Buffer Resistor Selection Questions / Experiment

[fuzzy645]

Looking to understand a few things about buffers, so I breadboarded this Orman version with a 2N3904 transistor and played around with using different resistors and the output caps.  I have some observations and some questions.  Here is the schematic for reference:



First, my  observations from this experiment

1. R1 and R2 form a voltage divider.  When R1 and R2 are equal, the output voltage at the base is exactly 1/2 the supply (therefore 4.5V), and the emitter voltage drop is approx .7 less than that, therefore 3.8V.  I tried and verified this with a pair of 220K resistors, a pair of 100K resistors, and a pair of 470K resistors.  Interestingly, the sound of the buffer was identical in all 3 cases.

2. I have heard some say it is best to shoot for 1/2 the supply at the emitter, so I changed values of R1 and R2 to 270K and 200K respectively, and this resulted in approx 5.1V at the base, and therefore a corresponding 4.4V at the emitter.  To my surprise, my ears did not hear any tonal difference when I made this change. I was expecting to hear some kind of difference (not sure what), but i couldn't detect any.

3. Lastly, I tried reducing  the output cap from the depicted 10 uf  to 2.2 uf and then increasing it to 22 uf.  Again, all 3 sounded exactly the same.

Now my questions:

1. Is it true that a desired voltage at the emitter of a buffer circuit should be approx 1/2 the supply, or does it really not matter?

2. If the answer to question 1 is yes, then why does every buffer schematic I run across (such as Ormans) show the R1 and R2 with identical values, which (as it forms a voltage divider) will leave output voltage at the base a 1/2 the supply, but  emitter below 1/2 the supply (approx 3.8V), or again does this totally not matter?

3. If the best practice is for R1 and R2  to have identical values, does it  then matter if one picks a pair of 100K vs. a pair of 220K vs. a  pair of 470K?  Like I mentioned above, it sounded the same in all 3 cases but perhaps there is a hidden reason. I would assume the 470K would give a higher input impedance so perhaps that is the way to go, since one  goal of a buffer is to provide a high input impedance.

4. Should I be hearing a difference between the output caps?  The value difference in my experiment seems huge (2.2 uf vs. 10 uf vs. 22 uf) so I thought I would hear an enormous difference, but I heard none.

5.  Lastly, if choosing a high gain bipolar transistor, is it true that approximate input impedance of the buffer is the HFE of the transistor multiplied times the emitter resistor?  So lets say a 2N3904 has a 500 HFE and the emitter resistor is 3.3K, the approx input impedance would be 1.6M?

Thanks!

Paul




QUESTION1:  To my ears, no mat

[ashcat_lt]

1)  You need to have room for the signal to swing both up and down without basing the transistor to provide voltage outside the bounds of the power supply.  If this bias voltage is too low then it will try to produce a voltage more negative than the bottom of the battery (or power supply) which is impossible.  Too high and it wants to put out more than the supply voltage which is impossible for the same reason.  In a unity gain buffer on 9V supply at typical guitar levels you've got a bit of wiggle room usually.  It won't likely swing more than a couple volts.  Anywhere close to the middle of the supply is good enough.

2)  In this situation the convenience of specifying two equal resistors outweighs the risks, as noted above.

3 & 5)  If I'm not mistaken, the two resistors are essentially parallel one another and that figure for the transistor itself.  So these two resistors help set the in-Z.  This will make a difference in the high frequency response of passive pickups connected directly to it.  The effect can be subtle, though the 100K as shown seems pretty low to me.

4)  The 10uF is way big for most loads to pass guitar or even bass frequencies.  You'd need to get much smaller to start to miss the bass response.

[fuzzy645]

4)  The 10uF is way big for most loads to pass guitar or even bass frequencies.  You'd need to get much smaller to start to miss the bass response.

Thanks for your answers.  With regards to question #4, would we then consider this output capacitor to be functioning as a kind of high pass filter?  So if the cap is large most all frequencies will pas, but if the cap is very small some of lows start to get filtered out?

[amptramp]

Thanks for your answers.  With regards to question #4, would we then consider this output capacitor to be functioning as a kind of high pass filter?  So if the cap is large most all frequencies will pas, but if the cap is very small some of lows start to get filtered out?

