resistance vs impedance and f_c

[electrolust]

Another question I can't find ready answers to on google.

I understand that a capacitor in series with a load forms a voltage divider and therefore a high pass filter.  The corner frequency f_c of that filter is determined in part by the resistance, at least if the load were a resistor.

But how does impedance interact with this?

Not that such a device exists, but imagine a load has a DC resistance R, and an impedance vs frequency that was flat above some f_c frequency and rolled off 20db/decade below f_c.  (A signal would be attenuated less and less below f_c.)  Then you put a capacitor in front of it with capacitance C such that the high pass filter formed between the capacitor and the load of 1/2*pi*R*C = f_c.  Would the filter formed with C and R, interact with the impedance to produce a flat frequency response?

[ejeffrey]

Yes.  For instance, if your 'load' is a capacitor, then the two capacitors form a voltage divider -- the division ratio is constant at all frequencies above DC because the two capacitors have the same frequency dependence.  Obviously non-ideal behavior will make it deviate from perfectly flat at extremely high or low frequencies, but over a large bandwidth they will match.

You can just use the standard voltage divider equations, but replace the resistances with complex impedances:

Vout/Vin = Z1(f)/(Z1(f) + Z2(f))

This also works if Z1 and Z2 are themselves networks containing Rs, Ls, and Cs.  This is how scope probes work.  They have a 9 meg resistor that makes a 10:1 voltage divider with the internal 1 meg resistance, but that only works at DC.  The scope has an input capacitance in parallel with its 1 meg, so the probe needs a capacitor in parallel with the 9 meg with the same ratio. If it is tuned exactly right, the probe and scope form a perfect 10:1 divider over all frequencies.  When you calibrate your probe, you are trimming this capacitance -- if it is too high or too low you will have a high-pass (overshoot, ringing) or low-pass (undershoot, slow response) behavior.

There is more to a scope probe to deal with the cable, but that is the general principle: you are matching impedance ratios to get a flat response to the highest bandwidth possible.

[T3sl4co1l]

Another question I can't find ready answers to on google.

Intuiting is good, but perhaps you should put some numbers to it?

The math isn't so bad, it's just algebra and rational equations.  No other functions, no sqrt (until you get into power, anyway), no exp or trig.

Cranking a big equation (say for >4 components in a non-trivial connection) is daunting, but at least in principle... if you know algebra, you know how to do it.  Doing it successfully is just a matter of scale, not of missing knowledge.

Not that such a device exists, but imagine a load has a DC resistance R, and an impedance vs frequency that was flat above some f_c frequency and rolled off 20db/decade below f_c.  (A signal would be attenuated less and less below f_c.)  Then you put a capacitor in front of it with capacitance C such that the high pass filter formed between the capacitor and the load of 1/2*pi*R*C = f_c.  Would the filter formed with C and R, interact with the impedance to produce a flat frequency response?

Actually, such a device does exist... put an R and L in parallel.   The slope below f_c is determined by the inductor (|Z| ~= w*L), and the flat "DCR" (EPR (equivalent parallel resistance), rather, since DCR won't have the same effect!*) is given by the resistor.

*We can, in fact, convert the circuit so that it has DCR (or ESR if you like) instead of EPR.  But it's a poor model for most purposes, because the conversion depends on frequency squared, so the ESR isn't a constant value, but a function.

Knowing this, we can put a capacitor in series with the R||L.  The impedance is:
Ztot = Z_C + Zeq
Zeq = Z_R || Z_L = j*w*R*L / (R + j*w*L)

so Ztot = 1 / (j*w*C) + j*w*R*L / (R + j*w*L)

Of course, using w = 2*pi*F for short (that's lowercase ω, except my keyboard doesn't have it, and the forum refuses to support it).  Or if you prefer Laplace domain, s == j*w.  j == sqrt(-1) (also 'i', but EEs use 'i' for current, so we use 'j' instead).

Now, your question was: will the impedance be constant?  I think you will see that it won't, because there's no minus that can cancel out.  Not that that's obvious, as written.  (There is cancellation possible, because of the multiple j's floating around.  But -- without getting too deep into this one equation -- this will only happen at one or two frequencies, which isn't "constant".)

