Why filters with current source behave differently?

[gauthamtechie]

A Low Pass RC filter behaves as a High Pass RC filter if the voltage source is replaced by a current source. What is an intuitive explanation to this phenomenon?

Also the current across the capacitor is a consequence of the Change in voltage as in I = C dv/dt. How does this go with the fact that we are supplying our sinusoidal source of current which has a constant RMS current ? Does this have any relevance to the filter behaving as high pass instead of Low pass in the case presented?

 

[simonbramble]

A Low Pass RC filter behaves as a High Pass RC filter if the voltage source is replaced by a current source. What is an intuitive explanation to this phenomenon?

Does it? With an RC filter, the impedance of the capacitor goes down with frequency, so you get more of a divider effect across the capacitor as frequency goes up (with a voltage source input). With a current source input, the voltage across the capacitor is dependent on the impedance of the capacitor and the current going through it (which is constant), so with a reducing impedance, the voltage across it goes down too.

I dont see how this is a high pass filter.

Regarding point 2, current flows through a capacitor, not across it. The RMS current is a single digit number and is representative of an ac signal (which is a varying current). The current going through the capacitor is proportional to the derivative of the voltage (with time). If you differential a sine wave, you get a cosine wave (... or another sine wave, just shifted in phase).The voltage and current are both sinusoidal. Forget RMS - this is just a single number (like the peak). it does not represent the varying nature of the waveforem

 

[MrAl]

Hi,

Are we talking about a regular RC low pass filter here with one resistor and one capacitor, or some other topology?

If it is a regular RC low pass then remember the resistor would have a constant voltage across it with a current source and that goes across all frequency. So the only thing left to change is the voltage across the capacitor, which would go down with frequency.

But one other little thing...the input was specified as a current source, but how should we interpret the output? Should we still take the voltage across the cap to be the output, or did you want us to take the CURRENT through the capacitor as output now? This question makes sense because if you changed the input type you may also be wanting to change the output type too. The answer to this one is even easier

 

[gauthamtechie]

Sorry about the First Point. I mistook the idea of replacing T networks with Pi networks et all while reading filters and how to replace voltage source with current sources.

And point 2 , do you mean since current across a capacitor is supplied by current source, it is the voltage across the capacitor which is a consequence of this current? Rather than I= C dv/dt, is it that the voltage now is just the integral of the current since the current is defined and kept at a certain peak amplitude by my current source?

 

[MrAl]

Hi,

Well if you take I=C*dv/dt you can rearrange to get dv=I*dt/C, no big deal right? And there we are solving for the change in voltage and we can easily see that as I gets bigger dv gets bigger too, however as dt gets smaller (faster frequency) then dv gets smaller too. That's a little indirect though.

Since we are working with AC here, it helps to look at it from the standpoint of impedance, where impedance Z=1/jwC with w=2*pi*f, so now we have an equation more directly in terms of frequency f. So the current I is:
I=E/Z
where E is the applied AC voltage, or if again we rearrange, we have:
E=I*Z

and substituting:
E=I/jwC

and just the magnitude:
|E|=I/wC

so we end up with:
E=I/(2*pi*f*C)

and here we see that the voltage E decreases with frequency because f is in the denominator of the right hand side.

So for constant I and increasing frequency E goes down.

 

[gauthamtechie]

thanks I get it now.

 

[Ratchit]

gauthamtechie,

thanks I get it now.

Do you? I hope so. Never forget that connecting a current source to a circuit means hooking up a large resistance to the circuit. After all, a current source is a voltage in series with a large resistance. That is bound to change the characteristics of the circuit. Although others have said otherwise. No current exists in a capacitor like it does in a resistor or coil. The current in a capacitor branch is a transient amount caused by the accumulation and depletion of charge on the capacitor plates.

Ratch

 

[Claude Abraham]

gauthamtechie,



- - - After all, a current source is a voltage in series with a large resistance. That is bound to change the characteristics of the circuit. Although others have said otherwise. - - -

Ratch

Hmmm, would you please elaborate? A current source may be equivalent to a voltage source and a resistance in series, but one can construct a true current source w/o a large series resistance. Take an inductor based LED constant current driver. When the inductor is energized with the FET turned on, the energy in the inductor is stored as L*I^2/2. Then the FET opens, and the inductor current free wheels through the catch diode and LED.

For high efficiency this approach is used. The inductor outputs a constant current. The voltage is determined by the LED and catch diode. There is no large series resistance. If there were the losses would be great, an undesirable thing.

