Reversed bias diode with applied voltage

[zak8000]

hi there

i just have a question about diodes say for example we have the following cct:
the circuit with the diodes with the 5V supply, one of the diodes is on and the other off and i understand why this happens but when dealing with the circuit the opamp and the diode i am confused. the output voltage of the diode and opamp cct is 5volts and this is because the positive terminal of the opamp is at a higher potential than the negative terminal and as a result the output of the opamp heads toward its positive rail which is 5 volts but then i assume that the diode would not conduct as the cathode is more positive than the anode and so the output voltage should be zero volts but the simulation says its 5 volts. so could someone please help me with this confusion and also if the output is 5 volts then what is the point of putting that diode in the cct when it doesn't change the voltage coming from the output of the opamp to the output of the cct?

 

[technician]

Are you certain you have drawn the circuit correctly. Are you sure you have the op amp + and - inputs the right way round?

 

[zak8000]

yes its drawn in the correct format. it would make sense to me if drawn the other way around

 

[technician]

I think I agree with you.
Don't see the point of it !

 

[skeptic2]

The problem is that there is no load shown in the diagram. Without a load on the diode, what is the voltage on the anode. When the anode is left open the voltage on it would have to be the same as the voltage on the cathode.

So what is the voltage on the cathode?

 

[sophiecentaur]

Before you decide what the output volts are you should give it a load (series resistance) and connect, via that resistance, to some potential. That is the only way that you can discuss what current may be flowing through that diode.

 

[zak8000]

the output end is connected straight into a micro controller model p80C557E (at the interrupt pin) i am not really familiar with micro controllers but should the load be the internal resistance of the microcontroller?

 

[skeptic2]

... but should the load be the internal resistance of the microcontroller?

Yes, but it is important to know if the input transistor to the interrupt pin an NPN or a PNP. I'm guessing a PNP because interrupts are usually active low. This means the interrupt pin will tend to pull the anode of the diode high which is the non-interrupt condition. Only when the opamp's output is low will the diode conduct and activate the interrupt.

 

[sophiecentaur]

Strictly speaking, if you don't know what the load resistance is (and it may be unspecified but highish) it would be better to define it by a particular resistor to Vcc. The spec would tell you (if you dig deep enough) what the input current is for 0 and 1 logic levels.

What is the function of the diode, though? Is it just protection for the gate input? Wouldn't a series resistor do as well? Or just define the output voltage range of the OP Amp circuit appropriately?

 

[NascentOxygen]

if the output is 5 volts then what is the point of putting that diode in the cct when it doesn't change the voltage coming from the output of the opamp to the output of the cct?

Without knowing the details of the following stage, we can but speculate. But if there were a number of these op-amp circuits to feed a common input pin of the following stage, then a diode after each op-amp, like this, would allow a wired-OR gate at the input to next stage. (It would be wired-AND, to be precise.)

I'm thinking TTL type input at the next stage, where it's active LO.

 

[sophiecentaur]

would allow a wired-OR gate at the input to next stage. (It would be wired-AND, to be precise.)

That's really nasty, you should wash your mouth out my boy. 'Proper' logic units are so cheap these days that this sort of thing is really not called for.

Also, most small scale integrated logic these days is surely CMOS - so you really would want to decide one way or another. It's always a problem when someone posts a bit of circuit that's been extracted from a larger circuit, expecting to get some sense. Then we all rush off with our own ideas and start squabbling.

 

[NascentOxygen]

It could be something as simple as being in parallel to a push button reset, the diode affording protection to the op-amp. Technically, it's still a wired-AND gate.

 

[skeptic2]

The diode would also be important in a situation where the opamp circuit was operating with a higher voltage supply than the microprocessor could tolerate.

 

[Kholdstare]

In Paul Falstad's simulator the diode just turns on even if one terminal floats. Use Cadence Orcad PSpice for glitch-free simulation. BTW the voltage at +ve terminal is 2.85V.