Assuming you mean that the potentiometer. Is replacing the 1K resistor. I.e. The potentiometer, is connected between the base of the transistor, and the positive +9V battery terminal. (I've looked at the pictures, and it looks like it might be, but I'm not 100% sure).
Then the LED would tend to switch off, when you set the potentiometer to around 0 ,
because you should really have an extra series resistor, between the potentiometer and the base, of e.g. 1K. Otherwise excessive current will try to flow between the base and +9V supply, because the diode junction of the transistors Emitter/Base will start to hugely conduct, at around 0.6V/0.7V.
The small 9V cell I can see, can probably only supply a few hundred milliamps (probably less), so the whole supply voltage will probably drop to around 0.8V (guestimate), and hence the LED will turn off. It may also damage the potentiometer and the small signal transistor.
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 ).
Assuming you mean that the potentiometer. Is replacing the 1K resistor. I.e. The potentiometer, is connected between the base of the transistor, and the positive +9V battery terminal. (I've looked at the pictures, and it looks like it might be, but I'm not 100% sure).
Then the LED would tend to switch off, when you set the potentiometer to around 0 ,I would have thought it would burn out the transistor, so this is surprising.
The observations regarding the lack of 1k resistor is good, also the internal resistance of the battery, however, someone is missing the elephant in the room.. this is not an incandescent bulb. It is an LED.
The observations regarding the lack of 1k resistor is good, also the internal resistance of the battery, however, someone is missing the elephant in the room.. this is not an incandescent bulb. It is an LED.
Just did a quick test. With a good alkaline 9V battery the short circuit current is about 400mA - giving the battery about 22.5 ohms of internal resistance.
That is enough to burn out an LED, but not enough to burn out a transistor (at best you might be able to get half a watt of heat outside of the battery).
Without the resistor (or the pot dialed to 0 ohm) there will be a few hundred mA flowing through the tranaiator's base, and due to the internal resistance the battery voltage has dropped to about volt or less, causing the LED to go out.
That is enough to burn out an LED,
Regulate current, not voltage.
Regulate current, not voltage.BINGO!
Where do you put the resistor on a transistor hookup for current regulation? Keep in mind that you will need to regulate from approximately 50 µA (not 0) to 100 mA. Still, not the method that I would recommend.
Thanks Everyone. I used a new 9v battery and it burnt the LED and got the transistor hot ( replacement LED also got burnt). It seems the dimming was probably a result of the LED glitching due to over-current.( New 9V Battery: 9.01 V and Old 9V Battery: 7.48 V ).
Regulate current, not voltage.BINGO!
Where do you put the resistor on a transistor hookup for current regulation? Keep in mind that you will need to regulate from approximately 50 µA (not 0) to 100 mA. Still, not the method that I would recommend.The resistor needs to go in the emitter side, to regulated the current, which can then be controlled by varying the base voltage with the potentiometer.
Also what's wrong with regulating down to 0mA?
And 100mA is too high for a small 5mm LED.
Okay, so if you want to dim an LED, your best option is a PWM (Pulse Width Modulator) circuit. The circuit allows the user to adjust the width of pulses, without changing the frequency of the pulses. This means that the effective voltage varies while the power supply voltage remains constant. This effectively regulates the average current applied. The LEDs will dim from dark to fully lit based on the regulating resistor in series with the LEDs or single LED and the voltage applied (power supply voltage).
Regulate current, not voltage.BINGO!
Where do you put the resistor on a transistor hookup for current regulation? Keep in mind that you will need to regulate from approximately 50 µA (not 0) to 100 mA. Still, not the method that I would recommend.The resistor needs to go in the emitter side, to regulated the current, which can then be controlled by varying the base voltage with the potentiometer.
Also what's wrong with regulating down to 0mA?
And 100mA is too high for a small 5mm LED.
I hope you don't mind me criticising your (apparently), adjustable 0 to 20 mA (approx), constant current circuit.
But, the way you have drawn it, the 100K pot, will probably not do anything, for around half its travel.
If, you changed it, so that the bottom lug of the pot, which currently goes to ground, was instead, connected to the point where the two diodes meet. Which should be around 0.6 to 0.7 volts.
Then most of the transistors Emitter/Base voltage drop will effectively be cancelled out.
Hence the adjustable pot, should vary over nearly all its range (rather than only half of it).
On the other hand, it is possible, my suggested change may reduce or cause instability. I would have to simulate it, or build it, or calculate hard, in order to find that out.
Thanks for your comment. I realised this later, when I was away from the computer.
Yes, connecting the resistor to the node where the diodes meet will help, but it still might not go all the way to zero. Another resistor, with a slightly lower value, than the pot, in series with the bottom side of the pot will fix it, although there will still be some dead-band.