[sdiy] Performance control - processing a control voltage

[Harry Bissell]

This is a tough call.  If you make a divider with a LDR, the resistance of the LDR will have
very finite values. It will have an 'on resistance' of some ohms to Kohms, and an off resistance
in the megohms.

Assuming the resistor is in the top of the divider, you need to work against a fixed load resistor.
lets say the resistances are from 1K (on) to 1MEG (off).

If you use a 1K resistor for the lower leg of the divider, you would go from 5mV to 2.5V. 
If you want to improve the high end, you will make the low end worse...

You might want to use an opamp buffer with a gain of 2, that would go from 10mV to 5V, assuming the opamp
was good for that output. A rail-to-rail opamp 'might' do the job if your load was not too large (low impedance)

You might opt to run the divider so that you could approach 5V on the high end, and chop off the excess voltage
at the low end by using a series diode.

Your circuit loading will be an issue in almost any case (except with the opamp buffer...)

does this help ?

H^) harry


----- Original Message -----
From: Matthew Smith <matt at smiffytech.com>
To: Synth DIY <synth-diy at dropmix.xs4all.nl>
Sent: Sat, 28 Jan 2012 19:10:33 -0500 (EST)
Subject: [sdiy] Performance control - processing a control voltage

I want to use LDRs for performance control, getting a 0-5V output 
depending on the amount of illumination they receive. (There is a 
feedback element here too, as there will also be light emitters in the 
system, but that's another story.)

If I make the LDR one half of a potential divider which has 5V across 
it, how exactly do I calibrate the system so that the maximum resistance 
of the LDR is, say 0V (depending on which end of the leg it is placed 
in) and a given value of illumination is 5V?

I realise that I'm going to need at least one op amp here and, if I can 
get away with it, I'd like to power it/them with just the single 5V 
supply going to the potential divider - so I can create a simple, 
three-connection module.

As far as I can tell, I need to both amplify and offset the voltage 
coming out of the potential divider, but not entirely sure of the 
optimum way of doing this.

Ideas?

Cheers

M

-- 
Matthew Smith

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Harry Bissell & Nora Abdullah 4eva