Regarding Darlington pair problem

I tried to do the steps... Ie = 1.98/100 = 0.0198A Beta for 1st transistor and 2nd one is 2 So total current gain= (beta1 +1)(beta2 +1) Ie/Ib = 3*3= 9 Ib= 0.0198/9 =2.2mA But answers are not matching... someone please help me what's wrong with my calculations

• The screenshot is a little cropped, is the beta value for the transistors 2? – alphasierra Sep 25 '20 at 9:20
• Yes... please help me out, I have no idea what's wrong with my calculations – Camila's voice Sep 25 '20 at 9:58
• Ie = 1.98/100. This is probably wrong. Isn't the voltage across the 100 ohm resistor is 500-1.98 ? If you use that you will get Ic, I think. – AJN Sep 25 '20 at 12:19
• The beta of your transistors does not appear to be 2. On the edge of the screenshot there are some pixels. Using a beta of 12 yields an answer from the list of available options. – alphasierra Sep 25 '20 at 17:23

The emitter current is not 1.98V/100Ohms. From the wording of the problem the darlington pair is sitting above the resistor (between the positive rail and the resistor). To get the emitter current Ie=Ib+Ic. To calculate the load current subtract the voltage drop from the darlington pair, and divide by the load resistance: I_Load=(V_Supply-V_DarlingtonDrop)/R_Load. That gives you the emitter current. From there you can calculate the base current using the usual darlington pair equations to write the collector current in terms of the base current Ic=(B1+B1B2+B2)Ie, and substitute into Ie=Ic+Ib. This site contains some useful tutorials on the subject.