Just looking at it geometrically -- as you started the question, after all -- suppose we had a component where the impedance is constant at low frequency, falling above f_c.  The left-to-right reverse of your start.  Why, this would be a capacitor in parallel with a resistor, or Z = R / (j*w*C) / [R + 1 / (j*w*C)].  If we add this instead, we get something that's still not immediately obvious, but we know, intuitively, that we should be able to equate something.

So, why not try to solve it?

Ztot = R1 / (j*w*C) / [R1 + 1 / (j*w*C)] + j*w*R2*L / (R2 + j*w*L)

(Note, I'm renumbering capacitor's R --> R1 and inductor's R --> R2.)

Let's clean up some of those denominators... yeh, right?

Ztot = R1 / (j*w*C) / [(j*R1*w*C + 1) / (j*w*C)] + j*w*R2*L / (R2 + j*w*L)

the j*w*C cancels giving

Ztot = R1 / (j*w*R1*C + 1) + j*w*R2*L / (R2 + j*w*L)

Common denominators,

Ztot = [R1*(R2 + j*w*L) + j*w*R2*L*(j*w*R1*C + 1)] / [(1 + j*w*R1*C)*(R2 + j*w*L)]

Now it might be tempting to continue simplifying and distribute and all, but this is a little better.  Here's a trick: you want to try to factor it into factors of the form constants, constants*j*w, (1 + j*w/constant), (1 + j*w/constant - w^2 / constant^2), etc.  Each factor of w is called a pole (if it's in the denominator) or zero (numerator).  If we can find pairs of poles and zeroes that we can make equal by adjusting the constants, we can cancel them out.  This is called pole-zero cancellation or compensation, and shows up in a number of places.  (A good example is adding capacitors in parallel with resistors to compensate for capacitive loading in a voltage divider, as used in oscilloscope attenuators.)

Ztot = [R1*R2 + j*w*L*(R1 + R2) - w^2*R1*R2*L*C] / [(1 + j*w*R1*C)*(R2 + j*w*L)]

Shifting around the constants...

Ztot = R1*[1 + j*w*L*(R1 + R2) / (R1*R2) - w^2*L*C] / [(1 + j*w*R1*C)*(1 + j*w*L / R2)]

At this point, it might be worthwhile to insert some other variables.  Anywhere we see L*C, we should be thinking: aha, resonant frequency!  So we will use w_0 = 1 / sqrt(L*C) there.  We might also combine (R1 + R2) / (R1*R2) = 1/Rpar, a nice observation.  Also, when we have a quadratic in w, we can label the coefficients more specifically: (1 + w*zeta / w_0 + w^2 / w_0^2), where zeta is the damping ratio.  This gets,

Ztot = R1*[1 + j*w*L / (Rpar*sqrt(L*C)*w_0) - w^2 / w_0^2] / [(1 + j*w*R1*C)*(1 + j*w*L / R2)]

It looks worse before it gets better.  That middle term is looking useless, right?  Hold on..

Ztot = R1*[1 + j*w*sqrt(L/C) / (Rpar*w_0) - w^2 / w_0^2] / [(1 + j*w*R1*C)*(1 + j*w*L / R2)]
Ztot = R1*[1 + j*w*zeta/w_0 - w^2 / w_0^2] / [(1 + j*w*R1*C)*(1 + j*w*L / R2)]

sqrt(L/C) / Rpar = zeta

This is important now!  sqrt(L/C) is a characteristic: the resonant impedance of the LC circuit (which you'll notice is series resonant, except with resistors in parallel with each element, rather than one in series).  The ratio of this against the equivalent damping resistance (i.e., the resistors act in parallel) is the damping ratio.

We're also in the final step, because we can factor this quadratic easily.  We know that if zeta > 1, the network is overdamped, and will factor into two real parts.  We can assign values of R1 and C to cancel the R2 and L parts, and achieve pole-zero cancellation.  If zeta < 1, the network is underdamped, and will ring when excited by a transient (and will have an impedance trough at w_0, the ratio between valley and flat-band being zeta).

