Dc and ac resistances of an ideal current source

[hanhan]

Hi,
I am really confused about current source resistance. An ideal current source has infinite reistance.
As my understanding, the resistance said above is AC resistance, Rac.
Rac = dv/di

(Courtesy of Jony130)
But how about DC resistance of an ideal current sourrce?
When you talking about ideal current source resistance, which kind of resistance do you imply, AC or DC resistance.
What is the exact definition of resistance? I don't know when we need to use DC or AC resistance. Please help.

 

[KeepItSimpleStupid]

OK, the real deal is you want to look at delta V/ Delta I. e.g. Suppose you built a 1 mA current source and with nearly the same components or say a voltage change, you could make that source output 1.1 mA.

So, you will look at the 0.1 mA and the difference in voltage. You could achieve, say 1e12 ohms of output Z, but it's not infinate.

I am still unclear when given a circuit, how one might calculate the expected output Z, but measuring it is clear.

When designing an I-V converter (The opposite of what you want), Vos and Ib are critical parameters. Ib also likes to vary a lot with temperature. I've heard the term "noise gain", but can't put my finger on the concept.

There is a standard trick. Say you wanted to measure the input impeadnce (resistance if you will) of an audio amplifier. You can inject a signal direct and then use a potentiometer in series and adjust it so the signal is 1/2 of what you had. Now measure the potentiometer. That's the input resistance of the audio amplifier.

You sould do some searching on how x10 scope probes work, In this case your scope is aprox 10 M ohm in parallel with say 22 pf and you have this scope probe with a resistor and a variable cap in it. You adjusting the cap to get a "perfect" impeadance divider and again, you could, in theory, measure indirectly the actual Z of the scope.

 

[rumpfy]

Anh,
In network analysis, the concept of 'Current source' and 'Voltage source' is used. A 'voltage source' is drawn as a battery as you have shown. However, a current source is shown as you have shown but without the battery.
In analysing a network, if it has a voltage source, the sum of all the voltage drops across resistors MUST add up to the source voltage.
In analysing a network, if it has a current source, the sum of all the CURRENTS MUST add up to the source current.
From an analysis point of view, it does not matter whether the source current or voltage is AC or DC.
Of course, to generate a current, one needs to have some kind of voltage supply and I think you are a bit confused by this.
The 'Voltage source' is the 'open circuit output voltage' of your energy supply; ie, no output current.
The 'Current source is the 'short circuit output current' of your energy supply; ie, no output voltage.
In your drawings, just remove the battery and replace it with a short circuit and then the drawing represents the situation.
Practically, if you have a 'voltage source', and you place a short circuit on the output, you get an infinite current.
If you have a 'current source' and place an open circuit on the output, you will get an infinite output voltage.
KISS is talking about practical things, but I think your question is related to the concepts of network analysis.
Hope this helps.

 

[MrAl]

Hi there anhnha,

In your first circuit, if you connect calculate the 'total resistance' 1v/1ma you get 1k.
So the current source sort of looks like a 1k resistor.
But wait, we can say the same thing about the voltage source because that also has 1v and 1ma.
So which is it that is the resistor, and which is the source?

Also, if we connect a 2k resistor in series with the two sources, we still have 1v and 1ma for the battery.
But now the current source has -1v and 1ma, and the 2k resistor has 2v and 1ma, so that's 2v across the resistor and -1v across the current source, so now the current source looks like a -1k resistor. So how did the current source go from being a 1k resistor to a -1k resistor?

More exactly, if we connect the voltage source and current source to a circuit containing a few resistors and measure the voltage at some node N and measure 2.3 volts, then if we open circuit the current source and measure that same node and see 1.8 volts and then short the voltage source and reconnect the current source we see that same node measure 0.5 volts, and 1.8+0.5=2.3 volts, the same voltage we get with both sources active.

What this means is that the current source acts like an open circuit impedance that has the ability to force a current through the external circuit.

So we really have two completely different animals here, a source is not the same as a resistor, and so they are not handled the same way mathematically either.

To sum up:
1. The voltage source acts like a zero impedance.
2. The current source acts like an infinite impedance.

For circuits with sources this also means that the calculations involve other parts of the circuit (other than the sources) in order to establish operating conditions. For example, a 2v battery in series with a 1k resistor gives 2ma, but that same battery with a 2k resistor only gives 1ma, which is entirely different because the 'external' circuit changed.

Another view of this is from the most basic circuit analysis idea using Ohm's Law. We have:
I=E/R
where
I is the current,
E is the voltage,
R is the resistance.

We see we have three different things here:
current,
resistance,
voltage.

They are not the same thing and are measured in different units, so we cant call a voltage a resistance for example.

