I have an inverting op-amp circuit (image attached)
The power supply voltage on the data sheet was +-15V.
Clipping has occurred at 14.0V on the inverting op-amp, before the connection of a diode.
Clipping has occurred at 15.0V after the diode connection. I'm not sure why this is?

Maybe - As the diode starts to draw the current the voltage across RF drops. When the diode is 0.7V, it is now opened and drawing all the current and voltage. So after the diode has been connected the signal clips at max voltage of 15.0V because all the current is being drawn by the diode.
The feedback voltage is drawn to the reverse bias of the diode which blocks it and so the negative voltage now clips at -1.0V

Also,
The op-amps saturation voltage is 1.0V. Does this mean that the amplifier is operating somewhere within its linear range and is not saturated until it gets to 1.0V from the input current. So it clips 1.0V from the input voltage?

thanks

 

I just spent 3 hours and fifteen minutes writing an explanation of this, and the website told me to refresh the page to get permission to send the "quick reply". (I was already logged in or this reply box would not have been available for me to type in.) I refreshed the page and hope the reply shows up when I click the "reply" button.

 

hit back unitl you see the reply you typed. copy and paste your post into a new post after you log back in.

 

Ahh! that sucks
Myself, on all of these site I always ctrlA and ctrlC (copy all the content) before I hit the submit button, especially when the write up takes some time.
This site sometimes gets the cursor stuck in the quick reply window for some reason. It is necessary to hit the refresh button at that stage, then click the post reply button, which should then work.
I'm sorry for your loss of time.

 

No need to refresh with the cursor stuck.

Press the down arrow to make sure your at the bottom (even though you cant see it)
then press [end] key to go to the end of the line. (that you cant see)
then press [enter]

wha-la.

 

I have an inverting op-amp circuit (image attached)
The power supply voltage on the data sheet was +-15V.
Clipping has occurred at 14.0V on the inverting op-amp, before the connection of a diode.
Clipping has occurred at 15.0V after the diode connection. I'm not sure why this is?

Are you referring solely to the clipping on the positive swing of the output voltage?

With the diode in place the negative excursions with be severely clipped once the diode conducts.

I can't account for the change in +ve cycle clipping onset - from 14V to 15V.

Are these results from a real physical set-up you are testing? Perhaps you could supply more details of the circuit construction & components if that's the case.

It must be a pretty good op-amp to provide full output swing to the +15V supply rail.

 

I'll try again, but I'm not going to spend hours, only to wonder if this site can allow me to send the answer.

First, the saturation voltage is a promise by the manufacturer that the amp will not use up more than 1 volt internally, so the maximum output will be at least Vcc minus one (if necessary), and the other polarity, too.

The + input of the amp is grounded, and the amp wants the other input to be the same voltage, so it will output whatever voltage is required to make both inputs be at the same voltage level. Another rule of op-amps is that they will not allow current to go in or out of the input pins, except for little tiny microamps needed to operate the input transistors.

For example: When your signal generator is delivering one volt positive, that volt pushes current through the 10 k resistor. The input terminal must be at zero volts for the amp to be happy, so one tenth of a milliamp (.1ma) must be going through the 10k resistor. The amp must neutralize the .1ma by sending a negative .1ma through the 100 k resistor. That requires negative 10 volts at the output, so that's what the amp does.

When you add the diode, the output only has to be negative .7 volts to shove .1ma through the diode, so that's what the amp does, it outputs .7 volts.

When the signal generator is delivering negative 1 volt, the amp has to deliver positive 10 volts at its output to neutralize the input current through the 100k resistor, so that's what it does.

When you attach the diode, the positive current can't get through the diode so nothing changes. The amp still has to output positive 10 volts to get .1ma through the 100k resistor.

The diode doesn't "draw" current. The current is always on its way to somewhere else. The diode either allows it, or it doesn't. Clipping happened because the input voltage was more than 1.4 volts and the amp ran out of voltage trying to neutralize the input current.