A potential problem is that an SMPS probably has a much higher output current capability than your little battery, so with your apparent inability to grasp the basics of driving LEDs safely the SMPS could fry a lot of LEDs in the blink of an eye if things aren't wired properly.
Nobody makes a 3V white LED. Some will be 2.8V and others will be 3.2V. If you use a 9V supply and connect three 3.2V LEDs in series and in series with a current-limiting resistor then THEY WILL NOT LIGHT! So you connect only two "3V" Leds in series and have a very hot current-limiting resistor.
I'm loosing the will to live..... Do you know the reason they call it a transistor!!!!
Transfer resistor.... You need to design the BJT to your circuit... If you fully turn on the transistor then there should be nearly 7.2v across your resistor... BUT if you work out your current gain you can partially switch on the BJT to give a smaller voltage across your resistor so the BJT dissipates some of the power...
At the moment ALL of the power is lost in the resistor... Using the BJT share the power loss between them..
Transfer resistor.... You need to design the BJT to your circuit... If you fully turn on the transistor then there should be nearly 7.2v across your resistor... BUT if you work out your current gain you can partially switch on the BJT to give a smaller voltage across your resistor so the BJT dissipates some of the power...
At the moment ALL of the power is lost in the resistor... Using the BJT share the power loss between them..
OK, using low value Resistance will BJT to diss less power rather than Resistance.
I am using 10k variable at base but i have found changing its value is not changing current passing from BJT, why at 0 it off only few 1 or 2mA changes why?
Transfer resistor.... You need to design the BJT to your circuit...
Are you still using a BC337 PNP transistor? What is driving its base?
Its datasheet shows that some of them saturate poorly (VCE=0.7V) when the collector current is 500mA and the base current is 50mA.
Are you still using a BC337 PNP transistor? What is driving its base?
Its datasheet shows that some of them saturate poorly (VCE=0.7V) when the collector current is 500mA and the base current is 50mA.
If the base current is 50mA then part of the 20k pot will have 8V across it (VBE is about 1V) and with 50mA then its value is only 8V/50mA= 160 ohms.
Then the portion of the pot is only 160/20k= 0.8% which is nothing. The tiny part of the pot will smoke when it tries to dissipate 8V x 50mA= 0.4W.
If the base current is 50mA then part of the 20k pot will have 8V across it (VBE is about 1V) and with 50mA then its value is only 8V/50mA= 160 ohms.
Then the portion of the pot is only 160/20k= 0.8% which is nothing. The tiny part of the pot will smoke when it tries to dissipate 8V x 50mA= 0.4W.
With a 20k base resistor at ground, the base current is (5V - 0.7V)/20k= 0.215mA. Then the maximum collector current can be 0.215mA x 20= 4.3mA, or maybe 0.215mA x 50= 10.75mA. Very low currents.
With the transistor turned off when its base resistor is +5V then you cannot use a 9V supply for the emitter.
With a 20k base resistor at ground, the base current is (5V - 0.7V)/20k= 0.215mA. Then the maximum collector current can be 0.215mA x 20= 4.3mA, or maybe 0.215mA x 50= 10.75mA. Very low currents.
With the transistor turned off when its base resistor is +5V then you cannot use a 9V supply for the emitter.
5V is the maximum allowed reverse-bias emitter to base voltage for most transistors. Exceeding this voltage causes the emitter-base junction to breakdown like a zener diode which damages the transistor.
But the base to emitter is almost always forward-biased in most circuits and 5V is NEVER used.
5V is the maximum allowed reverse-bias emitter to base voltage for most transistors. Exceeding this voltage causes the emitter-base junction to breakdown like a zener diode which damages the transistor.
But the base to emitter is almost always forward-biased in most circuits and 5V is NEVER used.
sorry for a hijack but can i ask a question please? why is 5v never used at the base? what about a pic pin?? or am i not understanding hmmmm i think i am missing the meaning of base to emiter rather than base?? sorry for the question
If the base current is 50mA then part of the 20k pot will have 8V across it (VBE is about 1V) and with 50mA then its value is only 8V/50mA= 160 ohms.
Then the portion of the pot is only 160/20k= 0.8% which is nothing. The tiny part of the pot will smoke when it tries to dissipate 8V x 50mA= 0.4W.
sorry one other thing i wanted to say, it would never have occured to me to work out what % of the pot was being used. its seeing info like this and real working out that makes threads like this real gem's ok i admit the OP drives a few mad but beginners like me take notes of this stuff is real good learning stuff thanks guy's
sorry for a hijack but can i ask a question please? why is 5v never used at the base? what about a pic pin?? or am i not understanding hmmmm i think i am missing the meaning of base to emiter rather than base?? sorry for the question