http://www.google.com/imgres?imgurl...m=isch&client=tablet-unknown&ved=0CCsQMygGMAY

I am wondering how applying a negative voltage to an op-amp's inputs has any effect on the output at all when both inputs are connected to NPN transistors which require a positive voltage to conduct. A negative voltage will just turn it off and more negative voltage won't make a difference if it is already 0 volts or below. Can someone explain how applying a negative voltage to the base of an NPN transistor causes anything at all to happen within an op-amp.

 

This is deep, meaning of life, type stuff.

In that pic the transistors are INSIDE the op amp. You seem to think that they are outside. It seems to me that the flaw in your argument is that you don't actually understand what you are looking at.

 

Sometimes a negative signal just means less positive. Does that answer your question?

 

I am wondering how applying a negative voltage to an op-amp's inputs has any effect on the output at all when both inputs are connected to NPN transistors which require a positive voltage to conduct. A negative voltage will just turn it off and more negative voltage won't make a difference if it is already 0 volts or below. Can someone explain how applying a negative voltage to the base of an NPN transistor causes anything at all to happen within an op-amp.

What you believe is 0V is really not the ground in the circuit (at least to the NPN transistors). The ground is at -V, so 0V is above the ground. In the middle, you see Q9, Q10 and a 5K resistor. Calculating the voltage the top of the resistor is (2*Vdd-1.4) volts above the bottom. It really doesn't care where 0V is.

 

http://www.google.com/imgres?imgurl...m=isch&client=tablet-unknown&ved=0CCsQMygGMAY

I am wondering how applying a negative voltage to an op-amp's inputs has any effect on the output at all when both inputs are connected to NPN transistors which require a positive voltage to conduct. A negative voltage will just turn it off and more negative voltage won't make a difference if it is already 0 volts or below. Can someone explain how applying a negative voltage to the base of an NPN transistor causes anything at all to happen within an op-amp.

Look at the bottom node in that diagram. See that it is labeled V-? That is the negative rail of the opamp, which is generally a negative voltage. What you are saying is basically true, but for the input voltage relative to the negative rail. In actuality, the inputs can't go all the way down to the negative rail and keep the opamp in its linear mode of operation since some overhead is needed for the current source and buffers that appear are beneath the input transistors.

 

So if the inputs are left floating at 0 volts with respect to positive and negative then both transistors are forward active and conducting somewhere between cutoff and saturation?

 

So if the inputs are left floating at 0 volts with respect to positive and negative then both transistors are forward active and conducting somewhere between cutoff and saturation?

If the inputs are floating, then the output usually oscillate between the minimum output and the maximum output. There is no control of the output.

 

So if the inputs are left floating at 0 volts with respect to positive and negative then both transistors are forward active and conducting somewhere between cutoff and saturation?

The statement, "left floating at 0V" is a contradiction. They are either floating, or they are being held at 0V. They can't be both. Also, you haven't defined what "respect to positive and negative" even means. Are you trying to say that the inputs are midway between the supply rails? If the supply rails are at, say, ±12V then if the inputs are at 0V relative to the common reference, they will be at +12V relative to the negative rail and at -12V relative to the positive rail.

If the inputs are floating, then there is nominally no source of base current and both transistors will be in cutoff. In practice, there will be some leakage current in both transistors and the output will often flail around very unpredictably. Floating inputs are almost never a good idea.

 

A floating input has no defined voltage, that's why it's called "floating".
Since the floating transistor inputs conduct no current and the input transistors are thus cutoff, the output will likely go to near one of the supply rails and stay there.

 

Thanks