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I used this for understanding why we should make a 2-transistor current source instead of a 1-transistor one.
Main results: 2 transistor is much less dependent on input voltages, and can more stably deliver power over a larger range of loads.
Use the switches on top and bottom to select one of two current sources, and notice the effect when switching supply voltage. The more complicated dual transistor circuit uses the 'about 0.7 volts collector to base voltage drop' and the same again for base to emitter of the rightmost transistor to regulate the voltage drop over the 100 ohm 'tuning' transistor. Thanks to that, we get a source current of 0.7/100 A = 7 mA. This is practically independent of the source voltage.
On the more simple one-transistor circuit, we have sized the voltage divider and regulating-resistor to also deliver a constant 7 mA. However, the 1.5:7.5 ratio only works with the 24 volts input. If we change over to 16 volts (keeping the 100 ohm load), we have very different results.
The rightmost is also seems less dependent on the load - it provides a little more current to the 3.5 kOhm load, 35x the 100 ohm load. This is simply Ohms law: e.g. at 16 volts, the left simple system sees 3500 + 450 ohms => 16/3950 = 4.0 mA, while the more complicated system only sees 3500+100 ohms = 4.4 mA.
Trying to size the simple source with a voltage divider that allows use of a 100 Ohm limiting resistor is very challenging as selecting proper values starts messing with the other properties of the transistor (current/voltage ratios) which are less relevant with higher collector/base voltage differences..
Finally, the right-most resistor on the base/collector of the complicated circuit's transistors is chiefly to 'weakly pull them to ground' and not leave them floating, but you will notice that changing it too much has a detrimental effect on the voltages and thereby the currrents. Too high = to much current, too low = too little current. It needs to be around 33 kOhms for these source voltages to get the proper voltage drop across the transistors.
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