Hello guys..

The text says:
"A resistor is the simplest I-to-V converter. However, it has the disadvantage of presenting a nonzero impedance to the source of input current; this can be fatal if the device providing the input current has very little compliance ."

Does the above phrase means that the device which has very little compliance has low output resistance which means that if it is cascaded to an output circuit which contain a resistor as a simplest I-V converter ...the resistor will going to load the device due to its little compliance..since resistor is showing a little resistance enough to load the circuit...


second ..
As the inverting input is maintained at virtual ground..then how can a current can flow in the photo diode branch..how does the circuit the acts as a I-V convertor..

please help...

 

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As the inverting input is maintained at virtual ground..then how can a current can flow in the photo diode branch..how does the circuit the acts as a I-V convertor..

please help...

Photons impinging on the detector create a minute current. That same current flows through the feedback resistor. The left end of the resistor is at virtual ground. What voltage is on it's right end?

 

The text says:
"A resistor is the simplest I-to-V converter. However, it has the disadvantage of presenting a nonzero impedance to the source of input current; this can be fatal if the device providing the input current has very little compliance ."

Does the above phrase means that the device which has very little compliance has low output resistance which means that if it is cascaded to an output circuit which contain a resistor as a simplest I-V converter ...the resistor will going to load the device due to its little compliance..since resistor is showing a little resistance enough to load the circuit...

No, that is not what it means at all. The phrase "has very little compliance" has a very specific technical meaning: it is the same as saying, "can only maintain the accuracy of its current output over a limited range of voltages."

You figure out the rest.

second ..
As the inverting input is maintained at virtual ground..then how can a current can flow in the photo diode branch..how does the circuit the acts as a I-V convertor..

please help...

Photovoltaic light sensors GENERATE current in response to light falling on them; that is how current can flow in the photodiode branch.

 

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The text says:
"A resistor is the simplest I-to-V converter. However, it has the disadvantage of presenting a nonzero impedance to the source of input current; this can be fatal if the device providing the input current has very little compliance ."
...

Do you understand what "compliance" means in the context of describing a current source?

 

Understand that a photo diode can generate current with zero voltage across the terminals (both sides shorted to ground) and will actually generate the maximum current under those conditions.
That is why photo diodes are often used in I-V (transimpedance) circuits.

 

Do you understand what "compliance" means in the context of describing a current source?

compliance means that the current from the source will vary significantly if the load voltage changes even a small amount....

 

compliance means that the current from the source will vary significantly if the load voltage changes even a small amount....

Just the opposite; the output voltage of a current source will do whatever it takes to keep the current constant. The range of voltages which a current source is capable of supplying while maintaining it's rated current is called its compliance.

Here is what Wiki says:
No physical current source is ideal. For example, no physical current source can operate when applied to an open circuit. There are two characteristics that define a current source in real life. One is its internal resistance and the other is its compliance voltage. The compliance voltage is the maximum voltage that the current source can supply to a load. Over a given load range, it is possible for some types of real current sources to exhibit nearly infinite internal resistance. However, when the current source reaches its compliance voltage, it abruptly stops being a current source.

 

Just the opposite; the output voltage of a current source will do whatever it takes to keep the current constant. The range of voltages which a current source is capable of supplying while maintaining it rated current is called its compliance.

Absolutely ..but what does the following phrase means.. "it has the disadvantage of presenting a nonzero impedance to the source of input current; this can be fatal if the device providing the input current has very little compliance ." I am unable to visualize it..

 

Three current sources:
First (green) is Ideal, maintains 1mA through any load resistance. Look at the voltage required as the load resistance gets higher. Its compliance is at least 1KV.

Second (red) is Non-Ideal due to finite output impedance. It does an ok job while the load resistance is less than about 1KΩ. What would you say its compliance is?

Third (blue) is Non-Ideal due to limited compliance (10V). It is capable of delivering its rated current to any load resistance less than 10KΩ.

 

Three current sources:
First (green) is Ideal, maintains 1mA through any load resistance. Look at the voltage required as the load resistance gets higher. Its compliance is at least 1KV.

Second (red) is Non-Ideal due to finite output impedance. It does an ok job while the load resistance is less than about 1KΩ. What would you say its compliance is?

Third (blue) is Non-Ideal due to limited compliance (10V). It is capable of delivering its rated current to any load resistance less than 10KΩ.

View attachment 91245

Ok iam thinking it this way....

A source with little compliance is equivalent to the source have a low value resistor in parallel - if the voltage across the source changes a small amount the current through the resistor changes and takes away some of the source's output current. In the circuit above the opamp doesn't allow any voltage across this parallel resistance and so no current flows through it. This ensures that all the current from the source flows into the feedback resistor and generates a voltage that can be measured at the output.

am i right??

 

Ok iam thinking it this way....

A source with little compliance is equivalent to the source have a low value resistor in parallel - if the voltage across the source changes a small amount the current through the resistor changes and takes away some of the source's output current. In the circuit above the opamp doesn't allow any voltage across this parallel resistance and so no current flows through it. This ensures that all the current from the source flows into the feedback resistor and generates a voltage that can be measured at the output.

Correct.

i.e the high input impedance of opamp does not let the source get loaded by the 1 mohm feedback resistor.
am i right??

No, the input input impedance to the opamp is zero (virtual ground). This makes it so none of the detector current is lost internally; all of the photo current flows in the 1meg feedback resistor...

 

Understand that a photo diode can generate current with zero voltage across the terminals (both sides shorted to ground) and will actually generate the maximum current under those conditions.
That is why photo diodes are often used in I-V (transimpedance) circuits.

Is it reasonable to believe that the current generated due to a photodiode is independent of the voltage across it.....if so then what initiates or provide a driving force for the electrons to flow in absence of potential gradient...

 

No, the input input impedance to the opamp is zero (virtual ground)

Now its for sure that opamp doesnt allow the current to flow across the parallel resistance...but for that it has to appear at a low resistance..

The effective input resistance will be the feedback resistor divided by the gain of the opamp - since that gain may be one million the input resistance can look like a 1 ohm resistor. If the effective output resistance of the source is 1 kilohm then 99.9% of the source current will go through the feedback resistor and thus generate the output voltage..

Can this be a possible explanation of opamp appearing at a low resistance....?

 

Is it reasonable to believe that the current generated due to a photodiode is independent of the voltage across it.....if so then what initiates or provide a driving force for the electrons to flow in absence of potential gradient...

It would if the diode had a high impedance but as the voltage increases the diode junction becomes forward biased, thus shunting away the current.

The driving force that generates the current is the energy from the light photons that are absorbed by the junction.