![]() This gets amplified so that 0.5 30=15mA flows into the collector and out of the emitter from the power supply. In this configuration lets say you put 0.5mA into the base of the first transistor. The second transistor is used to drive the load. Its collector is connected to Vcc, and its emitter is connected to the base of a second transistor. The base of the first is connected to the the arduino via a resistor to control the base current. Instead what if you used two transistors for each load. That would require you to source ~17mA form every pin which would go far beyond the maximum power dissipation of the chip. The arduino can source 20mA which means for this gain you can have a maximum collector current of around 600mA.īut what if you need to drive say 20 x 0.45A loads individually. This is fine for smallish control signals such as that from an arduino pin. What this means is that for every 1mA that flows into the base, 30mA can flow into the collector. Though this is somewhat meaningless without any specific part.Īn example - you haven't given any information on purpose, so I will make up a unrealistic example to demonstrate.įor a BJT (Bipolar Junction Transistor), the current that flows into the collector is proportional to the current that flows into the base.įor example transistor x has a Beta, or current gain, of around 30. Note : r′ e1 and r′ e2 have been assumed to be equal and each one equal to r′ e.Multiple transistor stages means more noise, and possibly a slower response. (slightly less than unity) as in case of emitter follower On an approximate basis voltage gain can be determined from the following relation also Voltage gain is closer in magnitude to one than to zero. That the output potential is the input potential less the base-to-emitter potential of each transistor. ![]() Voltage Gain:Applying Kirchhoff’s voltage law to the circuit shown in Fig. If high impedance source is directly connected to a low impedance load, most of the signal voltage will be dropped across the high impedance of the source and remaining source signal available may not be able to drive the load.Ĥ. Z out2 is quiet smaller than Z out1 i.e., output impedance is lowered.īecause of this, Darlington amplifier can be used to isolate high impedance source from low impedance load. The ac Thevenin’s impedance at the input r Th = R s||R 1||R 2 because by shorting all voltage sources R 1 and R s comes in parallel with R 2 Output Impedance: The output impedance Z out can be determined directly from the emitter equivalent circuits as follows:Īlternatively the output impedance Z out can be determined as follows : If there is a load resistance R L coupled to the emitter of second transistor, thenģ. This input impedance is very high because of the product of two betas, So input impedance looking into the base of first transistor, Z in2 is also the output impedance while looking into the emitter of first transistor. Where r′ e2 is the ac emitter resistance of second transistor. Since Z in2 = h fe2R E, and 1/h oe1 will appear in parallel in the small signal equivalent circuit, soĪlternatively input impedance can be obtained as follows: Input Impedance: Since Z in2 = h fe2R E is the emitter resistance of the first stage, the input impedance to the first stage is It means Darlington amplifier behaves like a single transistor having current gain equal to β 2.Ģ. We already discussed that, it was found that for single stage grounded emitter transistor amplifier, where 1/h oe was considered,Īpplying the above equation to this situation,įor h oeh feR E ≤ 0.1, a fairly good approximation (within 10%) isĪlternatively current gain can be obtained as follows: For the Darlington Amplifier circuit the input impedance Z in2 is close enough in magnitude to 1/h oe to necessitate considering the effects of h oe1. We have seen that 1/h oe, could be eliminated in the majority of the cases because the load impedance Z L << 1/h oe. ![]() The “fly in the ointment” is the closeness with which Z in2 compares with 1/h oe1. On a good approximate basis, these equations cannot be applied to the first stage. The Darlington amplifier circuit in its small signal ac form will appear, as shown in Fig. The biasing analysis is similar to that for one transistor except that two V BE drops are to be considered.
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