How can you turn perceived brightness (log scale) into a linear analog output using an LDR?

Question

This question is not about how to make illuminance (in lux) a linear function of voltage! It's about linearly representing the human perception of brightness, which itself scales logarithmic to the unit lux and is not equal to lux.

TL;DR: How can you transform the resistance of a light dependent resistor (up to 2MΩ) to an analog signal, so that it reflects the logarithmic growth in resistance as a (more or less) linear output?

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First of all: Although the relationship between Lux and LDR resistance is inversely proportional (see Figure 4 from datasheet), light intensity itself is perceived logarithmically.

• Night sky: ~0.1 lux (=1MΩ)
• Fullmoon: ~1 lux (=100kΩ)
• Dark indoors: ~80 lux (=10kΩ)
• Living room: ~600 lux (=2kΩ)
• Overcast sky: ~1.000 lux (=800Ω)
• Bright sunlight / clear sky: ~50.000 lux (=40Ω)

 

But so far the typical usage of an LDR in a voltage divider has served me well. I can measure quite a big range with a 10bit ADC. The only downside is that it has an "increased resolution" spot around resistances similar to the fixed R_S value, while magnitudes further out are getting more and more inaccurate.

This kind of "focal length" property might even be useful in situations where you want to specialize in a certain lux range (e.g. window light dependent LED dimmer). But as an academic question, I would like to know how to turn this curved measurement into a more linear output.

Answer 1

Consider something like this, which uses about 25 cents worth of parts: (as mentioned in the comments it would be better to keep the transistors together or use a co-packaged dual transistor, though only partial temperature compensation is provided).

^{simulate this circuit – Schematic created using CircuitLab}

It has an output that varies from about

500mV (630mV @ 0°C to 360mV @ 50°C) for 1M\$\Omega\$ resistance to

1.1V (1.18V @ 0°C to 1.01V @ 50°C) for 100K\$\Omega\$

1.6V (1.65V @ 0°C to 1.57V @ 50°C) for 10K\$\Omega\$

2.11V (almost zero and slightly reversed temperature coefficient) for 1K\$\Omega\$

2.62V (2.57V @ 0°C to 2.67V @ 50°C) for 100\$\Omega\$

2.82V (2.76V @ 0°C to 2.88V @ 50°C) for 40\$\Omega\$

Of course you could attempt to measure the temperature and correct for the varying drift in firmware, but maybe this is close enough (the LDR will have temperature sensitivity - I believe maybe 2:1 or more over the 0~50°C range, and tolerances in excess of +/-50% are not unusual to begin with- they are not precision devices).

Explanation- the R1/R2 divider maintains a constant 100mV (derived from the power supply) across the LDR. The voltage is chosen to keep the maximum current in the LDR reasonable (2.5mA) while still giving a decent signal.

The feedback current is passed through diode-connected Q1 which gives a logarithmic (though temperature-sensitive) voltage at the output of OA1.

Q2 and R3 produce a voltage to offset the output of OA1 and to provide partial temperature compensation, it is buffered by OA2 and then subtracted with gain via OA3 to give a high enough voltage that the ADC gives reasonable resolution without saturating the LM324s (which can reliably produce up to 3V output).

Diode-connected transistors are used rather than 1N4148 etc. because they have a more ideal voltage/current curve (\$\eta\$=1 rather than about 2 for silicon signal diodes).

Answer 2

The log-log characteristic of the LDR is mentioned as the problem to be solved but, within the question, the change in resolution seems to be the actual problem.

If linearity was really the main problem, the voltage divider approach would not be the best one, as it also adds another non-linearity. A possible solution to this would be an op-amp based current source.

If the application really demands better resolution throughout the 6 orders of magnitude input resistance range, an auto-range feature would probably work better. This could also be achieved with the current source solution and removing the log-log behavior in the software is a lot simpler and more reliable/stable.

Answer 3

LEDs are very linear in brightness vs current so you can use an LED and a second LDR with an op-amp to linearise the reading. Q1 here is boosting the output current to the led.

output is by sensing the voltage across R1, and hence the current through the led

^{simulate this circuit – Schematic created using CircuitLab}

if you want to go all the way to sun brightness you'll probably need a 3W led upgrade Q1 to something that can bolt to a heatsink, and use 1 ohm for R1.

Answer 4

Instead of linearizing, I'd use an analog to digital conversion with a wide enough range to cover what you need.

A good candidate would be a comparator relaxation oscillator that turns a resistance into a frequency. All you need is a comparator, two resistors, and an accurate cap.

