Hi,
If you want to use the original 100K/39K then you will have to get a 10meg LDR.

Note: in the circuit I have just posted I have misread your small Table of LDR values.
Its not a big deal, just swop the 5K pot and LDR positions on the circuit. Do you follow OK.
Use a LDR resistor combination to give ~8v on the Trig pin when the pot is at 50%.

I cannot say I have seen many 10meg LDR's, I will check for suppliers.
E
EDIT:
In the UK Rapid Electronics carry high value LDR's

 

OK understand, I will have to buy those parts.
So I might as well buy the 10M ohm LDR instead here my link to 10M LDR: http://www.jaycar.com.au/Passive-Co...stors/Light-Dependent-Resistor-(LDR)/p/RD3480
Thanks for your time.
Andrew

 

hi,
Here is the the corrected circuit for the Chinese LDR's.
There is no reason why you cannot use this circuit in place of the 10 meg LDR version.
If you do use the Chinese LDR's the original 100K and 39K are not used.
E

 

Hi Eric,
OK use #2 LDR and a 20K potentiometer, without the 100K and 39K are not used. got it.
Thanks again mate.
Andy

 

I just spend 10 minutes altering the circuit with recommend components, all working now
Thanks again Eric.