Tadde
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Greetings to everyone, this is my first post.
Sorry for my bad english, I'll do my best to be as clear as possible.
Question is: how can an op-amp (for example in the noninverting configuration) go in the linear range of its I/O characteristic after being powered up?
Experience clearly demonstrates that noninverting op-amps work obviously, but if I trace the (hypothetical) steps occurring immediately after the power-up, it seems to me that it should go into saturation and remain there.
So imagine an op-amp in noninverting config, and let's call V+ and V- the input and Vo the output voltages. Be R1 and R2 the two resistor forming the feedback so that voltage gain would be 10 (β=1/10). Power supply erogates +15V and -15V , so if output exits the interval (-15V, +15V), the amplifier saturates.
On the noninverting input V+ is a battery of +1V. So I expect that at the end of the transient following the power-up, I shall see +10V at the output.
So let's turn on the power and let's see what I imagine should happen : V+ = 1V and V- = 0V => Vo saturates at +15V;
the feedback divides this tension by 10 and so V- = +1.5V.
Now at the two op-amp input we have V+ = +1V and V- = +1.5V, and so the difference is -0.5V; this difference will drive the output of the op-amp in negative saturation (the open loop gain is so high that the voltage difference between input leads of an op-amp configured in negative feedback so that it should work in linear behaviour is of the order of tens of microvolts).
This negative output is divided by 10 and feed back to the V- terminal and so on.
So, as you can see if this analysis was true, the op-amp would never go in its linear zone.
Could someone of you please enlight me over this issue?
Thanks in advance.
Sorry for my bad english, I'll do my best to be as clear as possible.
Question is: how can an op-amp (for example in the noninverting configuration) go in the linear range of its I/O characteristic after being powered up?
Experience clearly demonstrates that noninverting op-amps work obviously, but if I trace the (hypothetical) steps occurring immediately after the power-up, it seems to me that it should go into saturation and remain there.
So imagine an op-amp in noninverting config, and let's call V+ and V- the input and Vo the output voltages. Be R1 and R2 the two resistor forming the feedback so that voltage gain would be 10 (β=1/10). Power supply erogates +15V and -15V , so if output exits the interval (-15V, +15V), the amplifier saturates.
On the noninverting input V+ is a battery of +1V. So I expect that at the end of the transient following the power-up, I shall see +10V at the output.
So let's turn on the power and let's see what I imagine should happen : V+ = 1V and V- = 0V => Vo saturates at +15V;
the feedback divides this tension by 10 and so V- = +1.5V.
Now at the two op-amp input we have V+ = +1V and V- = +1.5V, and so the difference is -0.5V; this difference will drive the output of the op-amp in negative saturation (the open loop gain is so high that the voltage difference between input leads of an op-amp configured in negative feedback so that it should work in linear behaviour is of the order of tens of microvolts).
This negative output is divided by 10 and feed back to the V- terminal and so on.
So, as you can see if this analysis was true, the op-amp would never go in its linear zone.
Could someone of you please enlight me over this issue?
Thanks in advance.