This can be input into a microcontroller timer to measure the frequency.

Answer 5

LDRs aren't logarithmic.

\$ R = R_{10}\left(\frac{10\ \mathrm{Lux}}{E}\right)^\gamma \$

\$R_{10}\$ is the resistance at 10 Lux, \$E\$ is the illuminance and \$\gamma\$ is a "constant" usually between 0.5 to 1 (citation needed).

To get a fairly linear response, you can apply a constant voltage across the LDR and measure the current through it.

^{simulate this circuit – Schematic created using CircuitLab}

Example design (with free bugs) - my website: https://oskog97.com/projects/light-sensor/light-sensor-part-1-design.pdf#page=39

Measurements - my website: https://oskog97.com/projects/light-sensor/light-sensor-part-2-testing.lowres.pdf#page=29 (really begins at page 24)

Answer 6

You could do it in software, or with hardware you could probably use an anti-log circuit. The problem would be where the curve changes direction.

I think that you could offset it to make the mid-point 0 V, have a circuit to detect the sign of the voltage and an absolute value circuit. Run the absolute value through an anti-log and then multiply by the sign.

Answer 7

OK, I guess what you want is an analog signal representing the logarithm of the illuminance. The graph indicates (conductance) ∝ (illuminance)^0.6, not linear. What I would do is bias one end of the detector with a fixed voltage, measure the current to "ground", and then extract the log of that current as a voltage using a diode. The power-law nonlinearity then simply becomes a scale factor multiplying the logarithmic output.

Sounds easy, but there are serious subtleties. Fortunately, this is a fairly common problem, and there are "logarithmic amplifier" chips that do this. An example is the LOG101.

On the other hand, if your ADC input has a nice high impedance, you could simply do the "poor man's" version:

^{simulate this circuit – Schematic created using CircuitLab}

This will add a bit of nonlinearity and temperature sensitivity. Unless you calibrate these out, it's probably only good to 30% or so as an absolute photometer. That's still far better than the human eye is at photometry.

Answer 8

Using WebPlotDigitizer to extract the data points from the graph I can plot the conductivity vs illumination. We see that the data is an extremely good fit for the conductivity being linearly proportional to LUX with a small offset of 40mOhms in the resistance.

But as an academic question, I would like to know how to turn this curved measurement into a more linear output.

In that case don't use the LDR as a voltage divider. Use it with an op-amp in a standard photodiode amplifier setup. You will then get a voltage that is linearly proportional to indecent light (See OA7 in the circuit below).

Human perception of brightness is logarithmic. So, if you wish to transform your linear sensor output so that it matches human perception, you would use a "LOG AMP".

Temperature stable log-amps can be purchased as off the shelf microchips.

The TL441CN is one example.

Or you can make one using standard op-amps.

^{simulate this circuit – Schematic created using CircuitLab}

The log amp consists of three parts.

1. A fixed voltage is placed across the LDR ot get a current proportional to LUX. OA7 converts that current into a voltage. Its a standard photodiode amplifier setup.

2. OA7 feeds a chain of non-inverting amplifiers that are cascaded together. Lets say we want to measure brightness between 0.1 LUX and 10K LUX (five orders of magnitude). We could then (arbitrarily) pick that we will use five stages each with a gain of 10. We could have also used 10 stages with a gain of sqrt(10). It just depends on how accurate you want to get.

3. Finally OA6 sums up the output of all stages.

Lets say we want 10,000 LUX to output +3.3V. Lets assume all the opamps are "rail-to-rail" output and powered by ±3.3V. Maximum output occurs when all five stages saturate and output -3.3V each. We have five stages going into the summing amplifier OA6. In that case we choose R18 = 20K so the summing amplifier has a gain of 1/5.

At 10,000 LUX the LDR has a resistance of 100 ohms. So the current in the LDR will be 3.3V/100 ohms = 33mA. Since this is the highest level we wish to measure, we want 33mA to generate an output on OA7 that just barely causes OA1 to start saturating. This will occur when the input is -3.3V/G = -0.33V (with G=10). Therefore R17 = 10 ohms.

The general equation for this circuit is VOUT = C * LOG10(LUX)

In this particular case 3.3V = C * LOG10(10000). So C = 3.3V / 4.

VOUT = 0.825V * LOG10(LUX)

Note that the op-amps in the circuit need dual supplies to work (+3.3V and -3.3V).

Also note that you must pick op-amps with very low input offset voltage. Ideally something much less than 3.3V / 100000 = 33uV.

Something like the TLC2652IN with a 3uV offset might be appropriate.