Doing the same for the denominator (and labeling the zetas and omega-naughts as 0 and 1) gives,

Ztot = R1*[1 + j*w*zeta/w_0 - w^2 / w_0^2] / [(1 + j*w*zeta_1 / w_1 - w^2 / w_1^2)]

w_0 = 1 / sqrt(L*C)
zeta_0 = Zc / Rpar
Rpar = R1*R2 / (R1 + R2)
Zc = sqrt(L/C)
w_1 = w_0 * sqrt(R2/R1)
zeta_1 = (Rav / Zc + Zc / Rav)
Rav = sqrt(R1*R2) [by the way, this is the geometric mean, hence 'av']

Now all we need to do is equate zeta_0 = zeta_1 and w_0 = w_1, and the poles and zeros cross out!

w_0 = w_0 * sqrt(R2/R1)

This is easy, just set R1 = R2!

Now that means R1 = R2 = Rav == R, just to simplify things a little, and Rpar = R / 2.

Zc / (R / 2) = (R / Zc + Zc / R)
2 * Zc = (R^2 / Zc + Zc)
Zc = R^2 / Zc
Zc^2 = R^2

Zc = +/- R

Noteworthy that the circuit will still be critically damped if we use a negative resistance (which can be synthesized with an op-amp).  But for almost all purposes, we pick the positive root.  So R = sqrt(L/C).

So, in conclusion, starting with an R||L circuit, we put in series with it, R1||C, where:
R1 = R
C = L / R^2

A lot of math, for such a painfully simple result, no?  But this is how things go.  There's a lot of things we can solve from those central equations, like how the impedance varies, or peaks (or troughs) with frequency, how wide the feature is (+/- 3dB points, say), what the voltage divider ratio is, what the impedance in the middle is, etc.  All those options, and more, are essentially present in that equation, there for the solving; we just have to take a direction that results in our desired solution.  So we should accept that the equation is much less simple, and that we will have some work to do, to find the one solution we're looking for.

Tim

[electrolust]

very nice, thanks!

[electrolust]

This is how scope probes work.  They have a 9 meg resistor that makes a 10:1 voltage divider with the internal 1 meg resistance, but that only works at DC.

On another, more careful reading, I caught this.

Lots of resources I can find say that resistors (unlike capacitors) work exactly the same in AC as in DC.  eg http://www.electronics-tutorials.ws/resistor/res_8.html goes into great depth to explain it.

I guess the reason for the parallel capacitors is because the probe cable itself has capacitance, enough to be significant at higher frequencies?

[T3sl4co1l]

Yes, but not just capacitance, but inductance as well, in fact it is a transmission line.

At high frequencies, if the line is terminated properly, then the input end has an ideal, constant impedance of e.g. 50 ohms (resistive only).

They lie a bit, when they say the scope is 1M || 20pF, or what have you: this is an adequate representation up to, say, 30MHz or so (where the capacitive reactance is still larger than the transmission line impedance), but for still higher frequencies (a necessity for any scope 100MHz or more with typical 1-2m long probes), signals would be utterly corrupted by ringing if steps weren't taken to address this.

So the trick is, it's not actually 20pF (which has a decreasing impedance at higher frequencies), but it's dampened with series resistance (which acts to terminate the probe cable at higher frequencies), and coupled into the input amplifiers, which are also not completely capacitive but resistive and sometimes inductive as well.

The equivalent circuit on the side of the probe tip looks more like, 9M into 1M at dc; but also a capacitor across that 9M, to compensate for cable capacitance; but further still, a resistor in series with that capacitor, so that, at very high frequencies, it looks something like a resistive probe (say, 450 ohms into 50 ohms).

Thus, the probe also presents a load to the circuit, which ranges from a high resistance at DC, to a small capacitance in the middle, to a modest resistance at high frequency.

I use the example of 50 ohms, but probe cables are typically a higher impedance, perhaps 93 ohms, and often made with lossy transmission lines.  This reduces the capacitance, and allows 1x probe settings -- where the probe tip connects directly to the transmission line.  The line is relatively unterminated in this use, so the loss serves to compensate for its inductance.  As a result, the frequency response is, very necessarily, cut short, at about the 1/4 or 1/2 wave frequency of the probe cable, typically around 10MHz.  (Keep this in mind next time you need yo do a sensitive, high speed measurement!)

Likewise, the scope front end isn't really properly terminated, into any particular load; if a lossy cable is used, such that high frequencies are uniformly attenuated by 10x, then the line resistance effectively replaces some or all of the series resistor I mentioned earlier, and termination (on either end) isn't really a big deal.

