Electronics > Beginners

Transistor Switching

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tpowell1830:
Hero999, PWM is a better choice because with a current circuit as you have shown, the full sweep of the pot will have to be precisely calculated as far as your limiting base resistance. This will be a hunting game every time that you add a new LED or simply replace the LED with the same type that you had working well with the circuit. Your circuit will work with some fidling around with the resistor values, but as soon as you replace the LED with another one of the same type or a different type, your limiting resistor will probably need to be changed again. It is very difficult to get the deadband out of the adjustment pot.

With PWM, the LED will change brightness from dark to bright every time, no fiddling around, no deadband. Change or add LED, no worries.

Hope this helps...

Zero999:

--- Quote from: tpowell1830 on July 09, 2018, 08:37:26 am ---Hero999, PWM is a better choice because with a current circuit as you have shown, the full sweep of the pot will have to be precisely calculated as far as your limiting base resistance. This will be a hunting game every time that you add a new LED or simply replace the LED with the same type that you had working well with the circuit. Your circuit will work with some fidling around with the resistor values, but as soon as you replace the LED with another one of the same type or a different type, your limiting resistor will probably need to be changed again. It is very difficult to get the deadband out of the adjustment pot.

With PWM, the LED will change brightness from dark to bright every time, no fiddling around, no deadband. Change or add LED, no worries.

Hope this helps...

--- End quote ---
I concede your point about the deadband, but the LED current will be fairly independent of the forward voltage. The current through the transistor is controlled by varying the base voltage. The voltage across the emitter resistor and therefore the current, is proportional to the base voltage, minus the base-emitter voltage. Even replacing the LED with a short circuit will not increase the collector current by much.

IC = (VB-VBE)/RE

The only issue I have with the circuit now is the potentiometer's value of 100k is be a bit high, as the impedance looking into the base will be about 7k, assuming an Hfe of just over 200. Reducing it and the resistor in the lower leg, by a factor of 20 would help, but the original poster might only have a 100k pot, which will probably do, even though it isn't optimal. It also doesn't need to be perfectly linear, as the operator will compensate to some degree, when adjusting the pot.

MK14:

--- Quote from: Hero999 on July 09, 2018, 10:35:27 am ---The only issue I have with the circuit now is the potentiometer's value of 100k is be a bit high, as the impedance looking into the base will be about 7k, assuming an Hfe of just over 200.

--- End quote ---

Two points.
One is the OP, seems to say he has a 10K pot, rather than a 100K one.


--- Quote from: Oslaw on July 07, 2018, 01:25:03 pm ---Both. I did just that and still same result. I am using a 9v battery. Below are pictures of the potentiometer tuned to both ends. At level 0 Ohm LED is dull and at 10k Ohms bright(at any resistance level above 0 ).

--- End quote ---

Secondly, the 7K impedance looking into the base, is only at full brightness, when the pot will be a very low resistance (as regards, the top two terminals of the pot, the other will be at nearly 100K, of course).
As the pot then decreases the current/brightness, into the base, the current flow reduces, so the effective impedance into the base, increases.
E.g. At 10% brightness = 10% of base current at full brightness = 10 x 7K = 70K very approximately, since Hfe will probably increase as the collector current drops.

tl;dr
Assuming OP has 10K pot, it should work reasonable well in that circuit, assuming there are not other issues.

Zero999:

--- Quote from: MK14 on July 09, 2018, 10:57:08 am ---
--- Quote from: Hero999 on July 09, 2018, 10:35:27 am ---The only issue I have with the circuit now is the potentiometer's value of 100k is be a bit high, as the impedance looking into the base will be about 7k, assuming an Hfe of just over 200.

--- End quote ---

Two points.
One is the OP, seems to say he has a 10K pot, rather than a 100K one.


--- Quote from: Oslaw on July 07, 2018, 01:25:03 pm ---Both. I did just that and still same result. I am using a 9v battery. Below are pictures of the potentiometer tuned to both ends. At level 0 Ohm LED is dull and at 10k Ohms bright(at any resistance level above 0 ).

--- End quote ---

Secondly, the 7K impedance looking into the base, is only at full brightness, when the pot will be a very low resistance (as regards, the top two terminals of the pot, the other will be at nearly 100K, of course).
As the pot then decreases the current/brightness, into the base, the current flow reduces, so the effective impedance into the base, increases.
E.g. At 10% brightness = 10% of base current at full brightness = 10 x 7K = 70K very approximately, since Hfe will probably increase as the collector current drops.

tl;dr
Assuming OP has 10K pot, it should work reasonable well in that circuit, assuming there are not other issues.

--- End quote ---
Good, I don't know why I thought 100k.  :palm: 10k and 8k2 would certainly be better.

The base impedance will probably drop, at lower currents. It's roughly equal to RE*Hfe. The 2N2222 data sheet gives a maximum Hfe, of about 400, with a collector current of 50mA, as the current is reduced further it drops. At 20mA, it's just over 200, hence why I rounded up to 7k but at 2mA, it'll be just over 150, so the input impedance will be around 5k.
https://www.onsemi.com/pub/Collateral/P2N2222A-D.PDF

MK14:

--- Quote from: Hero999 on July 09, 2018, 11:23:20 am ---Good, I don't know why I thought 100k.  :palm: 10k and 8k2 would certainly be better.

The base impedance will probably drop, at lower currents. It's roughly equal to RE*Hfe. The 2N2222 data sheet gives a maximum Hfe, of about 400, with a collector current of 50mA, as the current is reduced further it drops. At 20mA, it's just over 200, hence why I rounded up to 7k but at 2mA, it'll be just over 150, so the input impedance will be around 5k.
https://www.onsemi.com/pub/Collateral/P2N2222A-D.PDF

--- End quote ---

Don't worry about the 10K vs 100K mistake. It is terribly easy to make M1sTTaces.

I was getting mixed up as regards the Hfe changes with current. It is power transistors, whose Hfe usually drops as the current increases, especially at many amps. Small signal transistors can be different, sorry.

[Deleted next part of post - mistakes in it]

For now, I agree with you about the impedance. I will think about it further.

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