| Electronics > Beginners |
| Darlington pairs resistors. |
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| David Hess:
When driving a Darlington, the driver circuit can remove charge from the drive transistor but not the output transistor through the base-emitter junction so the base-emitter shunt resistor on the output transistor is absolutely needed both to remove charge and leakage. In practice Darlingtons are intended to operate with driver circuits which cannot remove charge so they include base-emitter shunt resistors on both transistors and really that is a good idea either way. |
| Simon123:
Well power you can also calculate, for example 2N3055 Current is about 1mA, Ube (which is constant 0.6V), so you do the math, 0.6V*1mA=0.6mW ... |
| exe:
--- Quote from: Simon123 on July 27, 2018, 04:42:45 pm ---2N3055 Current is about 1mA, Ube (which is constant 0.6V), so you do the math, 0.6V*1mA=0.6mW ... --- End quote --- That's just to remove leakage current. But to make it switch-off faster current should be bigger than 1mA. In practice resistor is about 100 Ohm or so (sometimes it's built-in 60 Ohm). So, it's a bit bigger than that. On my Sziklai pair it's noteceably warm (I'm building this PSU: http://www.bramcam.nl/NA/NA-01-PSU/NA-PSU-50.GIF). |
| sureshot:
Thank you for the help, I'm going to go with R1 5K and R2 60R that seems like a place to start. If it acts odd I might have to change something. That psu looks like a real task lol, not one for me at this stage. |
| David Hess:
--- Quote from: exe on July 27, 2018, 06:09:15 pm --- --- Quote from: Simon123 on July 27, 2018, 04:42:45 pm ---2N3055 Current is about 1mA, Ube (which is constant 0.6V), so you do the math, 0.6V*1mA=0.6mW ... --- End quote --- That's just to remove leakage current. But to make it switch-off faster current should be bigger than 1mA. In practice resistor is about 100 Ohm or so (sometimes it's built-in 60 Ohm). So, it's a bit bigger than that. On my Sziklai pair it's noteceably warm (I'm building this PSU: http://www.bramcam.nl/NA/NA-01-PSU/NA-PSU-50.GIF). --- End quote --- I usually figure on diverting at least 1/10th of the base current through the base-emitter shunt resistor. In a Darlington this just results in the drive transistor conducting that much more which is not a problem. |
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