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In HCNR200 Datasheet, the page 11, figure 13-A, there is a sample that confused me. In this sample, the Vin is applied to the inverting input and if the electric current passes through PD1, PD1 voltage goes to zero and finally when PD1 voltage reached to zero, the voltage on inverting input is around zero, but a bit more than non-inverting input voltage. So, how does op-amp output go to higher than zero and adjust the current of LED? And for right side op-amp, how does op-amp output go to more than zero while its non-inverting input is connected to ground and inverting input is going to higher than it?

*Op-Amps are powered by a single supply.

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  • \$\begingroup\$ Hello, and welcome to eesx. Can you post a picture of the figure you are referring to? Is it this? I found no trace of PD1, can you clarify? \$\endgroup\$
    – Vladimir Cravero
    Commented Feb 21, 2019 at 16:00
  • \$\begingroup\$ please check out this:https: //www.edn.com/Home/PrintView?contentItemId=4458406 \$\endgroup\$
    – Mohammad Hussain Asadzadeh
    Commented Feb 21, 2019 at 16:18
  • \$\begingroup\$ Figure 13-A isn't on page 11 and it doesn't have a part labelled PD1. Please edit your question to include the actual circuit you're asking about without relying on links to external sites otherwise it's likely to be closed as "unclear what you're asking". \$\endgroup\$
    – Finbarr
    Commented Feb 21, 2019 at 17:28

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I've copied the relevant figure from the data sheet so that I can speak to it.

The HCNR200 couples the LED output equally to the photodiodes. So assuming good matching, the current induced on the photodiodes is equal if their voltages are equal.

A photodiode is a current generator, so it will drive its cathode negative when light is shined on it. So in the circuit to the left, \$V_{IN}\$ tends to drive the op-amp positive input high, which lights the LED. This generates a current in PD1 which drives its cathode voltage down. Assuming stability (which you'll find to be a big assumption when you try to make this work), A1 will hold the LED current to whatever it needs to be so that \$V_{IN} - I_{PD1} R1 = 0\$.

This will induce a current in PD2. Again, the cathode of PD2 will tend to be driven more negative than its anode. A2 will counteract this by raising \$V_{OUT}\$. If PD1 and PD2 are nicely matched, then \$V_{OUT}\$ will be proportional to \$V_{IN}\$.

enter image description here

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  • \$\begingroup\$ That is Ok, but how does PD1 go to negative while Vin is positive? Suppose that PD1 is short circuit, so it can not goes to negative!! \$\endgroup\$
    – Mohammad Hussain Asadzadeh
    Commented Feb 21, 2019 at 16:14
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    \$\begingroup\$ @MohammadHussainAsadzadeh, the photodiode can turn the light energy hitting it into electrical energy, so it can operate in the II and IV quadrants. Another way of saying this is, it doesn't act like a short circuit, it acts like a current source. \$\endgroup\$
    – The Photon
    Commented Feb 21, 2019 at 17:17
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    \$\begingroup\$ If current from the Photodiode pulls down (goes INTO) the photodiode. \$\endgroup\$
    – analogsystemsrf
    Commented Feb 21, 2019 at 17:54
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    \$\begingroup\$ @MohammadHussainAsadzadeh - A photodiode is not like a resistor or transistor. It is, in fact, a generator. It produces voltage (and current) without any applied voltage. Light which is absorbed by the PD physically knocks electrons loose from the material which accumulate at one terminal. And, for what it's worth, ignoring offset voltages in the op amp, the + input never goes negative. It just gets closer to zero than it would if there were no PD. The proof of this is that the LED current is positive, which means the output voltage is positive, and this can only occur if the + input is pos. \$\endgroup\$
    – WhatRoughBeast
    Commented Feb 21, 2019 at 20:24
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    \$\begingroup\$ @WhatRoughBeast: Common mode on an LM358 goes to -VCC; it probably works enough below that to make the circuit work. Running it on dual supplies would be smarter, though. I wouldn't be surprised if there were parts out there that are specified to go slightly below the - input voltage. \$\endgroup\$
    – TimWescott
    Commented Feb 24, 2019 at 1:25

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