The 7mA driven into the base of the small transistor will be "wasted". Using a Darlington configuration, if the load draws 10A, 9.75A will flow through the collector of the large transistor and 250mA will flow through the small transistor into the base of the large one. To get that 250mA, one would need to drive the base of the small transistor with 7mA. If one needs to drive a 10 amp load and wants to avoid assuming a beta greater than 40, one will need to be able to drive the base of the large transistor with 250mA. In the Darlington configuration, the base current of the larger transistor helps to drive the load and is self-regulating. ULN2k series (I don't have the details of access any more, but I did see some of this way back when doing my studies in this stuff). I'm pretty sure this is what's done on the integrated multi-Darlington arrays e.g. Connecting the emitter of the smaller to the base of the larger is pretty trivial. Lastly, NPN Darlingtons can be easily constructed on an integrated circuit effectively as a single meta-transistor they share the same collector region but have different embedded base/emitter regions (with the size difference I mentioned earlier). This is without the need for pairing multiple transistors in parallel, which can cause uneven current splitting due to device differences, even on integrated circuits. Integrated Darlington pairs are configured in physical layout with a significant size differential, such that the main drive transistor has a much larger junction area than the first, allowing for much lower C-E on resistance for lower drive currents and much higher max current handling capability. The other transistor is therefore carrying 99%+ of the driven current. The one where the input signal arrives on the base carries maybe 1% of the current through it (assuming a beta of 100 most integrated circuit NPNs have betas much higher (~250) so the percentage is therefore even lower). Regarding other parts of your question: Darlingtons don't share the current between the two transistors 50-50. Regarding the use of NPN rather than PNP, Michael Karas answer is correct: you want ground-referenced control signals because the N-type transistors generally have better characteristics than the P-type equivalents. To clarify, one of the questions I'm asking is: Why can't we place this NPN transistor as-is above the load? Or, for that matter, place a PNP Darlington below the load? And also, why do Darlingtons even exist, when a complementary pair looks to be a cleaner solution? In the above example, it seems more sensible to either (1) place the load below the transistor (2) use a PNP Darlington or even better (3) use a complementary PNP pair as shown here: Granted, we might want to share the current between two transistors but in which case, please note that the second transistor is still carrying the full load (half via the C-E path, and half via the B-E path).įor that matter, why are transistors most commonly used for sinking current anyway rather than driving it? I've never understood that. Wouldn't it make more sense to use PNP for the purpose? This would avoid shunting the load current through both junctions at once. Very important: When we add the base-emiter voltage of the first transistor B1 to E1 (0.7 volts) plus the base-emitter voltage of the second transistor B2 to E2 (0.7 volts), we get as a result a voltage drop of 1.4 volts between the base and the emitter of the Darlington transistor. VB1E1 + VB2E2 = 0.7 + 0.7 = 1.4 volts.I notice that NPN Darlington transistors are commonly used to sink current. Darlington transistors are widely used in circuits where it is necessary to control large loads with very small currents. This seems to be a very big gain (the ideal one) but it is actually a lower gain. If we have two transistors with gain of 100 each (ß = 100), connected as a Darlington transistor using the above formula, the final theoretical gain would be: ß2 x ß1 = 100 x 100 = 10000. Darlington Current Gainįrom the Darlington transistor gain equation: IE2 = ß2 x ß1 x IB1, we can see that the current gain is much greater than the one of a single transistor, since it takes advantage of the current gain of the two transistors. Replacing the value of IE1 on the last equation (see equation (1)) we obtain the Darlington transistor gain equation: IE2 = ß2 x ß1 x IB1. Using the equation (2) and the equation (3) we obtain: IE2 = ß2 x IB2 = ß2 x IE1. From the image, we see that the emitter current of the T1 transistor is the same as the base current of the T2 transistor.
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