 

[hanhan]

Thank you, KeepItSimpleStupid, Rumpfy and MrAl for these helps.
KeepItSimpleStupid:

OK, the real deal is you want to look at delta V/ Delta I. e.g. Suppose you built a 1 mA current source and with nearly the same components or say a voltage change, you could make that source output 1.1 mA.

So, you will look at the 0.1 mA and the difference in voltage. You could achieve, say 1e12 ohms of output Z, but it's not infinate.

I believe this is the ac resistance of the current source.

In this case Rac = 10^12 ohms that is not infinite.

Rumpfy: Thanks for this but I think it is not really my question. Sorry I can't explain it clearly.
MrAl:

In your first circuit, if you connect calculate the 'total resistance' 1v/1ma you get 1k.
So the current source sort of looks like a 1k resistor.
But wait, we can say the same thing about the voltage source because that also has 1v and 1ma.
So which is it that is the resistor, and which is the source?

Also, if we connect a 2k resistor in series with the two sources, we still have 1v and 1ma for the battery.
But now the current source has -1v and 1ma, and the 2k resistor has 2v and 1ma, so that's 2v across the resistor and -1v across the current source, so now the current source looks like a -1k resistor. So how did the current source go from being a 1k resistor to a -1k resistor?

That is interesting. It is not really my question, though but I like to know that.

More exactly, if we connect the voltage source and current source to a circuit containing a few resistors and measure the voltage at some node N and measure 2.3 volts, then if we open circuit the current source and measure that same node and see 1.8 volts and then short the voltage source and reconnect the current source we see that same node measure 0.5 volts, and 1.8+0.5=2.3 volts, the same voltage we get with both sources active.

What this means is that the current source acts like an open circuit impedance that has the ability to force a current through the external circuit.

So we really have two completely different animals here, a source is not the same as a resistor, and so they are not handled the same way mathematically either.

To sum up:
1. The voltage source acts like a zero impedance.
2. The current source acts like an infinite impedance.

I am wondering if this is the way we use to prove that voltage source has internal resistance zero and current source has internal resistance infinity.
I often see that current source has infinite internal resistance. Then I tried to prove it but it is really confusing.
One of the answer that I received:

However, I then I feel more confused because there is another kind of resistance here, ac resistance.
I always why we don't calculate resistance of a current source by Ohm's law:

If we use this formula, the resistance of an ideal current source is not infinite but its value is a number.
What is the difference between resistances that are defined by two formulas above?
If I apply the two formulas above for an ideal resistance, the resistances computed from two formulas are the same. But when apply for an ideal current source, resistances calculated from each formulas are different. I really get stuck.
Why we don't apply the formula for an ideal current source and an ideal voltage source?

( I think the formula is not used because I always read that an ideal current source has infinite resistance and an ideal voltage source has zero resistance. Therefore, it is clear that the definition below is used in both cases.

 

[Ratchit]

Anhnha,

>I am really confused about current source resistance.

What are you confused about?

>An ideal current source has infinite reistance.

And the point is?

>As my understanding, the resistance said above is AC resistance, Rac.
Rac = dv/di

You say AC resistance. Don't you really mean incremental resistance? That is what Rac = dv/di means.

>But how about DC resistance of an ideal current sourrce?

You say DC resistance. Don't you really mean total resistance? That is what V/I means.

<When you talking about ideal current source resistance, which kind of resistance do you imply, AC or DC resistance.
What is the exact definition of resistance? I don't know when we need to use DC or AC resistance. Please help.

You seem to have a definition problem. Incremental resistance is a small change in voltage divided by the corresponding change in current. Total resistance is the total voltage divided by the corresponding value of current. Incremental resistance is the slope of the V_I curve. An ideal current source has the same value for both incremental and total resistance, namely infinity. Any questions?

Ratch

Addendum:
>I always why we don't calculate resistance of a current source by Ohm's law:
If we use this formula, the resistance of an ideal current source is not infinite but its value is a number.

You are not measuring the resistance of the current source that way. You are measuring the resistance of the load. You would get the same value if you used a voltage source. If you want to measure the resistance of the ideal current source, use Thevenin's theorem instead.

 

[hanhan]

Thank you, Ratch.

>I am really confused about current source resistance.

What are you confused about?
>An ideal current source has infinite reistance.

And the point is?

My confusion is about its definition:
I often read that an ideal current source has infinite resistance.
I wanted to know why and then I found an answer.

But when I see this definition, I am really confused about this definition of resistance. As you said incremental resistance.
Why we need that definition? Why don't only use total resistance?

This is said always in most textbooks:
1. The voltage source acts like a zero impedance.
2. The current source acts like an infinite impedance.

