# Questions obervations LDR circuit

Discussion in 'General Electronics Chat' started by lordeos, Sep 15, 2015.

1. ### lordeos Thread Starter Member

Jun 23, 2015
33
1
Hi all

I made a simple LDR (circuit diagram below) circuit and it works. But when i do some measurements and start experimenting with the circuit i did certain observations that i would like to check with someone and see if my reasoning is correct.

1) my first resistor is a +- 266 ohm resistor dropping 4V and allowing 20Ma ... when checking the voltage in light and dark the voltage dropped over that resistor remains +- 4v and the reading on my power supply stays at 20Ma (without series resistor wit the LED) , is it correct assuming i don't need a series resistor with my LED to control the current and voltage, since this job is already done by the first resistor ? I see some change in voltage when switching between light and dark but the difference is a max of 0.7 V on the LDR.

2) the voltage drop on my LDR is 1,5V in light and 2.0V in dark (lowest = 50 ohms highest = 200.000 ohms) ... i did check the LDR in full dark and it goes up to 1 Mega ohm. Is the 0.5V change in LDR caused by the fact the normal conditions (daylight and dark) is not enough to pump up the resistance of the LDR to megaohms ... if that would be the case then i would get almost supply voltage on the LDR, correct ?

3) Even when i put a black tape over the LDR the resistance becomes 1 Megaohm . When i do the calculation r2/r1+r2 i should have VS voltage which is 6v ... when i measure it on the circuit the max voltage over the LDR remains 2V, how come ? According to the calculations i should reach full vs voltage on LDR around 200K ...

4) When i do put a series resistor before my LED (example 1 Mega ohm) i see that in light conditions the circuit draws more Mili amps then in dark conditions (in dark circuit draws +- 3 miliamps, in light +- 20 Miliamps) ... Does this mean that in light conditions (LDR at lowest) the 20Ma goes directly through the LDR and in light the the 1megaohm series resistor limits the total circuit current and only 3Ma is the available current ?

5) The equivalent resistance with 2 resistors in parralel is lower then the smallest resistance. So by adding the 1 megaohm resistor in parralel to the LDR, is it correct that my Total resistance in the whole circuit does not change much (because the lowest resistance is 50 ohms of the LDR). it only influances the voltage and current in the branch of the LED ,however i do see that in that case in light or dark almost full supply voltage is dropped over the LDR propably caused by the fact that de resistance in LED branch is so high thats the current chooses the LDR path, correct ?

Thx for the help
Mike

File size:
32.4 KB
Views:
22
2. ### blocco a spirale AAC Fanatic!

Jun 18, 2008
1,440
368
Without R3 the voltage across the LDR cannot exceed the forward voltage of the LED and with R3 fitted the LED will still influence your readings. You must take this into account.

Last edited: Sep 15, 2015
3. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
You will get a much more abrupt turn-on of the LED with this simplified circuit:

The x-axis shows a log plot of varying LDR resistance, going from 20Ω to 2KΩ. (Independent variable of the simulation).

The left y-axis shows the forward voltage V(f) across the LED in volts Blue trace.

The right y-axis shows the currents in D1 Green and R1 Red, respectively.

Note that the LED just begins turning on as the LDR resistance gets to ~75Ω

4. ### dannyf Well-Known Member

Sep 13, 2015
1,835
367
Unless you tell your readers, they aren't going to know what exactly you meant by "some measurements" and "certain observations".

The first order of business, I think, is for you to measure the resistance of the LDR when lit and when dark. That information will answer many of your questions.

5. ### lordeos Thread Starter Member

Jun 23, 2015
33
1
Hi Dannyf

The sentence "But when i do some measurements and start experimenting with the circuit i did certain observations ..." is a proloque = the reason why i"m asking this questions.

- The actual measurements i did and specifications is explained in the 5 questions (in detail)

- In question 2 i mention "lowest = 50 ohms, highest 200.000 ohms) = light = 50 ohms , dark = 200.000 ohms.

- in question 4 i write "Does this mean that in light conditions (LDR at lowest)" --> lowest = 50 ohms

So i think i can't provide you with more details than what i wrote in the 5 questions

cheers Mike

Last edited: Sep 15, 2015
6. ### lordeos Thread Starter Member

Jun 23, 2015
33
1
How come the LDR doesn't exceed the forward voltage of the LED ?

I did the test with a series resistor and indeed the voltage of the LDR now exceeds the forward voltage of the LED and the remaining voltage is dropped over the series resistor with the LED.

Thx for the help

7. ### blocco a spirale AAC Fanatic!

Jun 18, 2008
1,440
368
Because the LED is a diode and behaves as such. As the resistance of the LDR increases; the voltage across it will rise until the diode starts to conduct i.e. when the voltage equals its forward voltage.

If you removed the LDR and put a link across R3 what voltage would you expect to see across the LED?

8. ### dannyf Well-Known Member

Sep 13, 2015
1,835
367
You can get a much sharper turn-on/turn-off with a TL431.

9. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
Even with a transistor...

The simplified one-resistor circuit is much better that what the OP started with...

10. ### ScottWang Moderator

Aug 23, 2012
4,858
767
This circuit will be waste the current for nothing when the LDR get the light, and then its resistance will become more less until the the current flows through the LDR big than led, the brightness of led will be more weak.