Ratch you post some stuff that I just cannot believe that you even believe. How can you think that an inductor supplying a current really consists of a voltage source plus large series resistance? Measurements confirm the opposite.

 

[Ratchit]

Claude Abraham,

Hmmm, would you please elaborate?

Certainly!

A current source may be equivalent to a voltage source and a resistance in series, but one can construct a true current source w/o a large series resistance. Take an inductor based LED constant current driver. When the inductor is energized with the FET turned on, the energy in the inductor is stored as L*I^2/2. Then the FET opens, and the inductor current free wheels through the catch diode and LED.

For high efficiency this approach is used. The inductor outputs a constant current. The voltage is determined by the LED and catch diode. There is no large series resistance. If there were the losses would be great, an undesirable thing.

Let's take a simpler, more familiar implementation. Specifically, a linear power source with a current limit feature. Crank up the voltage to the max and limit the current to say, 1 ma. That makes it effectively a 1 ma current source. You have in effect an electronic load in series with a voltage that looks to the driven circuit like a large value resistor. There is even a method in circuit analysis called "substitution" that allows one to simplify circuits this way. https://books.google.com/books?id=-...epage&q=electric substitution theorem&f=false .

Ratch you post some stuff that I just cannot believe that you even believe. How can you think that an inductor supplying a current really consists of a voltage source plus large series resistance? Measurements confirm the opposite.

You are confusing the method with the implementation. There are lots of ways to make a current source, but it is going to look to the driven circuit as a large resistance in series with a voltage. In other words, a current source.

 

[Claude Abraham]

I am quite familiar with the Norton-Thevenin equivalence principle. I was just pointing out that such equivalencies are functional, not absolute. Take an example, a 1.0 ohm load is driven by a 1.0 amp current source with 9 ohms of shunt resistance. Of the 1.0 amp generated by the current source, 0.90 amp reaches the 1.0 ohm load, and 0.10 amp is lost in the internal shunt resistance of the CCS. Thus the power in the load is 0.81 watt, with 0.09 watt lost in the CCS shunt resistance. This configuration is 90% efficient.

If we replace the CCS having a shunt resistance of 9.0 ohms, with a CVS (constant *voltage* source) of 9.0 volts having a 9.0 ohm series resistance, the circuit is externally identical as far as I & V measurements go. If 2 black boxes were measured with volt-meters and ammeters, one can not distinguish which black box has the CVS or the CCS. But there is a difference thermally, i.e. they are not equivalent regarding power.

A 9.0 volt CVS w/ 9.0 ohms internal series R is loaded w/ 1.0 ohm as was the CCS above. The current in the load is still 0.90 volts as before. THe voltage at the load is still 0.90 volts as before. Of course the power is identical being I*V = 0.81 watt as before. But look at the source power. The power lost in the 9,0 ohm source resistance is 7.29 watts! The efficiency is a measly 10%!

I just wanted to convey that any CCS can be regarded as functionally equivalent to a CVS plus proper R value in series. From an I & V perspective, they are equivalent. But not from a power perspective at all. If the load resistance is smaller than that of the source, CCS is more efficient. Likewise, if Rload > Rsource, CVS is more efficient. I just wanted to let readers know that there are applications such as LED drivers, where true constant current drive is much better than a voltage drive plus a large series resistance. I would hope that you agree. BR.

 

[Ratchit]

Claude Abraham,

I agree with everything you said in your previous post. I found out recently that one cannot use equivalent impedances for power calculations. See post #11 at https://www.electro-tech-online.com/threads/kirchoffs-law.133655/ .

However, power and efficiency is not what the OP was asking about. He was asking about why his filter changed frequency response when he switched from a voltage driven to current driven input. My answer was that one of the frequency determining elements was the source impedance that changed when he switched from voltage to current drive.

Ratch

 

[MrAl]

Hi guys,

The functional approximation to the current source can be said to be a voltage source in series with a high resistance, but in most cases that voltage will have to be high if the resistance is high. For example, if we want a 1 amp current source this way we could use a 1M resistor but then we'd need a 1Mv voltage source! And since the power is I^2*R we'd see a power dissipation of 1e6 watts (1 megawatt). That's totally and completely unacceptable if we are building a practical circuit, but in theoretical exercises it's totally and completely acceptable. That's because in some analysis we dont care if something outside of the main circuit draws a lot of power, even a megawatt, because that is not included as part of the analysis. It's like, do we care if the power company right now is producing 100 megawatts at some power station down the road? No, because we mostly care what happens in our own house.

