- Joined
- Apr 1, 2011
- Messages
- 15,221
- Helped
- 2,901
- Reputation
- 5,814
- Reaction score
- 2,990
- Trophy points
- 1,393
- Location
- Minneapolis, Minnesota, USA
- Activity points
- 114,064
I'm picturing how a simulator algorithm models an op amp. I think it has a hard time figuring out how to present 0V output.
When it is at V+, it is easy to suppose it presents a low impedance path to V+.
When it is at V-, we think of it presenting a low impedance path to V-.
But what about somewhere in between? How can it present 0V?
At first we might think it would be a disconnection from either supply polarity. But that isn't right. The output cannot present high impedance.
But how much impedance should it present? And should it present the same impedance to both the positive and negative supplies?
A terminal at 0V should become ground for both V+ and V-.
It should provide a path to V- to some extent.
And it should provide a path to V+ to some extent.
It needs to be dynamic. Say you hook up a 500 ohm load. The other end of the load goes to V+. The output terminal should not change. So to do that, the op amp finds that it has to provide 10k internal resistance to V+, and 470 ohms internal resistance to V-, in order to provide 0V to the load.
Now say the load changes. Or the output changes. The op amp must calculate new internal resistances.
To simulate all this has got to be a chore.
Then there is the current in the supply legs. To calculate this, the algorithm has to know what is the current in the load, or loads, and how much current it should source, and how much it should sink. Etc.
The process cannot be easy to model.
When it is at V+, it is easy to suppose it presents a low impedance path to V+.
When it is at V-, we think of it presenting a low impedance path to V-.
But what about somewhere in between? How can it present 0V?
At first we might think it would be a disconnection from either supply polarity. But that isn't right. The output cannot present high impedance.
But how much impedance should it present? And should it present the same impedance to both the positive and negative supplies?
A terminal at 0V should become ground for both V+ and V-.
It should provide a path to V- to some extent.
And it should provide a path to V+ to some extent.
It needs to be dynamic. Say you hook up a 500 ohm load. The other end of the load goes to V+. The output terminal should not change. So to do that, the op amp finds that it has to provide 10k internal resistance to V+, and 470 ohms internal resistance to V-, in order to provide 0V to the load.
Now say the load changes. Or the output changes. The op amp must calculate new internal resistances.
To simulate all this has got to be a chore.
Then there is the current in the supply legs. To calculate this, the algorithm has to know what is the current in the load, or loads, and how much current it should source, and how much it should sink. Etc.
The process cannot be easy to model.