The output should be "V+ minus V-" * Open-loop-gain, and in my case where an ideal opamp is used, the output would be saturated at one of the voltage supply rail.
Sorry... and even if there is no time difference, I try to always remember that any of us may not be able to reply rather quickly. That is why I used the word 'postponed' in my previous post... assuming you might be busy
I think you agree with me that your answer is wrong since we have here a well defined circuit with its sources (after removing C1).
Just to be sure that we think alike here is what I mean:
V2 (IN-) = 1V
V3 (IN+) = 0V
Therefore:
V2 (IN-) > V3 (IN+), and the output V1 knows very well at which supply rail should be saturated.
You were right, V1 = V- (or close to it)
Let us continue to the next step and bring C1 back while the circuit is running.
A capacitor means a storage device (like a battery if you like). So we can't be sure if C1 still has some charges in it or not. In case there are, there would be a voltage on its two terminals.
Since you as I like that everything is well defined, we try to be sure first that Vc1 = 0V (In practice, we discharge it through a resistor, or by shorting it if you like high current spikes :twisted: ).
Now we have an opamp with:
V2 = 1V
V3 = 0V
V1 = V- (I hope you agree that V+ = - V- = 5V to get numerical results later, otherwise let me know your choice)
And C1, with an initial voltage of 0V, will be added between V1 and V2 (yes, we are now watching a real test ;-) )
What will happen?
(1)
You found out in advance that the steady voltage of C1 when you first simulated the circuit is V1-V2, that is:
Vc1 = -5 - (+1) = - 6V
Here we assume that the opamp output is of rail to rail type (otherwise the output V1 would be a bit higher as -4.7 V for example).
(2)
Therefore when C1 is placed between V1 and V2, the circuit starts to charge it from 0V to -6V. But what is about the charging current from t=0 to infinity?
Well, as soon as C1 is in place, V1-V2 becomes 0V (much like if we place a small empty battery of 0V).
Fortunately this insertion helps the opamp to leave saturation and be in the linear region. I guess you know why. The reason is that at t=0 V1=V2=1V which means the ouput V1 is no more saturated and the opamp can restore its full control on its three terminals (V2, V3 and V1). We know when the opamp is linear V2 becomes close to V3 (due to the very high gain), and here V3 = 0V so yes... V2 which was 1V follows quickly (say at t=0.00000001
) its brother IN+ and be close to 0V too. But V1-V2 at this moment is 0V (forced by C1), the output had no choice but to be 0V too. All this will happen at time 0 :grin:
(3)
To be continued...
(3)
Now we have:
V1=V2=V3=0 and the opamp is linear since its output is not saturated.
Also V4 (the input source) = 1V
Therefore a ‘constant’ current can pass in the resistor R1 from V4 to V2:
I(R1) = (V4-V2) / R1
I(R1) = (1 - 0) / 1e6 = 1e-6 A
At node V2, the sum of currents should be zero and since the opamp input current is negligible:
I(C1) = I(R1) = 1e-6 A
Where I(C1) is the current from V2 to V1.
And, as you know, the voltage on C1 starts to rise, based on:
Vc(t) – Vc(0) = I(C1) * t / C1
And since Vc(0) = 0
Vc(t) = I(C1) * t / C1 = V2 – V1
But as long the opamp output is not saturated, V2=0 in this circuit, so:
V1 = - I(C1) * t / C1 = - 1e-6 * t / 1e-6
V1 = - t , a very nice looking result
But this can’t last for too long because V1 has a lower limit at -5V which means after 5 sec the opamp will lose again its control on (V1, V2 and V3).
(4)
After the first 5 seconds, V1 has no choice but to stay calm at -5V and V3 is glad with its stable connection to ground hence V3=0.
But V2 has to quit now its brother (at IN+) and follow the rules of its surroundings instead (that is R1 and C1).
The nodes at V4, V2 and V1 form a simple RC circuit so the voltage on C1 will continue to rise but based on the general exponential formula:
Vc(t) – Vc(0) = [ Vmax – Vc(0) ] * [ 1 – e^(-t/RC) ]
Where
Vc(0) = 5V
Vmax = V4 – V1 = 1 – (-5) = 6 V
That is:
Vc(t) = [ 1 – e^(-t) ] + 5
So after the first 5 sec:
Vc(t) = V2 – V1 = [ 1 – e^(-t) ] + 5
Where:
V1 = - 5 V
Therefore:
V2 = [ 1 – e^(-t) ]
And for t = infinity
V2 = 1 V as expected
(5)
To be continued...
General note:
My intention is to present what is happening in real step by step, which may remind us what we learnt on each component and how it works. I believe a simulator can help much better if one already knows how his circuit runs in general if real. That is why I started with a real circuit here. But I also agree that this way is seldom recommended in the today's very fast world :grin: