AC bypass capacitor + resitors - how to switch to the same DC gain?

[midwayfair]

"schematic" of sorts:
   C
---B
   E
   |___
   |     |
   1K   470R
   |     4.7uF
   |___|
      |
      G

I have a chart that shows the frequency cutoff of an AC bypass cap:
http://m647.photobucket.com/albumview/albums/tonmann/Circuit%20Workshop/EmitterCap.jpg.html?o=0

And the 470R is limiting the AC gain of the 4.7uF.

If I want the 4.7uF cap to have almost NO effect (i.e., flat response), without removing any components or disconnecting anything, is it sufficient to put a second 1K in parallel with the 1K? Then the resistance would be 500R and close to the same resistance for the DC and AC sides.

Something tells me it's not that simple, though ... for one thing, I'm not sure if the 470R always divides with the 1K and any reduction in the 1K, to a point, will make the 470R smaller, so maybe there will always be a midrange bump as long as that cap is in circuit ...

Can someone point me to some required reading on this subject or a formula I can use to make the calculations?

Thanks!

[induction]

Could you just treat the cap as a frequency-dependent resistor with resistance 1/(2*pi*f*C)?

[R.G.]

Put a 1M in series with the cap. Make your switch short across the 1M.

Put a 1M in parallel with the 470R. Make your switch OPEN the 470R.

[GGBB]

I use this calculator: http://www.muzique.com/schem/filter.htm.  Easier than 1/(2*pi*r*c).

To lower the knee frequency of an RC pair, you need to either raise the cap value or raise the resistor value.  So if you paralleled the 4.7µF cap with a 15µF, it will go from 72Hz to 17.2Hz.

[midwayfair]

These are both practical solutions, but not quite what I'm asking... I was more wondering if it could be done solely by changing the emitter resistor and not touching the 470R+4.7uF at all.

Put a 1M in series with the cap. Make your switch short across the 1M.

Put a 1M in parallel with the 470R. Make your switch OPEN the 470R.

But that will leave me with less gain when the cap is out of the equation (though I recognize that your method is a good way to prevent a pop). I want the same gain when the bypass cap "boost" is out of the equation. For that I probably have to go with a combination of yours and GGBB's methods, with a 1M+47uF switched in parallel with the 470R+4.7uF.

[GGBB]

These are both practical solutions, but not quite what I'm asking... I was more wondering if it could be done solely by changing the emitter resistor and not touching the 470R+4.7uF at all.

Beyond me really, but I suspect the answer is no - at least not perfectly - as you alluded in your OP.  Unless it becomes zero.  Your piggyback idea probably gets close though (I realize that probably isn't much help - sorry).  But I have to ask - is this for something other than guitar?  Because the 4.7µ/470R knee is already below the low E.

[R.G.]

These are both practical solutions, but not quite what I'm asking... I was more wondering if it could be done solely by changing the emitter resistor and not touching the 470R+4.7uF at all.

Sorry -  I read your question wrong.  Going back to the start:

I have a chart that shows the frequency cutoff of an AC bypass cap:

The problem with charts is that you have to be really sure you have all the resistors and caps that will have a significant effect all accounted for. This is something you're intuitively uneasy about, hence the question. Good intuition!

And the 470R is limiting the AC gain of the 4.7uF.

Not exactly. the 1K, 470R and 4.7uF form a single network with a varying impedance with frequency. At frequencies well below the frequency where the cap's impedance is comparable to 470 ohms, the network is effectively 1K. Note that the 470/4.7u crossover is at 72Hz, and you don't get "free" of this until down around 7.2Hz. At frequencies well above the 72Hz point, say, above 720Hz, the gain is determined by the value of 1K paralleled with 470R. The 470R is not completely in or out between 7.2Hz and 720Hz.

The R-C cutoff point is a rule-of-thumb fiction. A useful one, to be sure, but a fiction; at least an arbitrary decision point. At 72Hz, the impedance of a 4.7uF cap is 470 ohms - plus a capacitive phase shift - so that's the half-power point for only those two components. You can't really completely ignore the cap or the resistor until you're 10:1 above or below the turnover point. That was actually a question on my linear controls course midterm exam. I guess the professor did a good job teaching me. 

If I want the 4.7uF cap to have almost NO effect (i.e., flat response), without removing any components or disconnecting anything, is it sufficient to put a second 1K in parallel with the 1K? Then the resistance would be 500R and close to the same resistance for the DC and AC sides.

No. Because of the long frequency interaction, and the way that the 470+4.7u are always in parallel, you need to switch to a pure resistance that's 1K paralleled with 470R, or 320 ohms. If you switch from 1K paralleled by 470R in series with a 4.7uF to a 320R, the gain set by the 320 will be the same as the gain above 720Hz, where the capacitor's effects have settled out. This will be an ugly switch, because the DC conditions set by the 1K for biasing will be dramatically changed if you switch on the fly.

Something tells me it's not that simple, though ... for one thing, I'm not sure if the 470R always divides with the 1K and any reduction in the 1K, to a point, will make the 470R smaller, so maybe there will always be a midrange bump as long as that cap is in circuit ...

Can someone point me to some required reading on this subject or a formula I can use to make the calculations?

You have just graduated to needing to know how to solve networks, not simple R-Cs.

What you really need now is some practice in learning what vector impedance is - a capacitor actually shifts phase, the current through it leading the voltage across it, as well as changing impedance with frequency. You need to know how to write down or estimate the impedance of R-C networks of more than one pair, which is not tricky, but to do it right you need to do some algebra right. And you need to - in this case - understand what the DC versus AC side effects of the network are.

The transistor gain, if it's a simple setup with a collector resistor, is Zc/Ze+ze, where Zc is the impedance on the collector, Ze is the external impedance on the emitter, and ze is the internal impedance at the emitter, generally just the Shockley resistance.

The impedance at the collector is Rc in parallel with the series combination of any output capacitance and any load resistance - which may itself be a biasing network in series with the input impedance of another transistor base, or something else. The impedance Zc is the parallel/series combinations of the entire network on the emitter. In this case it's a 1K in parallel with a 470R and a 4.7uF in series. "ze" is pretty close to 26mV/Ie in most cases, and can often be ignored.

You have hit a case where the impedance aspects of Ze cannot be ignored, since what you're really asking is to make the reactive part of the impedance disappear and the resistive part suddenly be the same as the value far away from the RC turnover points.

Again, this is not terribly complicated, but you're starting to hit sophomore-level EE stuff. Or what was sophomore level back in 1970. It would really be best for your long-term learning if you got the whole story and not just a few quick tricks.

[midwayfair]

RG, thanks for the pretty epic response. I at least know the terms I need to search for now, and I have a lot of reading to do.

[ashcat_lt]

You have just graduated to needing to know how to solve networks, not simple R-Cs.

What you really need now is some practice in learning what vector impedance is - a capacitor actually shifts phase, the current through it leading the voltage across it, as well as changing impedance with frequency. You need to know how to write down or estimate the impedance of R-C networks of more than one pair, which is not tricky, but to do it right you need to do some algebra right.
...
Again, this is not terribly complicated, but you're starting to hit sophomore-level EE stuff.

And that's where you lose me.  Apparently if two or more reactive components interact in certain ways a wormhole opens and everything turns imaginary!  I can handle vectors, but the when and why part is beyond my grasp just yet.

[R.G.]

And that's where you lose me.  Apparently if two or more reactive components interact in certain ways a wormhole opens and everything turns imaginary!  I can handle vectors, but the when and why part is beyond my grasp just yet.

It's probably not beyond your grasp - you just haven't seen it presented the right way. In a capacitor, you have to fill it with charge before the voltage across it can rise, and suck charge out of it before the voltage can drop. The voltage change is time-delayed from the current change. If you're feeding it a sine wave, the current goes in and out first, and the voltage across the cap is always chasing it up and down the slopes of the sine wave.

You can actually set up an oscilloscope with one channel reading the current through a cap with some kind of current probe and a differential probe reading the voltage across the cap, and the current waveform will be ahead of the voltage across the cap by 90 degrees at every frequency where the parasitics don't mess things up.

In a resistor, the current is always, instantly in phase with the voltage across it.

Neither the current nor voltage on a cap are imaginary -they're just phase shifted from what a resistor of the same impedance would be doing at that frequency. The use of complex algebra with imaginary numbers is actually a mathematical convenience to force a consistent way to represent numbers which have some always-in-phase component (like the current through a resistor) and some always-out-of-phase component, like a capacitor's current. The use of the term "imaginary" throws people. "Imaginary" numbers are in fact as real as the "real numbers" - it's a notation to describe vector directions with numbers that works out well for how we use numbers.

But it does take a while to get comfortable with that.

[thelonious]

So glad you posted this question, Jon. I have a lot of reading to do myself now. Thanks RG!

[R.G.]

This is one of those "next step deeper" things.

Once you get comfortable with capacitors versus resistors, inductors get easier as well.

Inductors have a similar story, just mirror image. For inductors, you put a voltage across them and the current builds up slowly as you "charge up" the magnetic field. The current lags the voltage in increasing. If you put a square wave of voltage across an inductor, the current can be seen to increase as a linear ramp. In fact, the "pulse inductance test" is one of those ways people have figured out to measure inductance. The slope of the current ramp and the voltage tells you the inductance directly. If the ramp is not linear, the inductance is changing, and this one of the ways to tell where an inductor starts to saturate without un-necessary fireworks.

Resonance is interesting. Put a current into a cap, the voltage builds up linearly, funneling energy into the electrostatic field. Charge up a cap, then connect an inductor across it. The cap instantly puts a voltage on the inductor, which starts funneling energy into its magnetic field, and the current building up linearly with the instantaneous voltage. As the energy in the cap decreases, the energy in the inductor increases until the cap can't send any more.

The inductor then PULLS on the cap to keep current flowing, and sucks the cap voltage negative until the inductor gets all the energy it can pull out. This charges the cap to the other voltage polarity, and when the inductor can't suck any more, the cap starts charging the inductor the other way. When this gets going, the cap and the inductor play see-saw with the energy, shoving it back and forth.

This would go on forever except that both the cap and the inductor have some resistance, and resistance always eats up energy, not storing it at all. So the small amount of resistance eats a little of the energy on each swap back and forth, until there's none left. This is ringing, and damping.

[bluebunny]

Thanks RG!

+1 1 1 1 1  1   1 

Have you ever considered putting on EE summer schools, R.G?  I would happily take two weeks off work and come to Texas to sit in a classroom...   

[tubegeek]

Highly recommended (don't be put off by the title) Visual Complex Analysis by Tristan Needham.

A summary:

The "imaginary" number line runs at right angles to the "real" number line. That's it. That's all you need to know, except that i times i = -1.

Plot any complex numbers as points on the coordinate plane using the x-axis for  "real" and the y-axis for "imaginary."

Once you do that, everything you need to know or to do using the complex number space is just coordinate geometry, exactly as you learned in high school algebra and a little pre-calc and trig.

It's SUCH a relief!
And, RG, I don't want to hear about j - ok?

[R.G.]

"i", "j", whatever. That's the dodge I mentioned -

The use of complex algebra with imaginary numbers is actually a mathematical convenience to force a consistent way to represent numbers which have some always-in-phase component (like the current through a resistor) and some always-out-of-phase component, like a capacitor's current.

It is exactly as you described - imaginary numbers are presented at right angles to real numbers, and "complex" numbers are always in the form of R+jI, where R is the real part, I is the imaginary part, and "j" (sorry - I already used "i" for the imaginary coefficient) . This is the graphical "convenience" that lets us visualize it.

The use of complex numbers as an orthogonal description of numbers in a plane instead of a line of numbers gives us a way that is not only visualizable, but also seems to fit the way the universe really works. In effect, the imaginary "i" or "j", the square root of negative one, seems to force a fit to a second dimension for numbers on a plane. It's generalizable to N dimensions, which are of course not visualizable beyond three or four. At least I can only dimly visualize four. Maybe some people can do more.

Complex algebra is the entre to all the higher regions of math. But fortunately, we don't have to get into that numerical stratosphere to work on pedals. Simple complex numbers of R+jI will do fine for audio work.   

[bluebunny]

Have you ever considered putting on EE summer schools, R.G?

And a math class too??   

[R.G.]

 Guys, central Texas is 100F and 80% RH in the summer. *I* look for educational places to go that are not Texas in August.   

Something in the Rockies or the Sierras is about right. San Francisco is the right temp, but there are all those *people*. 

[bluebunny]

Guys, central Texas is 100F and 80% RH in the summer. *I* look for educational places to go that are not Texas in August.  

Good point, well made.

.
.
.

I'll bring a fan and a crate of chilled beers.  

--

Seriously R.G. - you've managed in a handful of paragraphs to explain inductance in a refreshingly simple way that no one has ever done for me before.  And then you go and do the same for complex arithmetic.  Thank you.  

[R.G.]

You're more than welcome to any help I've provided.

For a lot of these things, the math is a necessary first step before you can construct a mental image of what's happening - literally a feel for what the energy is doing inside the parts. But the math is what enables the understanding. You need both the math and the mental image.

