Join Date: Feb 2010
Originally Posted by Mooly
The 22k's need to allow sufficient base current to flow in order to allow the series pass transistor to turn on hard enough to deliver the load current.
The voltage across the 22k is equal to the supply minus the two 0.65 volt B-E volt drops of the two transistors. For supply voltages over 10 volts we can disregard these and simply say that the 22k resistor will see the full supply, 110 volts in this case.
The gain of the transistor determines how much base current we need.
We know are current source can only deliver 30 milliamps (because of the 0.65v across the 22ohm) and so given an approximate DC current gain for the pass transistor we can work back. The data sheet for the 2SA1837 says DC current gain can be from 100 to 320.
Pick the lower value or 100 and divide the 30 milliamps by that. This gives 0.3 milliamp.
The required resistor would be 110 (voltage across the resistor) divided by required current (0.3 milliamps) which is 366k
Why has the designer chosen 22k ?
We play safe to ensure it will work under all possible conditions, perhaps in temperatures as low as -25C and perhaps with lesser specced (lower gain parts).
Normally I would go for something like an 82k resistor in a situation like this and for that particular part. That's decreased the calculated value by a factor of four and ensures any device of that type will work correctly under any possible usage condition for this particular circuit.
Based on the above, the 22k seems excessively low to me, and as a consequence dissipates more heat than we need.
Wattage of the resistor needs to be (110*110)/R which for an 82k would mean a 0.14w rated part. Again, for reliability we would choose a larger part, perhaps a 0.5 watt.
There is 45V voltage drop on 1k5 resistor, you must calculate with 65V instead 110V rail voltage.