So in these cases the analysis is broken into two parts, the part we are interested in and the part we are not interested in. Case in point, since we are only interested in the response of the circuit to a current source, we dont care where that current source came from or how it was made because we wont look at that in detail, and we wont use this type of current source in the lab unless it is more restricted to what we can actually do in the lab.

I would assume that Ratch introduced this concept so that we could look at the circuit as a RC circuit with very very large R driven by a voltage source, but we still dont look at the additional R as a power eater, only as a functional mechanism that may help us understand the circuit.

 

[Claude Abraham]

Of course, Mr. Al. that is what I stated in my posts. For functionality, the 2 are equivalent. I just had to respond to the comment that " a current source is just a voltage source plus series R". Had the statement read as follows "one can replace a CCS plus shunt R with a CVS plus series R for circuit analysis" I would whole-heartedly concur. Still your point is well taken. So we appear to have a consensus on this issue. Thanks.

 

[Winterstone]

"...but one can construct a true current source..."

Just one comment from my side: I think, this statement must be elaborated as well.
1.) What is a "true current source"?
2.) Is there any voltage source "behind"?
Remember: What is this thing called "current"? It is movement of charges, is it not?
And which force is it that causes the charges to move? Isn`t it a voltage (resp. the corresponding field) ?

W.

 

[Claude Abraham]

"...but one can construct a true current source..."

Just one comment from my side: I think, this statement must be elaborated as well.
1.) What is a "true current source"?
2.) Is there any voltage source "behind"?
Remember: What is this thing called "current"? It is movement of charges, is it not?
And which force is it that causes the charges to move? Isn`t it a voltage (resp. the corresponding field) ?

W.

1) A true current source outputs a fixed current regardless of load impedance. As Zload varies, a CCS will output whatever V is needed to maintain fixed current.

2) There is an associated voltage, but refrain from using the phrase "voltage source". By definition a voltage source maintains fixed voltage by outputting variable current per load impedance variation. A current source outputs a voltage, but it is not a voltage source and vice-versa.
Which force causes charges to move is Lorentz force, Voltage is not a "force". Before this issue was well understood, the terms "magnetomotive force" and electromotive force" were coined. Later we learned that neither is really an actual force.

Voltage is work done per unit charge, joule/coulomb, current is charge flow rate in coulomb/second. One does not "cause" the other. Charges move when in the presence of E & B fields per Lorentz force. But to create fields, work must be dome moving charges to separate them resulting in a field. It's chickens and eggs. A current through a resistor will encounter collisions w/ lattice ions. Some electrons will fall from conduction into valence, releasing energy as photons. The heat from a resistor is this photonic emission. The current resulted in voltage drop. But an E field, or B was needed to sustain said current. A current source or voltage source can provide this field. But a battery which is a voltage source produces a terminal voltage and E field by separating ions, moving them AGAINST the E field. So the battery through chemical reaction forces ions to move against the field, which is current, in order to maintain a voltage. You cannot make one w/o the other, neither is the "cause" nor the "effect" of the other.

A true current source needs voltage compliance to work. A true voltage source needs current compliance to be useful. Both are needed but only one is held fixed. For a current source the voltage varies with load impedance in order that current be held at fixed value. I will elaborate if needed.

 

[Winterstone]

Hi Claude,

thank you for the detailed answer. Of course - I must agree, no doubt about the definitions you have mentioned.
My only concern was the formulation "...one can construct a true current source" (construct !).

W.

 

[MrAl]

Hi Claude,


Well it just struck me as a little strange that you would want to talk so fervently about an issue that really had nothing to do with the original request. When we talk theory we talk theory, and dont care about all the practicalities. I would think that maybe we might mention the power consumption in passing, but you seemed to focus on this with all your might That's one reason why i brought up the "break the circuit into two parts" idea which is how we handle things like this. Dont get me wrong, it's just fine that you brought htis up, no problem there, but it's just a side issue i would think. Maybe you were worried that the OP would take the meaning to be that we somehow gained power in the circuit using a resistor and voltage source, a power that shouldnt be there.

But anyway, the current source being thought of as a voltage source in series with a very large resistance is often brought up and quite common to talk about. Maybe the more common level of current would be less and thus not present such a seemingly bad practice, such as a 1v source followed by a 1 megohm resistor, for 1ua 'constant' current.
Interesting though, we start to immediately see the error in this madness. If the load contains pure resistance we're ok, but if the load contains a voltage source, all bets are off because even 1/2 volts would lower the current from 1ua to 0.5ua, a huge difference in many circuits. So a high voltage along with that high resistance is sort of also implied i think. If we used a 1000 volt source and a 1000 megohm resistor, then with that same 1/2 volt internal source we'd see a minor change in current level.