The in-depth analysis of probe construction, from simple or contrived examples like these, up to full lossy models, would be quite involved, but it should at least be a help that: when they give an equivalent or effective circuit, it need not be consistent with what they actually used, and it's very likely there are subtle omissions, like capacitors with resistance, or the distributed inductance or characteristic impedance of transission lines.

Tim

[electrolust]

Tim thanks again.  It will take me some time to digest all this but I do have a quick question.

So, in conclusion, starting with an R||L circuit, we put in series with it, R1||C, where:
R1 = R
C = L / R^2

And here you are trying to choose C in order to get a flat response, yes?

[T3sl4co1l]

Tim thanks again.  It will take me some time to digest all this but I do have a quick question.

So, in conclusion, starting with an R||L circuit, we put in series with it, R1||C, where:
R1 = R
C = L / R^2

And here you are trying to choose C in order to get a flat response, yes?

Yes, by matching the w_0, zeta parameters of the numerator and denominator, we know that they will only differ by a constant, which happens to be R if you work it out.

Tim

[rx8pilot]

Actually, such a device does exist... put an R and L in parallel.   The slope below f_c is determined by the inductor (|Z| ~= w*L), and the flat "DCR" (EPR (equivalent parallel resistance), rather, since DCR won't have the same effect!*) is given by the resistor.

*We can, in fact, convert the circuit so that it has DCR (or ESR if you like) instead of EPR.  But it's a poor model for most purposes, because the conversion depends on frequency squared, so the ESR isn't a constant value, but a function.

Knowing this, we can put a capacitor in series with the R||L.  The impedance is:
Ztot = Z_C + Zeq
Zeq = Z_R || Z_L = j*w*R*L / (R + j*w*L)

so Ztot = 1 / (j*w*C) + j*w*R*L / (R + j*w*L)

Of course, using w = 2*pi*F for short (that's lowercase &omega;, except my keyboard doesn't have it, and the forum refuses to support it).  Or if you prefer Laplace domain, s == j*w.  j == sqrt(-1) (also 'i', but EEs use 'i' for current, so we use 'j' instead).

Now, your question was: will the impedance be constant?  I think you will see that it won't, because there's no minus that can cancel out.  Not that that's obvious, as written.  (There is cancellation possible, because of the multiple j's floating around.  But -- without getting too deep into this one equation -- this will only happen at one or two frequencies, which isn't "constant".)

Just looking at it geometrically -- as you started the question, after all -- suppose we had a component where the impedance is constant at low frequency, falling above f_c.  The left-to-right reverse of your start.  Why, this would be a capacitor in parallel with a resistor, or Z = R / (j*w*C) / [R + 1 / (j*w*C)].  If we add this instead, we get something that's still not immediately obvious, but we know, intuitively, that we should be able to equate something.

So, why not try to solve it?

Ztot = R1 / (j*w*C) / [R1 + 1 / (j*w*C)] + j*w*R2*L / (R2 + j*w*L)

(Note, I'm renumbering capacitor's R --> R1 and inductor's R --> R2.)

Let's clean up some of those denominators... yeh, right?

Ztot = R1 / (j*w*C) / [(j*R1*w*C + 1) / (j*w*C)] + j*w*R2*L / (R2 + j*w*L)

the j*w*C cancels giving

Ztot = R1 / (j*w*R1*C + 1) + j*w*R2*L / (R2 + j*w*L)

Common denominators,

Ztot = [R1*(R2 + j*w*L) + j*w*R2*L*(j*w*R1*C + 1)] / [(1 + j*w*R1*C)*(R2 + j*w*L)]

Now it might be tempting to continue simplifying and distribute and all, but this is a little better.  Here's a trick: you want to try to factor it into factors of the form constants, constants*j*w, (1 + j*w/constant), (1 + j*w/constant - w^2 / constant^2), etc.  Each factor of w is called a pole (if it's in the denominator) or zero (numerator).  If we can find pairs of poles and zeroes that we can make equal by adjusting the constants, we can cancel them out.  This is called pole-zero cancellation or compensation, and shows up in a number of places.  (A good example is adding capacitors in parallel with resistors to compensate for capacitive loading in a voltage divider, as used in oscilloscope attenuators.)

