- #1
PainterGuy
- 940
- 70
Hello everyone,![Wink :wink: :wink:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
I am having difficulty understand the concept of current source and what it really is. I have written down below what I think of current source. Some things I said will be wrong and there could be possibility that some are not wrong!
Now you help me out please. Much grateful.
Please have a see on these two pages:
https://docs.google.com/viewer?a=v&...0MGMtYzAxMjBjZDRhYWU0&hl=en&authkey=CMKo1rUF"
https://docs.google.com/viewer?a=v&...0NTYtZmRmZjYyN2ZlNmJk&hl=en&authkey=CIrjo6wP"
I think 'in reality' every power supply such as batteries are voltage sources with a almost fixed voltage and almost fixed power which they can deliver. There would be internal resistances in these power supplies which is mostly shown as Rs in the diagrams.
On the "Page 1" above I think Rs is not the internal resistance which stands for internal losses. Perhaps it has simply been put there to derive an expression for a current source. A current source is a device which can deliver a fixed current value with variable voltage. I don't think 'in reality' there exists a current source but you can get one by manipulating the voltage source. Even in the current source, Rs, does not stand for internal resistance representing internal losses. Rather is has been used to derive an expression.
In Fig 8.6 on Page 1 if you had a variable Rs then you can have a variable voltage source. In the same manner if have variable Rs in Fig 8.7 then you can have variable current source.
But what would happen if we change RL? Keeping the Rs unchanged and increasing the value of RL to RLL would make the value of current passing through RLL less than the current IL which was passing through RL. So how can we keep the value of current still same, out aim is to retain the IL although now we have bigger resistance RLL. Simply answer is increase the voltage in the constant current source? How are we to do it?
Cheers
I am having difficulty understand the concept of current source and what it really is. I have written down below what I think of current source. Some things I said will be wrong and there could be possibility that some are not wrong!
Please have a see on these two pages:
https://docs.google.com/viewer?a=v&...0MGMtYzAxMjBjZDRhYWU0&hl=en&authkey=CMKo1rUF"
https://docs.google.com/viewer?a=v&...0NTYtZmRmZjYyN2ZlNmJk&hl=en&authkey=CIrjo6wP"
I think 'in reality' every power supply such as batteries are voltage sources with a almost fixed voltage and almost fixed power which they can deliver. There would be internal resistances in these power supplies which is mostly shown as Rs in the diagrams.
On the "Page 1" above I think Rs is not the internal resistance which stands for internal losses. Perhaps it has simply been put there to derive an expression for a current source. A current source is a device which can deliver a fixed current value with variable voltage. I don't think 'in reality' there exists a current source but you can get one by manipulating the voltage source. Even in the current source, Rs, does not stand for internal resistance representing internal losses. Rather is has been used to derive an expression.
In Fig 8.6 on Page 1 if you had a variable Rs then you can have a variable voltage source. In the same manner if have variable Rs in Fig 8.7 then you can have variable current source.
But what would happen if we change RL? Keeping the Rs unchanged and increasing the value of RL to RLL would make the value of current passing through RLL less than the current IL which was passing through RL. So how can we keep the value of current still same, out aim is to retain the IL although now we have bigger resistance RLL. Simply answer is increase the voltage in the constant current source? How are we to do it?
Cheers
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