The main point though is that the current source is viewed as external to the system, while the circuit in question is considered internal, yet we can still combine elements of the external circuit with elements of the internal circuit as long as we limit the view point we take.

This turned out to be an interesting discussion and it's good to hear your ideas too

 

[Claude Abraham]

Hi Claude,

thank you for the detailed answer. Of course - I must agree, no doubt about the definitions you have mentioned.
My only concern was the formulation "...one can construct a true current source" (construct !).

W.

Agreed. We can construct a true current source just as we can and do construct a true voltage source. We use feedback, and control the output for fixed current or voltage. I don't know if you're implying that current sources are "constructed", but I think it's worth mentioning that Mother Nature offers 2 devices that are natural sources, not "constructed". An inductor is a true current source, and a capacitor is a true voltage source. Of course they give up their energy eventually but their behavior is such. Likewise a photodiode terminated in a very low Z exhibits true current source behavior, while a PD in a high Z exhibits voltage source behavior.

Anyway, your point as well as that of Mr. Al is well taken. I hope I didn't come across as being too preachy, but I guess I felt that it should be emphasized that power is important. Nowadays driving LEDs is quite common. A large voltage source and large resistance is viable, the LEDs would not care, but efficiency does deserve some due consideration. Thanks to all.

 

[MrAl]

Hi Claude,

Well it was good that you brought it up though, in case the OP didnt realize the big difference when we look at the practical side.

Just for the fun of it, here's a little theory for the approximate current source....

To be exact, we have two different circuits and we want to see how different or similar they are. One is the true current source powering the RC LP filter, and the other is a high voltage source with a large resistance in series with it and that is what powers the LP filter. We are looking at the output of the filter, across the cap.
The LP filter is made up of R1 and C.

The response with the current source I is:
[LATEX]\[\frac{{I}^{2}}{{w}^{2}\,{C}^{2}}\][/LATEX]

and the response with the voltage source V1 and large series resistance Rs is:
[LATEX]\[\frac{{V1}^{2}}{{w}^{2}\,{C}^{2}\,{Rs}^{2}+2\,{w}^{2}\,{C}^{2}\,R1\,Rs+{w}^{2}\,{C}^{2}\,{R1}^{2}+1}\][/LATEX]

equating the two we have:
[LATEX]
\[\frac{{I}^{2}}{{w}^{2}\,{C}^{2}}=\frac{{V1}^{2}}{{w}^{2}\,{C}^{2}\,{R1}^{2}+2\,Rs\,{w}^{2}\,{C}^{2}\,R1+{Rs}^{2}\,{w}^{2}\,{C}^{2}+1}\]
[/LATEX]

and from this we can see that if we divide the top and bottom of the right hand side by Rs^2 we can immediately equate the squared current source on the left with V1^2/Rs^2, so we have:
[LATEX]\[{I}^{2}=\frac{{V1}^{2}}{{Rs}^{2}}\][/LATEX]

so the equivalent current source is:
[LATEX]\[{I}=\frac{{V1}}{{Rs}}\][/LATEX]

and that leaves us to equate the two denominators:
[LATEX]w^2*C^2=(w^2*C^2*R1^2)/Rs^2+(2*w^2*C^2*R1)/Rs+w^2*C^2+1/Rs^2[/LATEX]

Now if we keep R1 small in comparison to Rs by making R1=Rs/A where A is a large number, we get:
[LATEX]\[{w}^{2}\,{C}^{2}=\frac{2\,{w}^{2}\,{C}^{2}}{A}+\frac{{w}^{2}\,{C}^{2}}{{A}^{2}}+{w}^{2}\,{C}^{2}+\frac{1}{{Rs}^{2}}\][/LATEX]

And here we see that if we make A large (as planned) the first two terms get very small and we are left with:
[LATEX]w^2*C^2=w^2*C^2+1/Rs^2[/LATEX]

and when we see Rs large that term gets very small too so we end up with:
[LATEX]w^2*C^2=w^2*C^2[/LATEX]

So when Rs is large compared to R1 we get an approximate equality. As mentioned before, the voltage has to be large too.

 

[Winterstone]

An inductor is a true current source, and a capacitor is a true voltage source.

Perhaps a cause of misunderstanding: What is the definition of a "true" source in comparison to an "ideal" source?


Likewise a photodiode terminated in a very low Z exhibits true current source behavior

This example could reveal the difference between "true" and "ideal". The properties of an ideal source must not depend on load conditions, do they?