Following up this question: Thévenin equivalent: is this resistor useless?

I've now changed the voltage source to a resistor since this is a photoresistor. However I have no idea now how to represent the output voltage? How does it make sense to have a circuit without sources? I am supposed to have an output voltage in terms of illumination of the photoresistor. I need some kind of hint because I have no idea how to analyze this circuit.

enter image description here

I mean, if I did what I did before I would have:



  • 1
    \$\begingroup\$ What is V1? That looks like a source to me. \$\endgroup\$
    – Hearth
    Jan 2 at 18:21
  • \$\begingroup\$ @Hearth I have just edited. But this doesn't make sense, how can I obtain a voltage proportional to the voltage across the LDR which I will then amplify? \$\endgroup\$ Jan 2 at 18:34
  • \$\begingroup\$ You have to add an EXTERNAL supply \$\endgroup\$
    – jp314
    Jan 2 at 18:38
  • 1
    \$\begingroup\$ This is starting to seem like an XY problem. Can you please tell us what the purpose of this problem is? Or post the original problem description from your lab assignment? \$\endgroup\$
    – Carl
    Jan 2 at 18:40
  • 2
    \$\begingroup\$ @Chu: as I read it, the upper terminal of V1 is -5V, despite the "+" sign in the battery symbol. \$\endgroup\$ Jan 3 at 0:53

2 Answers 2


As Hearth points out, an LDR doesn't generate a voltage. It's a highly non-linear light-dependent resistor.

I took this "fuzzy" curve from this datasheet, which gives an idea about various LDRs:

enter image description here

These curves are not strictly linear in the log-log-domain. But they are close enough that for circuit simulation using a strictly linear log-log-domain approximation should be sufficient.

What I did to simulate them was to create a symbol and a Spice model for it that allows me to input a linear voltage source. This model can be used for both the LDR resistance as well as for generating a curve representing the illumination in LUX:

enter image description here

A control voltage that spans from \$0\:\text{V}\$ to \$1\:\text{V}\$ is then applied. For now, I do have to use the formulas shown in order to compute the proper value for the chart. But all I'm doing here is showing how an LDR sensor might be simulated.

  • \$\begingroup\$ Hi @jonk! Thank you for showing me how the simulation of a LDR can be done. I have posted another question showing the big picture of what I am doing as apparently my questions are not leading anywhere. I need to arrive to an expression of the output voltage in terms of the LDR sensor illuminance, but I am stuck. Hope you can give it a look: electronics.stackexchange.com/questions/602755/… \$\endgroup\$ Jan 3 at 0:34
  • \$\begingroup\$ @GrangerObliviate Looks like no one is providing you with sufficient answers. Glancing over your questions, anyway, and seeing how many (almost none) have been selected. Are you having difficulty getting useful answers here? If so, just say so. I'll take a peak. Maybe something will come to mind. \$\endgroup\$
    – jonk
    Jan 4 at 4:11

For one thing, your circuit does include a voltage source, you've called it V1 and it's a -5 volt source in series with the LDR.

For another thing, the output of a circuit doesn't have to be a voltage. In most cases with LDRs, the output is a resistance, which you can measure by applying a known voltage and measuring the current, or by applying a known current and measuring the voltage. For such a circuit, the source is in the readout device, not in the sensor. An LDR is, physically, a chunk of intrinsic semiconductor (usually cadmium sulfide or cadmium selenide) that changes resistance in response to light striking it (any semiconductor does this); there's no voltage involved unless you apply one externally.


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