Why they don't state clearly that these are incremental resistance not total resistance? I confused because when I apply the definition of total resistance for an ideal current and voltage source they are not infinite or zero.

>As my understanding, the resistance said above is AC resistance, Rac.
Rac = dv/di

You say AC resistance. Don't you really mean incremental resistance? That is what Rac = dv/di indicates.

Yes, I mean incremental resistance.

>But how about DC resistance of an ideal current sourrce?

You say DC resistance. Don't you really mean total resistance? That is what V/I means.

Yes, I mean total resistance.

You seem to have a definiton problem. Incremental resistance is a small change in voltage divided by the corresponding change in current. Total resistance is the total voltage divided by the corresponding current. Incremental current is the slope of the V_I curve. A ideal current source has the same value for both incremental and total resistance, namely infinity. Any questions?

The bold part is what I am confused. You said with an ideal current source, the incremental and total resistance are the same but I don't see it.

In this picture, incremental Rac = infinity but the total resistance is totally different.
In the left picture:
Total resistance: Rtot = V/I = 1V/1mA = 1K ohms
In the right picture:
Rtot = V/I = 2V/1mA = 2K ohms.
=> They are not the same.
When we need to use each definition?
With an ideal resistor incremental and total resistance are the same but some other components such as voltage source or current source, they are different.

 

[Ratchit]

Anhnha,
Too bad you answered so fast. See the addendum to post #6. The incremental resistance of a ideal current source (ICS) is delta V/delta I . Since delta I does not change, its value is 0. So, delta v/0 is infinity. Using Thevenin's theoren for the total resistance of the ICS we have the open circuit voltage of infinity, and the closed circuit current of some finite value. Therefore, infinity/finite value is infinity. Do you now see why the incremental value and total value of a ICS is the same?

A V-I diode curve is not straight. Therefore the incremental resistance at a point of the curve is not the same as the total resistance at that point. Just apply the definition of the two types of resistances, and you can easily see the difference.

Ratch

 

[KeepItSimpleStupid]

Ratch:

While we are on this subject, how might one computer the output Z of Fig 5 in this Application Note: https://www.google.com/url?sa=t&rct...1wpdwVR2JtwQo0A&bvm=bv.52164340,d.dmg&cad=rja

 

[hanhan]

Thank you, Ratch.

Too bad you answered so fast. See the addendum to post #6. The incremental resistance of a ideal current source (ICS) is delta V/delta I . Since delta I does not change, its value is 0. So, delta v/0 is infinity. Using Thevenin's theoren for the total resistance of the ICS we have the open circuit voltage of infinity, and the closed circuit current of some finite value. Therefore, infinity/finite value is infinity.

Interesting point. I feel a bit relieved now.

. Do you now see why the incremental value and total value of a ICS is the same?

Yes, as above.

You are not measuring the resistance of the current source that way. You are measuring the resistance of the load. You would get the same value if you used a voltage source. If you want to measure the resistance of the ideal current source, use Thevenin's theorem instead.

Why we can't use the method to measure resistance of current source? Do we have to use Thevenin's theorem to calculate resistance for active devices?
And I wonder why Thevenin resistances are the resistances of these active devices.

A V-I diode curve is not straight. Therefore the incremental resistance at a point of the curve is not the same as the total resistance at that point. Just apply the definition of the two types of resistances, and you can easily see the difference.

Ah, with components that their voltage-current characteristics are linear then two types of resistance are the same. I can easy see it now.

 

[Ratchit]

KISS,
I would say that its total resistance is 5 volts divided byu whatever amps it is set to sustain. Its incremental resistance will be very high because it is electronically controlled to not change its current during incremental voltage variations.

Ratch

 

[Ratchit]

Anhnha,

>Why we can't use the method to measure resistance of current source? Do we have to use Thevenin's theorem to calculate resistance for active devices?
And I wonder why Thevenin resistances are the resistances of these active devices.

As I said before, you are measuring the voltage across the load and the current through the load. That gives you the resistance of the load. You cannot measure the resistance of the current source while it is connect to a load by just measuring the voltage and current of both the load and source when they are the same. Study Thevenin's theorem to see why.

Ratch

 

[hanhan]

As I said before, you are measuring the voltage across the load and the current through the load. That gives you the resistance of the load. You cannot measure the resistance of the current source while it is connect to a load by just measuring the voltage and current of both the load and source when they are the same. Study Thevenin's theorem to see why.