Ztot = [R1*R2 + j*w*L*(R1 + R2) - w^2*R1*R2*L*C] / [(1 + j*w*R1*C)*(R2 + j*w*L)]

Shifting around the constants...

Ztot = R1*[1 + j*w*L*(R1 + R2) / (R1*R2) - w^2*L*C] / [(1 + j*w*R1*C)*(1 + j*w*L / R2)]

At this point, it might be worthwhile to insert some other variables.  Anywhere we see L*C, we should be thinking: aha, resonant frequency!  So we will use w_0 = 1 / sqrt(L*C) there.  We might also combine (R1 + R2) / (R1*R2) = 1/Rpar, a nice observation.  Also, when we have a quadratic in w, we can label the coefficients more specifically: (1 + w*zeta / w_0 + w^2 / w_0^2), where zeta is the damping ratio.  This gets,

Ztot = R1*[1 + j*w*L / (Rpar*sqrt(L*C)*w_0) - w^2 / w_0^2] / [(1 + j*w*R1*C)*(1 + j*w*L / R2)]

It looks worse before it gets better.  That middle term is looking useless, right?  Hold on..

Ztot = R1*[1 + j*w*sqrt(L/C) / (Rpar*w_0) - w^2 / w_0^2] / [(1 + j*w*R1*C)*(1 + j*w*L / R2)]
Ztot = R1*[1 + j*w*zeta/w_0 - w^2 / w_0^2] / [(1 + j*w*R1*C)*(1 + j*w*L / R2)]

sqrt(L/C) / Rpar = zeta

This is important now!  sqrt(L/C) is a characteristic: the resonant impedance of the LC circuit (which you'll notice is series resonant, except with resistors in parallel with each element, rather than one in series).  The ratio of this against the equivalent damping resistance (i.e., the resistors act in parallel) is the damping ratio.

We're also in the final step, because we can factor this quadratic easily.  We know that if zeta > 1, the network is overdamped, and will factor into two real parts.  We can assign values of R1 and C to cancel the R2 and L parts, and achieve pole-zero cancellation.  If zeta < 1, the network is underdamped, and will ring when excited by a transient (and will have an impedance trough at w_0, the ratio between valley and flat-band being zeta).

Doing the same for the denominator (and labeling the zetas and omega-naughts as 0 and 1) gives,

Ztot = R1*[1 + j*w*zeta/w_0 - w^2 / w_0^2] / [(1 + j*w*zeta_1 / w_1 - w^2 / w_1^2)]

w_0 = 1 / sqrt(L*C)
zeta_0 = Zc / Rpar
Rpar = R1*R2 / (R1 + R2)
Zc = sqrt(L/C)
w_1 = w_0 * sqrt(R2/R1)
zeta_1 = (Rav / Zc + Zc / Rav)
Rav = sqrt(R1*R2) [by the way, this is the geometric mean, hence 'av']

Now all we need to do is equate zeta_0 = zeta_1 and w_0 = w_1, and the poles and zeros cross out!

w_0 = w_0 * sqrt(R2/R1)

This is easy, just set R1 = R2!

Now that means R1 = R2 = Rav == R, just to simplify things a little, and Rpar = R / 2.

Zc / (R / 2) = (R / Zc + Zc / R)
2 * Zc = (R^2 / Zc + Zc)
Zc = R^2 / Zc
Zc^2 = R^2

Zc = +/- R

Noteworthy that the circuit will still be critically damped if we use a negative resistance (which can be synthesized with an op-amp).  But for almost all purposes, we pick the positive root.  So R = sqrt(L/C).

So, in conclusion, starting with an R||L circuit, we put in series with it, R1||C, where:
R1 = R
C = L / R^2

A lot of math, for such a painfully simple result, no?  But this is how things go.  There's a lot of things we can solve from those central equations, like how the impedance varies, or peaks (or troughs) with frequency, how wide the feature is (+/- 3dB points, say), what the voltage divider ratio is, what the impedance in the middle is, etc.  All those options, and more, are essentially present in that equation, there for the solving; we just have to take a direction that results in our desired solution.  So we should accept that the equation is much less simple, and that we will have some work to do, to find the one solution we're looking for.

Tim

Bravo Tim.....fantastic presentation. Thank you!