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Already, apparently from the feedback I got from this: How to model the output signal of this circuit?

It is better to just explain the whole thing. I am doing a project involving a LDR sensor that is being illuminated by a lamp therefore varying its resistance. Now this LDR goes into a signal conditioning module. I am supposed to estimate the conditioned voltage (the output of the model) as a function of the illumination.

Here is the full circuit:

enter image description here

Any tips on how I should do that? Thank you!

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  • \$\begingroup\$ Without the LDR datasheet, there's no way to justify a quantitative relationship between illumination and voltage output. Even with it, it's only "typical" at best and even then only given the optical circumstances they used in the datasheet to get their typical values. If this is homework, they should be providing the details they expect you to use. \$\endgroup\$
    – jonk
    Jan 3, 2022 at 0:46
  • \$\begingroup\$ Oh yeah the datasheet is this one: kth.se/social/files/54ef17dbf27654753f437c56/GL5537.pdf. Is the quantitive relationship given by the gamma parameter? \$\endgroup\$
    – Granger Obliviate
    Jan 3, 2022 at 0:49
  • \$\begingroup\$ Well, what do YOU think that parameter means from the datasheet? They aren't entirely descriptive there, but if you think a bit and look at the chart, the meaning should arrive. And by the way, which of these devices (it is a series of such) are you to simulate? \$\endgroup\$
    – jonk
    Jan 3, 2022 at 0:53
  • \$\begingroup\$ The datsheet is not good no. We were not told which one of the devices we would be using. Is the gamma the "slope" of the linear (in log-scale) curve? EDIT: I was able tocheck and it is the GL5528 \$\endgroup\$
    – Granger Obliviate
    Jan 3, 2022 at 1:24
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    \$\begingroup\$ Let's take your device. \$\gamma=0.6\$. This means that at 100 lux you should expect to see \$10\:\text{k}\Omega\:\cdot\:\left[\frac{10}{100}\right]^{0.6} \approx 3.162\:\text{k}\Omega\$. (Or twice that if you use \$20\:\text{k}\Omega\$, instead.) That's all they are telling you. They are not telling you what happens at 1 lux or 0.1 lux or 1000 lux. You only get two data points from that datasheet. You could assume it is linear in the log-log-domain. And if so, you have the entire spec and could make a complete curve for the circuit. Otherwise, you can only do so for a short range. \$\endgroup\$
    – jonk
    Jan 3, 2022 at 4:52

1 Answer 1

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Your device is the second one in this table:

enter image description here

Which means that you use the second entry in this table:

enter image description here

Pick a value from between \$10\:\text{k}\Omega\$ and \$20\:\text{k}\Omega\$ from the Light resistance column above and call it \$R_{_\text{CAL}}\$. I don't know which to use or if you should use something somewhere in between. It's your call. But you do have to pick a number.

The formula to use is:

$$R_{_\text{LDR}\left(x\right)}=R_{_\text{CAL}}\cdot\left[\frac{10}{x}\right]^\gamma$$

Where \$x\$ is the illumination of interest, in LUX.

Since their \$\gamma\$ column is only about \$10\:\text{LUX}\$ and \$100\:\text{LUX}\$, all you can reasonably conclude is that the curve follows the given value of \$\gamma\$ within that range. Beyond that you are told nothing.

But at least within that range you can now pick a value for \$R_{_\text{CAL}}\$ that you can defend and then compute the value for everything from \$10\:\text{LUX}\$ to \$100\:\text{LUX}\$ and work out the output voltages from the schematic and those values. That much is doable, given the data sheet.

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