![]() We will calculate the base current for switching the load.Īs we know, when the load will be switched on, the collector current will be the load current. The first Transistor gain will be 30 and the second transistor gain will be 95. So, the total collector current IC is a combinational gain of individual transistors gain.Ī 60W load with 15V input voltage needs to be switched using two NPN transistors, creating a Darlington pair. Now as previously, we have seen that IC = β 1IB 1 + β 2IB 2 As, IB2 or IE2 = IB1 (β1 + 1) We can change this relationship further with IC 1 + IB 1Ĭhanging the IC1 as we did previously, we get β 1IB 1 + IB 1 IB 1 (β 1 + 1) IB2 is controlled by the emitter current of T1, which is IE1. Let’s see how the current gain is the multiplication of the two transistor’s current gain. In this situation, the current gain is unity or greater than one. Now as discussed above, we get the Collector current β * IB 1 So, the total current gain ( β ) is achieve, when the collector current is β * IB as hFE = fFE 1 * hFE 2Īs two transistors collector is connected together, Total Collector current (IC) = IC1 + IC2 So, the base current IB1, which is controlling the T1 is controlling the current flow at T2’s base. The first transistor T1 providing the required base current (IB2) to the second transistor T2’s base. The two NPN transistors T1 and T2 are connected together in an order where, T1 and T2’s collectors are connected. ![]() In the above image, two NPN transistors created a NPN Darlington configuration. The overall current gain of the Darlington pair will be- Current gain (hFE) = First transistor gain (hFE 1) * Second transistor gain (hFE 2) In the below image we can see two PNP or two NPN transistors are connected together.
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