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# Role of diodes

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#### Saravana Vilashini M

##### New Member
What is the role of the diode in the circuit attached?

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What is the role of the circuit in your attached drawing? Is it a BS circuit from a class or is it something useful?

It supposedly forces current to flow in a single direction, and forces about 0.6v potential across each capacitor.

I am attaching the full circuit. The input currents get subtracted due to opposite polarity and then current is converted to a voltage. But my question is output remains the same with and without diode. why?

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• LTC6269-10.PNG
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Try asking yourself if it seems that I1 + D1 + C1 does anything meaningful? WIll anything pass to the opamp?

I1 and I2 is meaningful as it is has opposite polarity it gets subtracted and (165u - 150u) = 15uA passes to the opamp

Both current sources are considered to be floating so I1 only forms a circuit through diode D1 (And C1 during transitions.) There is no way it can provide any input to the op amp.
For the second point you need to read up on op amps to understand how they behave. As the non inverting input is connected to ground it attempts to keep the inverting input at the same potential. (I.E ground) An op amp will be considdered to have an infinite input impedance so no current flows in or out of it's input. the only place I2 can go is through is through R3. So for that to happen the op amp output must go to a negative voltage which produced a current of 150 uA (I don't understand the terminology of your pulse description but I am assuming the current is 150 uA) So this voltage must be - 150 mV. Diode D2 does not pass any current as there is zero volts across it. The capacitors will have to be taken into consideration at the transitions of the pulse. If you understand this description of the DC behavior you can then do the calculations which takes into account the reactance of the capacitors.

Les.

I have modified the circuit. Please let me know if this is fine?

I have modified the circuit. PLease anyone explain how does the diode and capacitors behave and flow of currents I1 and I2 ?

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• LTC6269-10.PNG
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The net current difference between the two sources is 15uA.
That current is produced by just 15 millivolts across the 1K resistor.

The signal voltages never get anywhere near the 600mV or so needed for the diodes to conduct.
With steady-state conditions, the capacitors do nothing at all.

If i give I1 - 2A and I2 - 5A. Will the diodes conduct?

If you have to ask that question you have not understood my reply in post #6
What are you actually trying to do ?

Les.

Sorry, Iam new in designing. Yes, true. I felt it difficult

The highlighted wire keeps everything at ground. You'll have current flow but no potential (voltage) difference.

I just want the currents to subtract using two equivalent circuit of the photodiodes. Will the current direction which i have drawn is right?

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• LTC6269-10.png
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If the photo diodes behave as current sources then you will get an output in proportion to the difference in currents What is the purpose of the diodes and capacitors ? If you tell us exactly what you are trying to achieve we will be able to assist you better.

Les.

I want to replace photodiodes with the equivalent circuit of the photodiode.

I want to replace photodiodes with the equivalent circuit of the photodiode.

What is the goal of your circuit?

Noise cancellation

When D1 and D2 become forward bias and reverse bias?
Photodiode operates in reverse bias. So, Diodes (D1 and D2) are in reverse bias condition? How to say theoretically?

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