I know Thevenin's theorem very well but I don't see why Thevenin's resistace is the resistance of current source.
By using Thevenin's theorem, an ideal current source can be converted to an voltage source with infinite voltage, Vth = infinity and a resistor Rth.
Rth = Vth/Rth = infinity/I = infinity
The resistace Rth = infinity but why you can say that this is the resistance of current source?
I think this is the case if resistance of voltage source is zero. I know it is really zero but assume that you are now don't know what is the resistance of an ideal voltage source, then can you conclude that the resistance of current source is Rth and it is infinite.
Similarly, by using Thevenin's theorem for an ideal voltage source I can easily see that Rth = 0 ohm but how ca you know that the resistace is the resistance of the voltage source?

 

[hanhan]

Ratch,
I have just read Thevenin's theorem again and I think I get it now. Thank you for the help.

 

[MrAl]

Hello again,

The simple answer is to try to construct a current source out of a high voltage source and high value resistance. If the current source really looks like a high impedance, then we should be able to use a high value resistor and voltage source to construct a current source, at least to some very good approximation.

Starting with 1v and 1 ohm, we try to power a 1 ohm resistor with that, and we get 1/2 amp. Then we try to power a DIFFERENT resistor like 2 ohms, but when we do that we see the current goes down to 1/3 amp. So we have a current source with the 1v and 1 ohm, but it's not that accurate yet. Note that it may actually work in some circuits that dont demand great accuracy over a wide range of loads though.

So now lets try a 10v source and 10 ohm resistor. Note we just scaled everything up by 10. Now we power the 1 ohm resistor and we get 0.91 amps, and we power the 2 ohm resistor and we get 0.83 amps. So we've improved the current source a little.

Now we scale by another 10 times, to 100v and 100 ohms. Now we power the 1 ohm load and we get 0.99 amps, and power the 2 ohm load and get 0.98 amps. So now our two load extremes see current levels that are within 1 percent of each other. Much better than before.

One more time we go to 1000v and 1000 ohms, and powering the 1 ohm load we get 0.999 amps and the 2 ohm load we get 0.998 amps, a great improvement over the starting point.

So what made the big difference? The series resistance of the source was higher so with a change of load the current looked almost the same, which is EXACTLY what a current source does.

But lets take this all the way. Each time we scaled up by a factor of 10 we multiplied the voltage source by 10 and also the series resistance by 10 (that make up the simulated 'current source'), and then to calculate the actual current we divided the voltage source by the total resistance, which is the sum of the series resistance and the load resistance RL. We can state this mathematically as:
I=V*n/(Rs*n+RL)

where I is the actual current that would be measured, and Rs is the series resistance, and 'n' is the multiplying factor (we used 10 each time above but we did it three times so 'n' would have been 1000). We want to see what would happen if we scaled this up by an infinity scaling factor, so we take the limit of that above as 'n' goes to positive infinity:
limit(I)=limit(V*n/(Rs*n+RL)) as n --> +infinity

and lo and behold we get:
I=V/Rs

and note that the load resistance RL has disappeared, which is exactly what happens with a true current source because the load resistance does not change the current at all. So we end up with an ideal current source!

 

[hanhan]

Thank you, MrAl.
Interesting. The type of current sources you said are originated from voltage sources. But I can see that this type of current source is only accurate
if RL <<Rs.
Are most of current sources in real life created this way?

 

[MrAl]

Hi again Anhnha,

Some are but most are not created that way. I showed that method because that takes us from a zero impedance device (voltage source) and gets us to an infinite impedance device (current source).

The idea was that as the Rs resistance increases we see the approximation to a current source get better and better. The higher Rs gets, the more the circuit load RL looks like it is being powered by a current source, and the impedance of a votlage source in series with a resistance (like Rs) is equal to Rs alone.

We could also look at it from the standpoint of a control circuit similar to what we might find in real life. We "regulate" the current so that it stays constant even when we change the load resistance RL. We could use a sense resistor to sense current and large open loop gain AOL and then let the gain to to infinity and maybe let the sense resistance go to zero. We'd get a constant current output again. But that's a bit more involved and what we really wanted to see was why the impedance of a current source appears to be infinite, not how real life constant current generators work.

Of course real life current generators are not perfect, so there will be some slight variation in output current. This is actually necessary. And there is a top limit based on how much voltage we apply to actually run the circuit, as if we apply a load resistance which is too high the required constant current might not appear because there's simply not enough voltage coming in from the power supply to be able to supply that resistor with the controlled current level.
For example, if we have a 1ma constant current and we connect a 1k resistor, we see 1v across that resistor, but if we connect a 1Meg resistor we'd need 1000 volts to get that same 1ma of current. So that would not work, but in theory a constant current generator always provides the right current, and we solve for the max voltage as well to make sure the real life circuit can actually work correctly.