I have a newbie question. I get how the attached LDR circuit works. During daylight the LDR is conducting robbing the transistor of current and keeping the LED off. When it gets dark current is applied to the transistor and the LED lights. My project is to build a solar charged bug zapper circuit. My question is regarding the impact on my little 1.2v rechargeable battery during daylight when I'm trying to charge the battery. If the LDR is conducting, robbing the transistor of power, is this not 'shorting' the battery through the LDR and resister? I need to charge that battery during the day so as to power my bug zapper circuit at night. If I'm shorting my battery during the day I won't be able to charge it successfully right? Any experienced insight would be much appreciated. I don't want to settle on this as my 'turn on' solution if it's going to defeat the charger part of my project.
Thanks for any guidance!
Grant

 

Your thinking needs some adjustment. The primary current path in your circuit is through the emitter to base and then through R2. If there was no LDR, the circuit would be always on. (ps, there is a danger: If R2 is adjusted to zero ohms, the transistor will smoke.) With the LDR installed and activated with light, the LDR provides the current through R2 and so the transistor contributes nothing. No base current means no collector current, and the LED remains off.

As for, "shorting", this is a matter of, "how much". One hundred thousand ohms will allow 90 millionths of an amp to flow out of your battery. R1 will allow .0171795 amps (17.18 milliamps) to flow when the LED is on. You can expect a minimum current of 17.27 milliamps (total current) when the LED is on.

Check the ability of your solar panel. I bet it produces at least 5 times as much current as the alleged "short" (17.27 ma).

Is your sense of proportions getting adjusted?

 

Too cool. Thank you for taking the time to give such a clear and concise answer. I was hoping it would be a matter of how material the 'drain' would be but I haven't gotten my head around the math well enough to be comfortable yet. I'll use your answer to help verify my walking through the numbers.
Thanks again and Happy Thanksgiving.

 

ps, you need at least 390 ohms in series with the base of the transistor so you can adjust R2 to zero ohms and not smoke the transistor. Not that you're going to actually USE it adjusted to zero, but you will adjust it to zero accidentally in the process of getting the whole thing adjusted properly. Somewhere from 390 ohms to 4700 ohms will work as the sum of the 390 plus the pot.

All these numbers are based on using a 9V battery, not the 1.2V you typed in.

Too cool.

That's what the, "like" button is for. That's how we get paid.

I have some doubts about the 100K potentiometer. If you need 390 ohms to 4700 ohms to allow proper triggering, what are you going to use the other 95,300 ohms for? LRDs don't usually get to a couple of thousand ohms. Measure yours. I'm thinking you will need a Darlington transistor to get this sensitive enough. Hang on while I make a drawing.

 

I was looking for some help on the concept, which you clearly gave and so sent a 'similar' circuit. Actually, I am powering this on a 1.2v rechargeable. I need the entire output passing through this to 'load' when the LDR turns off. The load is a combination of a low to high voltage A/C-transformer-capacitor to handle the bug zapper mesh in parallel with a joule thief circuit to power a UV LED. So, I need the LDR to do it's thing but need this portion of the whole project to take away as little power as possible. I've already gone ahead based on your feedback but have removed the pot and, so far, have gone with a 10k resistor in it's place. I need to 'play' with that as I'm also using a 3906 transistor, not the 557 in the diagram. Sorry if I have led you into taking time but again, I was looking for basic guidance as to whether, regardless of application, this was in fact a potential solution for my 'turn on at dark' need.
Thank you very much for your time and guidance.

 

The transistor will be anything on this page:

http://www.mouser.com/Semiconductor...Z1z0z60lZ1z0y4cnZ1z0j3xxZ1yzp3ai&Ns=Pricing|0

Now we're getting numbers that make sense with an LDR and a 100k pot!

and now we're working with a very different circuit.

 

The Darlington will use up too much voltage so I would try a double invert circuit if you must stick with bipolar transistors. The mosfet version wastes the least amount of battery voltage and has a very high input impedance to better match the LDR, so much that you wouldn't need the double configuration. 2N7000 is a popular number.

Right now I'm getting a bit brain fried. If you're versatile enough to work from what I gave you, fine. If not, be honest next time. There's a butt load of wizards here, but we don't have crystal balls.

Edit: The mosfet version needs way more than 1.2 volts to control its gate. I guess you're stuck with the bipolar version.

 

Right now I'm in that uncomfortable 'not knowing what I don't know' space. You've gotten me past a true hurdle here. I have all of the circuit components I need for my project including high voltage circuit, joule thief, solar charger circuit and (now) how to turn it all on at night. I suspect my problem will come up when I try to tie them all together. The guidance you've given me let's me pass the entire 1.2v through with the single 3906 and 10k resistor I mentioned under the daylight control of the LDR. I'm not sure why I would need a double invert circuit at all but suspect that I'm going to run into problems with amperage, not voltage, when I try to tie everything together. I've got a good solar charger option so I can run the mah values of the 1.2 up but now I'll need to be more concerned about the health of the individual components.
Thanks again for your patience and help. I truly appreciate it and will try to blend in with the norm regarding future questions.
Grant

 

The double invert circuit is like a programmable Darlington. You can get a current gain out of it higher than 10,000 to one and only use up about 0.3 volts, collector to emitter, on the output transistor. Less if you're willing to search datasheets for lowest saturation voltage in a switching transistor. None of the resistors are labeled (they range from zero to infinity) so you can optimize it for best current gain, least saturation voltage, widest frequency response, high input impedance, etc.

There is also a transistor called a Sziklai pair which works like a double invert circuit, but I'm not familiar enough to even know if you can buy them.

 

You're way beyond me at this point but I'm taking notes. Again, I'm past the roadblock of powering my project at night based on your help. Now, I have to put all four components into parallel and see if I can power everything with my little 1.2v battery which is looking smaller every time I look at it. I may be back......
Thanks again.
Grant

 

If you in series an 1n4148 with b of bjt in the post #1, it can be increase the sensitive of cut off time for bjt, you can try to only using bjt, in series with one 1n4148, in series with two 1n1418 to check the sensitive.

 

I'm sorry but I don't understand what you are saying. I see you referring to using a switching diode in series with the transistor but I don't understand why you are suggesting adding that. Where are you suggesting this would be added and more importantly, why?

 

The LDR is the shunt resistor of bjt, why the bjt will cutoff, because when the brightness of sunlight strong enough to some degrees and that will cause the current flows through the LDR more and make the Ib not enough to turn on the base of bjt, so the Ic of bjt will turn off.

So here I will make example then maybe you will understand, assuming that the degree of sunlight have ten degrees from 0 to 10, now if you didn't add the 1n4148 then when the sunlight lighting up to degree 3 the Ib will cutoff, if you adding the 1n4148 then the Ib will cutoff when the sunlight lighting up to degree 2 the Ib will cutoff, the voltage of common point of LDR and R2 will rising to 1.4V, that will caused the Ib cutoff earlier.

No 1n4148:
The current of shunt is I_LDR = Vbe/LDR = 0.7v/LDR

1N41418 added:
The current of shunt is I_LDR = Vbe+0.7/LDR = 1.4V/LDR