Take transformers. Very few people have an idea how they really work. Yeah, the coupled magnetic fields, V = N* d(phi)/dt, all that stuff. What's really going on is that the primary *voltage* pumps energy into the magnetic field, and that field is a coupling port between the primary and secondary. Whatever energy you take out of the secondary is instantly replaced by energy flowing in from the primary. There are no losses in the field when that happens, and there is no time lag. Anything taken out of the M-field bucket is INSTANTLY replaced through the primary.

That's why you cannot saturate a transformer from the secondary, only the primary. For a given voltage and frequency into the primary, the mag field goes to some limit, and no more. Nothing you can do on the secondary will affect that. You can burn it up from the secondary by making too much current flow and the wire resistances overheat it, but that's a different way of destruction.

All of the imperfections of transformers are a result of how well or poorly every single stinking turn of wire relates to the M-field inside the volume of the core, or to the losses in the resistances of the wire or eddy currents in the core material or the external capacitances.

It just takes a lot of math to get to that.

[tubegeek]

In effect, the imaginary "i" or "j", the square root of negative one, seems to force a fit to a second dimension for numbers on a plane.

It was, as I mention, such a relief once I got that picture in my head. And there are tricks specifically for complex algebra that result from that x/y setup which A) work perfectly, B) are pretty easy, and C) have explanatory power.

I have one minor clarification for you:

"complex" numbers are always in the form of R+jI, where R is the real part, I is the imaginary part, and "j" (sorry - I already used "i") for the imaginary coefficient .

In the quote above I moved a parenthesis to make it say what I *think* you meant, R.G.

In other words, the imaginary part has two parts: there is some scaling factor that is a real number (in math & physics, this is called a scalar.) That number, a plain, ordinary real number, gets multiplied by the imaginary number i (in mathematics) or j (in physics and electronics, because i is already used for the electrical property of current.)

So - the part I want to correct:

What you called I (the scalar coefficient of the imaginary part) is normally shown as X in textbooks etc. as its general abbreviation. This also is the abbreviation for the reactance of the component. Reactance is therefore shown using the letter X and that is nice and consistent.

So the general complex impedance is usually represented as R + jX

(In math the standard convention is a + bi to represent a general complex number. For a=0 the number is purely "imaginary," and for b=0 the number is a "real" number. "real" and "imaginary" just means they are found along one or the other axis of the complex plane, and do not shift out into the rest of the complex plane off the axes. a and b are always real numbers and i the square root of -1.)

In electronics what that translates to is that, when we look at a complex impedance R + jX we see that if X = 0 we have a purely resistive impedance (R only, plotted point found on the x axis.) If R= 0 we have a purely capacitive and/or inductive reactance (X only, plotted point found on the y axis.)

There are two ways to have X=0 in electronics: first, we can have a pure resistance with no inductance and no capacitance. Second, we can have any value of capacitance or inductance we like, with the restriction that the capacitive reactance X_C and the inductive reactance X_L are equal but opposite. Then they will total to X=0 and that is also a zero reactance. This is the condition of RESONANCE described (beautifully!) above by R.G.

Multiplying two complex numbers is not too hard. It involves only the FOIL method for multiplying binomials that may be familiar from high school algebra, and the fact that i squared = -1. But it's even easier if you do it graphically on the complex plane.

If we take two complex numbers p and q and multiply them, the result (the product) r is as follows:

- the distance of r from the origin will be: the distance for p times the distance for q

and

- the angle where you will find the point (measuring all angles counterclockwise starting from the positive x axis) will be: the total of the angle for p plus the angle for q.

So complex multiplication can be boiled down to one multiplication of two positive real numbers, and one addition of two angle measurements.

These are the kind of simplifications that are found and extended in the Needham book. If I wasn't using my wife's limited cell phone data plan and her computer right now, I'd upload the book to the documents topic for everyone. I will do that when I get back. It's an enormously helpful book. Every single trig identity you ever learned can be solved in a few steps with this method, for example. AND THERE'S MORE!! ACT NOW!!

Finally, if you are interested in the underpinnings of the stuff above, note that it is all explained by Euler's formula, which you can find good info about here

This formula is better known by its special case, e^iπ + 1 = 0 which is also called Euler's identity.

I'm sorry to go on so long: this is what you get when a math/electronics teacher goes on vacation and has unlimited time to be pedantic.