One stop reference for all design related stuff.
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First up, the most simple, Saturated switch using BJT and Darlington. Notes in the attached figure.
(a) Saturated Switch using BJT
(1) The ratio of Ic/Ib < hfe (min) of the selected transistor.
(2) take care that calculated Ic is less than Ic max which can be found in the datasheet.
(3) How did we determine ic/ib < hfe for saturation?
Consider the simple circuit shown in figure, with one base resistor Rb, One collector load RL ,
Let's assume for example RB = 1M, RL = 1k, B1 = 9V, hFE = 300 and VBE = 0.6V.
The voltage across RB will be VS-VBE=8.4V, so IB=8.4/1M=8.4µA. IC=IB·hFE=8.4µA·300=2.52mA. So the voltage across RL will be 2.52V.
Now, What will happen if in the example above, RB=100k instead of 1M?
IB=8.4/100k=84µA. You may expect that IC will be 84µA·300=25.2mA, but that isn't possible since the voltage across RL would be 25.2V which is more than B1.
IC,max in this circuit is B1/RL=9/1k=9mA. So even if IB=84µA, IC will be 9mA. IC/IB=107, which is less than hFE. In such a case, when IC/IB < hFE, we say that the transistor has become saturated and can be considered as a closed switch (between C and E).
(b) Saturated Darlington
As explained in figure attached, if Ib loads the weak base supply, then a more powerful high beta transistor called darlington can be used. This will saturate even with tens of microamps of base current and won't load the weak base power supply.
Post all design related articles in continuation by editing this post
First up, the most simple, Saturated switch using BJT and Darlington. Notes in the attached figure.
(a) Saturated Switch using BJT
(1) The ratio of Ic/Ib < hfe (min) of the selected transistor.
(2) take care that calculated Ic is less than Ic max which can be found in the datasheet.
(3) How did we determine ic/ib < hfe for saturation?
Consider the simple circuit shown in figure, with one base resistor Rb, One collector load RL ,
Let's assume for example RB = 1M, RL = 1k, B1 = 9V, hFE = 300 and VBE = 0.6V.
The voltage across RB will be VS-VBE=8.4V, so IB=8.4/1M=8.4µA. IC=IB·hFE=8.4µA·300=2.52mA. So the voltage across RL will be 2.52V.
Now, What will happen if in the example above, RB=100k instead of 1M?
IB=8.4/100k=84µA. You may expect that IC will be 84µA·300=25.2mA, but that isn't possible since the voltage across RL would be 25.2V which is more than B1.
IC,max in this circuit is B1/RL=9/1k=9mA. So even if IB=84µA, IC will be 9mA. IC/IB=107, which is less than hFE. In such a case, when IC/IB < hFE, we say that the transistor has become saturated and can be considered as a closed switch (between C and E).
(b) Saturated Darlington
As explained in figure attached, if Ib loads the weak base supply, then a more powerful high beta transistor called darlington can be used. This will saturate even with tens of microamps of base current and won't load the weak base power supply.
