I looked over the web for an answer for this question, but everybody tells that if you don't that your op amp will not work, So I will rephrase my question here: why the op amp will not work if you don't provide a path for input bias current.

I found an answer that refers the reason to coupling capacitor on the power supply of op amp, where the input bias current will full charge it(at least that what I understood) but assume that you have a signal : A + sin(t), then I expect that the DC part (A) will full charge any capacitor that pass it, which is not make sense.

 

Show a schematic of what you are talking about.

 

Show a schematic of what you are talking about.

I'm talking about Rb and Vbias, why we should provide a DC return path, and why there is a voltage source not groung.

 

All op-amps use a transistor for their input. It might be bipolar or it might be a mosfet. The bipolar transistor leaks current through its base, either into or out of the op-amp. That current must find a path to some voltage between the collector and the emitter voltages or the transistor will just lock up at fully on or fully off. The mosfet leaks a lot less current through its gate, but the gate voltage still has to be between the source voltage and the drain voltage or it will lock up either fully on or fully off. If all you want is an op-amp with its output stuck to Vcc or ground, you don't need a DC path. If you want your op-amp to operate in the linear range, it must be connected to some voltage between the supply rails and that voltage source must be more powerful than the input bias current.

 

You can answer your own question with a little thought experiment.
What happens to the input node voltage (which has some small stray capacitance) if the bias current has nowhere else to go?

 

I looked over the web for an answer for this question, but everybody tells that if you don't that your op amp will not work, So I will rephrase my question here: why the op amp will not work if you don't provide a path for input bias current.

I found an answer that refers the reason to coupling capacitor on the power supply of op amp, where the input bias current will full charge it(at least that what I understood) but assume that you have a signal : A + sin(t), then I expect that the DC part (A) will full charge any capacitor that pass it, which is not make sense.

Same word "bias" but used two different ways, I think. "Bias" as used in the schematic is not the same "bias" referring to some character of the op amp. "Bias", as used in the schematic is a reference voltage compared to the input signal.

 

Because a BJT only conducts current in one direction, which means it must be biased so that the AC current doesn't attempt to go below 0 (in the collector to emitter direction in an NPN). And since there's a relationship between base current and collector current, if there's no path for the transistor to draw DC current, it can't develop the proper collector-emitter current to function as you want it to.

See below. The input signal is biased so that Q1 has current flowing from collector to emitter at all points of the signal (red trace). Q2 has the DC level blocked, causing the signal at its base to be centered at (about) 0V, so only the portions above 0V turn on the transistor.