Simple LDR Circuit problem....

[bradzyc]

Hi guys,

Right a really simple question here and I'm missing something really basic.

I have a simple LDR circuit that I can't seem to get working. I'm completely adept at Electronics and it's use is just for a small project I'm working on in my spare time.

Attached below is a picture of the circuit I have now. I have previously had this working with 200mA and 12V. The bulbs are both 6V and the transistor is a TIP120. I previously used a 2.2k Ohm fixed resistor followed by a 10K Ohm (Max) variable resistor set at around 5K Ohm. The LDR goes from 3K Ohm uncovered to 65K Ohm covered.

Since I had this working with a power pack on 200mA, I then changed this to a set of 8 x AA batteries which gave me 12V at 4A, and this is where I got most of my problems. I've used Ohm's law numerous times getting various figures, however I really don't think I'm doing it correctly.

I've been trying to figure out the current going through the circuit using Ohm's however, reading up online, I know I get the current across resistors, does this mean what is coming out after the resistor or does this mean I take this from the battery current?

Again, I apologies for the real simple question but if anyone could help me calculate the resistance I need to enable my circuit to work and how it is done I would really be in your debt!

Many thanks!
Brad

 

[Jeff Rosenbury]

AA batteries are nominally 0.24 amps. They can draw a lot more, but then they will have more internal resistance.

Add an internal battery resistance. Then the circuit analysis becomes difficult since the battery resistance depends on the current draw which in turn depends on the battery resistance. A fresh AA has an internal resistance of about 0.15Ω/amp, but this quickly goes up as the battery discharges. Also the batteries will become less efficient at higher discharge rates.

I'm pretty sure you can't get 4A at 12 volts without more or bigger batteries.

 

[bradzyc]

AA batteries are nominally 0.24 amps. They can draw a lot more, but then they will have more internal resistance.

Add an internal battery resistance. Then the circuit analysis becomes difficult since the battery resistance depends on the current draw which in turn depends on the battery resistance. A fresh AA has an internal resistance of about 0.15Ω/amp, but this quickly goes up as the battery discharges. Also the batteries will become less efficient at higher discharge rates.

I'm pretty sure you can't get 4A at 12 volts without more or bigger batteries.

I'm not too sure on what amp each battery is but I do know that I'm reading 4 Amps on the multimeter. This did seem very high to me to be honest.

 

[mr166]

With the light shining on the LDR you should have less than 1v from the base of the darlington transistor to the emitter in order to keep it turned off. With no light the base to minus voltage needs to be as high as possible. Thus the total resistance of the pot and series resistor needs to be about 60K to keep the transistor from conducting. When dark the base to minus voltage will increase to 6v causing about 5v to be across the bulbs and 7 volts to be wasted across the transistor. The circuit would work much better if you put the bulbs in the collector circuit ie from +12 to the collector. Then you would need 12v bulbs or put the 6v bulbs in series. Then shining the light on the LDR would turn the bulbs off. You could move the LDR to the +12 side and make the resistor 1.5k and move it to the -12 side of the supply to reverse the on off of the bulbs. This is much more efficient than your original circuit.

 

[Jeff Rosenbury]

I'm not too sure on what amp each battery is but I do know that I'm reading 4 Amps on the multimeter. This did seem very high to me to be honest.

I don't doubt the 4 amps. But if you check the voltage while drawing 4 amps, I doubt it will be 12V for more than a minute or so.

 

[bradzyc]

With the light shining on the LDR you should have less than 1v from the base of the darlington transistor to the emitter in order to keep it turned off. With no light the base to minus voltage needs to be as high as possible. Thus the total resistance of the pot and series resistor needs to be about 60K to keep the transistor from conducting. When dark the base to minus voltage will increase to 6v causing about 5v to be across the bulbs and 7 volts to be wasted across the transistor. The circuit would work much better if you put the bulbs in the collector circuit ie from +12 to the collector. Then you would need 12v bulbs or put the 6v bulbs in series. Then shining the light on the LDR would turn the bulbs off. You could move the LDR to the +12 side and make the resistor 1.5k and move it to the -12 side of the supply to reverse the on off of the bulbs. This is much more efficient than your original circuit.

I see.

Just a question though, how did you come to the 60K resistance value number?

I will try this and see what it yields.

 

[bradzyc]

I don't doubt the 4 amps. But if you check the voltage while drawing 4 amps, I doubt it will be 12V for more than a minute or so.

Apologies, I got you!

 

[mr166]

I see.

Just a question though, how did you come to the 60K resistance value number?

I will try this and see what it yields.

Well with light shining on the LDR you want the base to -12v ( let's call it ground to make things easier ) to be less than 1.4 v in order that the darlington transistor does not conduct. When the LDR is conducting you said that it was 3K ohms. So, 12v x3K/60K=.6v. This enough to keep the transistor off. If it was a regular transistor it would start to conduct at about .6 volts.

 

[mr166]

BTW in your circuit the voltage at the light bulbs equals the base to ground voltage minus the base to emitter voltage which in the conducting state is about 1.2v to 1.4 volts. The remainder of the 12v just turns into wasted heat in the transistor. That is why the circuit I suggested is better since about 95% of the power goes to the bulbs when the transistor is turned on.

 

[mr166]

If you are using flashlight bulbs then each one should be drawing about .75A or about 1.5 amp for the pair in parallel. The transistor will need a heat sink to handle the 9 watts it will be dissipating in your circuit. If you don't use one it will overheat and blow out.

 

[mr166]

Excuse me for being a serial thinker and posting multiple times but 2 6v flashlight bulbs in parallel will not work with only 200 ma available as in your first circuit. Could you be using some sort of 6v led assembly?

 

[Baluncore]

In your design the globe voltage is decided by LDR and resistance ratios. I usually make a couple of changes that improve performance. See attached schematics.

1. Replace TIP120 darlington with an N-channel MOSFET. Threshold voltage is then referenced to negative rail. Full supply voltage appears across the load so globes can be in series and use half the current. A heatsink may be needed on the MOSFET for slow changing light levels.

2. Use an LM555 timer as a Schmitt-trigger with thresholds at 1/3 and 2/3 Vcc. Those will track the LDR : R ratio and provide hysteresis. The 555 output can sink 100 mA. As before use a MOSFET as the power switch for higher currents. A lower power solution, depending on supply voltage, might be to use a CMOS LMC555. With a schmitt-trigger a heatsink will not be needed.

Adjust R1 to set threshold switching points.

MOSFETs and LM555s invert. Swap R1 Adj and R2 LDR to reverse output state.

 

[bradzyc]

Many thanks for all the helps guys.

I have put the bulbs in series after the emitter of the transistor and used the same position transistors. I tried my previous set of resistors (stated in the original post) on a breadboard and the circuit works. I used a matrix board previously but for some reason the strips weren't conducting at all as I wasn't getting any voltage drop across the resistors so now I know that the problem was with the actual matrix board.

The circuit now works.

Many thanks for all your help.

Baluncore - I will try these options to get a more efficient circuit. Thanks for the help.

Brad