resistor, R E,Īllows for stability of the gain of the transistor, despite fluctations in the β values.During the positive peak of the signal, i C = 1 + 1 = 2mAĪnd during the negative peak (point B), i C = 1 − 1 = 0 mA Q2. A transistor employs a 4 kΩ load and V CC = 13V. The resistors help to giveĬomplete control over the voltage and current that each region receives in the transistor. Advantages of the Voltage Divider BiasĪgain, voltage divider bias is the most popular and used way to bias BJT transistors. So there must be balance in theĭesign of the value of R E. However, realize that the larger R E is, the moreįor amplification in the circuit. Is even more stability against the β of transistors. Milliampere difference in the output current, I E. Even though there is a difference of 50 of β in the transistors, there is less than a half of The answer is, the transistor output current doesn't vary much despite the large difference in β values of the In changes to the β or will it vary little and still be stable? What will the new calculation be, and will the output current of the transistor swing largely But let's say we have to swamp out that transistor and the In gain of the output current of a transistor circuit.įor the above transistor, the β was equal to 100. We do this by carefullyĬhoosing the emitter resistance, R E, which provides stability against differences in β. We must design them in a way that produces the same gain despite fluctuations in the β values. Therefore, when we are designing transistor circuits where we want roughly the same gain in all of them, There is no way to replicate the same exact βs across of a transistor, its gain or amplification factor, can vary by large amounts from transistor to transistor,Įven if they're the same exact type from the same batch.
The R E provides stability in gain of the emitter current of a transistor circuit. I C≈ I E How Emitter Resistor, R E, Fights Against the Instability of β The collector current I C is approximately equal to the emitter current. Then, we calculate for the emitter current using the following formula: We calculate R B below, which we will use the next calculation for I E. The base supply voltage, V BB, is calculated by: Below is a typical BJT receiving voltage divider bias:įor the circuit above, we're going to assume that β=100 for the transistor.
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