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QuoteYour LDR's are of a higher resistance when in light, so the circuit have to be adapted for this.Either you swap the trimmer potentiometers for some 200k to 500k, or you keep the ones you have and add a resistor to each trimmer where it connects to the +9V line. Break the connection and insert the resistor in between.The trimmer will have much less influence if you just add a resistor, but it will be cheaper and easier.Start with say a 100k resistor and test if the trimmer can now get it in range.Measure the voltage over the LDR with the trimmer in each outer position and report back your findingsI have changed all my trimmer potentiometers to another ones with 250K.I measured the voltage over the LDRs. Here are the readings.right LDRs : 7.38 and 5.84left LDRs : 5.87 and 7.29I measured the voltage of the battery and it was at that time 9.09
Your LDR's are of a higher resistance when in light, so the circuit have to be adapted for this.Either you swap the trimmer potentiometers for some 200k to 500k, or you keep the ones you have and add a resistor to each trimmer where it connects to the +9V line. Break the connection and insert the resistor in between.The trimmer will have much less influence if you just add a resistor, but it will be cheaper and easier.Start with say a 100k resistor and test if the trimmer can now get it in range.Measure the voltage over the LDR with the trimmer in each outer position and report back your findings
The datasheet for your LEDs will list their operating voltage.
Use the LED's in parallel with a resistor for each...
(I assume the readings are with the trimmers in each outer position)
I need to buy me more 1k resistors.
Quote(I assume the readings are with the trimmers in each outer position)I don't think I understand what you mean by that!!!
I measured the voltage over the LDRs. Here are the readings.right LDRs : 7.38 and 5.84left LDRs : 5.87 and 7.29I measured the voltage of the battery and it was at that time 9.09
while we are at it, I have another question.I have asked this question before in this thread, but no one has answered it yet.When I turn the resistance of the trimmers down to 0, aren't my Red LEDs suppose to go off?Well, they are not. They are not even going dimmer or brighter. Any suggestions on that?!!!
Quote from: Aber on December 11, 2010, 02:45:55 PMQuote(I assume the readings are with the trimmers in each outer position)I don't think I understand what you mean by that!!!These measurments:QuoteI measured the voltage over the LDRs. Here are the readings.right LDRs : 7.38 and 5.84left LDRs : 5.87 and 7.29I measured the voltage of the battery and it was at that time 9.097.38V and 5.84V represents what then?
Quote from: Aber on December 11, 2010, 02:40:07 PMI need to buy me more 1k resistors.Not quite... They shouldn't be more than 270 Ohm maximum.
With a full 9V battery and 20mA through the LED's, you need: (9V-3.6V)/0.02A = 270 OhmWhen the battery is down to 7V (and thus still have power left) you need: (7V-3.6V)/0.02A = 170 OhmAt the end-of-life, the battery supplies 5.4V and then you need: (5.4V-3.6V)/0.02A = 90 Ohm...
A much better way is to make a 21mA constant current circuit like this, to replace the resistors. That way the current will stay the same over the battery life.
BTW. You need to use 6 AA-cells or similar, as a PP3 (box shaped) 9V battery just won't have the power to drive it.
WOOW, ok now this is too advanced for me. I thought this project is supposed to be easy and not complicated for beginners. From what I have been reading through threads, I believe some beginners have built the circuit with the exact parts mentioned in the tutorial, and it worked for them.
QuoteWith a full 9V battery and 20mA through the LED's, you need: (9V-3.6V)/0.02A = 270 OhmNow this is advanced and new to me.
With a full 9V battery and 20mA through the LED's, you need: (9V-3.6V)/0.02A = 270 Ohm
This is way advanced for me to understand. First, it looks like I need to make a big change in my circuit. Second, I don't quit understand this schematic. I have spent hours searching the net to find out what of these schematics mean. Now I understand that Q1 and Q2 refer to the collector lead of a transistor.
But there are things I don't understand about it:-Is B+ means the positive side of the battery? If so, I don't see the B- . What is 0V-You are using BC547 and BC337, different than what the tutorial is suggesting which is BC538 for all four transistors.-R1 and R2, are these resistors? If so, what is 3k3 and 33R?-LED- and LED+ are for Red LED?
I am using 9V Ni-MH 250mAh Rechargeable battery. Doesn't this battery have the power to drive it?