Current source, voltage source

[PainterGuy]

Hello everyone,

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!:tongue: 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 R_s in the diagrams.

On the "Page 1" above I think R_s 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, R_s, 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 R_s then you can have a variable voltage source. In the same manner if have variable R_s in Fig 8.7 then you can have variable current source.

But what would happen if we change R_L? Keeping the R_s unchanged and increasing the value of R_L to R_LL would make the value of current passing through R_LL less than the current I_L which was passing through R_L. So how can we keep the value of current still same, out aim is to retain the I_L although now we have bigger resistance R_LL. Simply answer is increase the voltage in the constant current source? How are we to do it?

Cheers

 

[viscousflow]

In a current source, Rs would be on the order of 10 MOhms while RL on the order of 10 Ohms. Therefore, varying the load resistance does next to nothing to the current source in terms of demanding a voltage change.

On the voltage source, vice versa is true. With an extremely small resistance it doesn't matter what RL is, practically all the current will pass through the load.

Is all a game of magnitudes and approximations.

 

[I_am_learning]

After you include internal resitances in ideal Voltage Source (or in Current source), than that makes them non-ideal and hence practicaL. Their Terminal Voltage (or the current they provide) are no longer constant. They varry with variation of Load (RL to RLL .)
So we see that,
the current provided by and terminal voltage of practical voltage source changes with load.
Ditto for the current source.

So, what makes a practical voltage source different from practical current source, if both's voltage and currents aren't constant?

The answer lies in quantity. A practical voltage source's voltage varry slightly with change in load while current varies greatly. A practical current source load current varries only slightly while voltage varies greatly. (how a current sorce achive this is a different story).

So, how do you calculate by how much does the output voltage of voltage source (or output current of current source) varries with load?
You use Voltage divider formula for Voltage source and Current divisor formula for current source.
And remember Rs of voltage source is very small and Rs of current source is very large in compared to 'general' or 'normal' load resistance Rl.

 

[viscousflow]

And remember Rs of voltage source is very small and Rs of current source is very large in compared to 'general' or 'normal' load resistance Rl.

You have that backwards there. Rs needs to be large for the Voltage source and Rs needs to be small for the current source in comparison to RL.

 

[I_am_learning]

Hi, viscousflow,
Say, I have a 100 V voltage source. I am going to connect it to a 100ohm lamp.
Assume the internal resistance of the source to be
1. 1 ohm
2. 10000 ohm
Calculate output voltage and current in each case.
Pherhaps then we can know, who was correct.

 

[viscousflow]

Hi, viscousflow,
Say, I have a 100 V voltage source. I am going to connect it to a 100ohm lamp.
Assume the internal resistance of the source to be
1. 1 ohm
2. 10000 ohm
Calculate output voltage and current in each case.
Pherhaps then we can know, who was correct.

Yes you are correct, thanks for the mental problem. I guess I was thinking about it backwards. Replaced a few words, now my first post is correct.

 

[PainterGuy]

Many thanks thecritic and viscousflow.

I still have many questions to ask and will ask them soon. Be there to help me out please!

Cheers

 

[Bassalisk]

How do you build a current source :S how do you make a source always give out a constant current :S

 

[sophiecentaur]

You need lots of volts available and what is, effectively, a variable series resistor which is adjusted to give the wanted current into whatever load you present it with.
There are 'active' circuit elements that behave more or less that way
The Anode of a Thermionic Triode is a pretty good current source, actually and so is the Collector of a transistor. With a bit of feedback, however, you can make the current much more 'constant' over a limited range of load values.

 

[Averagesupernova]

I always thought of a current source as a very very very very high voltage supply with a very very very very large resistor in series. The larger the series resistor the less effect the change in load resistance has on the load current.
-
The main reason a collector in a transistor is a good current source is because of the feedback caused by the emitter resistor. Most people are at first unable to see that an emitter resistor causes negative feedback. Sophie, don't get me wrong, I'm not correcting you, just a little elaboration.

 

[cabraham]

A current source can be realized w/ a common generator. If the generator's torque is regulated to constant value, then the current will remain constant w/ varying load resistance. If the generator's speed is well regulated, the voltage (& frequency) will remain constant w/ varying load.

The power company maintains a very constant turbine speed. This produces constant voltage as well as constant frequency. They could provide the grid w/ a source of constant current instead, but there are difficulties. Losses due to current are I^2*R, vs. losses due to voltage which are V^2*G. It happens that insulators are near ideal at any temperature, & losses are very low. But conductors are further from ideal at common temps & losses are much greater.

Superconductors are lossless, but until high temp superconductivity is realizable & affordable, generating & transmitting at full voltage w/ variable current per load demand, is much more efficient than the alternative. So the power company operates their generators in the constant voltage mode. Constant current would be too lossy.

Furthermore, in addition to using constant voltage mode exclusively, the generated voltage is stepped up via transformers. Long distance transmission utilizes high voltage & low current to minimize resistive losses in the conductors.

Batteries can be constructed for either constant voltage or constant current operation. With primary cells like NiMH, Li ion, alkaline, CZn, etc., constant voltage mode is much better. No time to elaborate. But nuclear batteries, still in development, work way better as constant current mode of operation. Do a search using "nucell" as a key word for more details. Did I help?

Claude