Can someone show me an effective circuit that will turn on at night and off during the day.
I have been playing around with one circuit I found but the problem was that the LEDs always drew some power from the batteries even when in full light (they should've been off). Then when it was dark they were never bright enough. So in order to increase the intensity of the LED I increased the source Voltage but this also increased the voltage drain when it was off (in full light) .
If you use solid-state technology there is always leakage currents. You might not be able to do anything about it unless you use a relay, but then what happens is you need an active to control the relay- which means you need a circuit that will amplify the photo-readings you get to drive the relay. What will happen is you get zero current consumption when off, but when it's on you have the to power the LED and the relay, and the relay probably consumes way more power than the LED. So the power you end up saving when off ends up being used up (and then some) to power the relay when the LED is on.
My batteries are putting out 7.24VDC
The Potentiometer is dialed to a point where I get good illumination when on (in Dark) and apears off when light hits the Photoresistor.
Question 1.
When I try to read the resitance of the pot I get 1.14 when the multimeter is set to 20K, how should this be read is it 1.14 Kohms?
Question 2.
I am taking my multimeter readings from A and B (from the pos & neg of the LED). Is this the correct place to take the reading from ?
Question 3.
The LED Load when on is 5.85V This seems a bit much for a LED that is supose to light at only 3V.
Do you agree ? if so how can I reduce my voltage to the LED and still get the illumination I want ?
Question 4.
LED Load when off is 1.3V
This is the problem because I would expect no voltage to be read when the LED is not lit, can you sugget why it is drawing so much voltage when it apears off and how I can solve this ?
My batteries are putting out 7.24VDC
The Potentiometer is dialed to a point where I get good illumination when on (in Dark) and apears off when light hits the Photoresistor.
Question 1.
When I try to read the resitance of the pot I get 1.14 when the multimeter is set to 20K, how should this be read is it 1.14 Kohms?
Question 3.
The LED Load when on is 5.85V This seems a bit much for a LED that is supose to light at only 3V.
Do you agree ? if so how can I reduce my voltage to the LED and still get the illumination I want ?
Put a resistor in series with the LED. Use Ohm's law based on the current spec of the LED.
mettam said:
Question 4.
LED Load when off is 1.3V
This is the problem because I would expect no voltage to be read when the LED is not lit, can you sugget why it is drawing so much voltage when it apears off and how I can solve this ?
It seems the bjt is slightly on. I'm not sure I understand why the LED is in parallel with the collector resistor. Why not eliminate the collector resistor and put a series resistor in line with the LED based on Q3 above?
Changing the transistor to an NPN confuses understanding the circuit.
But it looks like the original circuit would work backwards, the LED would light when there is light.
It seems the bjt is slightly on. I'm not sure I understand why the LED is in parallel with the collector resistor. Why not eliminate the collector resistor and put a series resistor in line with the LED based on Q3 above?
Reading the pot while in-circuit is not just the pot resistance, it is with the surrounding circuit, and could be way off - even different when power is applied. The bjt is short for bipolar junction transistor. Try eliminating the collector resistor and put a small series resistor with the LED.