![]() ![]() Post added at 11:11 - Previous post was at 11:10. Thank you for the reply you made easy to understand but i am looking forward for your next explanation So I'll continue tomorrow unless someone else had already done the job by then. But it's way past midnight here now and I've been missing a lot of sleep lately. Regarding your last question, I don't want to give you some offhand - and possibly misleading - answer. You will gain a better understanding as you get more and more involved in electronics. The above explanation is a condensed and somewhat simplified one. For switching applications, taking it as 0.7V for Ic of several mAs and 0.8-1.0V for some hundreds of mAs is usually close enough in practice. Besides, base current is usually supplied from a much higher voltage than Vbe so that variations in the actual value are swamped by the overall voltage. Fortunately, it does not vary too much from unit to unit. For practical designs (as opposed to textbook derivations which are often not of much use), base-emitter voltage is best estimated by experience gained from looking at various datasheets that include graphical presentations. Vbe and Vbe(sat) have a more complex relation to Ic and Ib. Scale that down to 20mA and you get about 50mV max for a 2N2222 and 40mV for a 2N2222A. So, for your circuit, deduce Vce(sat) from the value closer to your 20mA load, that is, Vce(sat) for Ic = 150mA. Vce(sat) is roughly proportional to collector current. It is an inherent characteristics of the particular transistor and cannot be reduced. The voltage drop is significant and, depending on how heavy the load is and other circuit conditions, it can be a substantial portion of the total supply voltage. However, when a transistor, especially a bipolar junction transistor (BJT) like the 2N2222 is used as a switch, it's not as close to perfect as a mechanical switch. It always has a small resistance and the load current, in flowing through the switch contact point, causes a small voltage drop in the switch so that not quite 100% of the supply voltage reaches the load. But a real-world switch is never quite perfect. This is very nearly true of a switch in good condition. ![]() That is, when the switch is closed (on), 100% of the voltage goes to the load. When we use a mechanical switch to control a voltage source to a load, we generally think of the switch as perfect. First of all, you should understand that the voltages given on that part of the datasheet you reproduced are saturation voltages Vce(sat) and Vbe(sat), not just Vce and Vbe. ![]()
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