The low-frequency turnover is set by the capacitor and the impedance of the stage it is feeding.  If you are feeding 10,000 ohms through 10 µF, the -3db point is at 1.6 Hz.  Most amps and effects have a higher input impedance than that.  If you want to automate the calculation, you can use this:

http://www.muzique.com/schem/filter.htm

There is an approximation here: the resistance is the load resistance plus the output impedance of the emitter follower, but that will always be lower than the emitter resistor, so it will only make a few percent difference in most cases and it works to reduce the gain slightly below unity but extend the bass response.

[thelonious]

In addition to what amptramp said, I believe the main reasons people tend to use large output caps are: 1) Without knowing the input impedance of the pedal or amp that will follow the buffer, people tend to plan for the worst and provide a big output cap when possible. 2) If the output cap is an electrolytic, it will introduce some phase distortion near its corner frequency, so people tend to make the caps large enough to avoid that problem at any frequencies of concern (which for guitar is usually 82Hz and up)... but that kind of stuff probably matters more for hifi than for guitar anyway. 3) You never know if someone is going to plug a ______ into your pedal (bass, keyboard, bassoon, whizzbang atomic octave down pedal), so maybe frequencies lower than 82Hz do matter.

If the input impedance is 10k, as in amptramp's example, you would have to have a 0.22uF or smaller capacitor to start hearing a big difference, I think. And if the input impedance is higher, as it usually is, you'd have to use an even smaller cap to hear a large difference.

But I am a noob, so perhaps you should wait for confirmation on those things, and maybe I will learn something, too.

[R.G.]

1. Is it true that a desired voltage at the emitter of a buffer circuit should be approx 1/2 the supply, or does it really not matter?

As noted before, it matters only if the signal you're buffering is getting near the size of the power supply. The emitter follower circuit has its emitter one silicon-diode-drop below its base, about 0.5 to 0.7V. With ideal, zero-tolerance resistors the emitter output could only swing down from one diode drop below half the supply, and could, if properly driven, swing up to nearly the plus power supply. If you're amplifying 4.2V peak (double that for peak-to-peak) signals on a 9.00V supply, it probably matters. The closer you get to using up the whole power supply voltage, the more it matters. For signal much less than 2v-3V peak, it's only an academic issue.

Or - approximately half the supply is good, not mandatory, unless you are buffing big signals.

2. If the answer to question 1 is yes, then why does every buffer schematic I run across (such as Ormans) show the R1 and R2 with identical values, which (as it forms a voltage divider) will leave output voltage at the base a 1/2 the supply, but  emitter below 1/2 the supply (approx 3.8V), or again does this totally not matter?

It is much more convenient to use two identical resistors than to slightly offset one of them. Resistors only come in fixed values, and the ideally correct value to get the base one base-emitter-drop higher may not be available. It's convenience instead of theoretical perfection.

3. If the best practice is for R1 and R2  to have identical values,

"Best" has many meanings. It often means "simplest to buy and use" instead of "highest performance" as in 2 above.

does it  then matter if one picks a pair of 100K vs. a pair of 220K vs. a  pair of 470K?  Like I mentioned above, it sounded the same in all 3 cases but perhaps there is a hidden reason. I would assume the 470K would give a higher input impedance so perhaps that is the way to go, since one  goal of a buffer is to provide a high input impedance.

Details matter. As noted in the earlier answers, the two resistors are effectively in parallel as regards input impedance. Two 100ks give an equivalent 50K, two 220ks give 110k, and two 470ks give 235K. Each of these appears in parallel with the base input impedance of the transistor, which for the emitter follower connection can be approximated by current gain times the emitter resistor and any AC loading put on it. Assuming that the loading will be no worse than the emitter resistor itself and a current gain of 300, the input impedance of the transistor is 300*16.5K, or 4.95M. Big. A transistor gain of 100 gives about 1.65M.
So the input impedance is dominated by the bias resistors until they get up to about 3M each.

Also, it matters A LOT whether it's some other electronics driving it or a guitar pickup. Pickups have a huge inductive component, and that is what makes them have bigger SOURCE impedance at high frequencies. The source impedance of a guitar pickup may be up in the 100K range at treble frequencies. To avoid losing those high frequencies, you need input impedances in buffers **for guitar pickups as a signal source** of significantly larger than 100K, and "significantly larger" usually means 10:1 bigger. Hence the 1M nominal input in most guitar input circuits.

4. Should I be hearing a difference between the output caps?  The value difference in my experiment seems huge (2.2 uf vs. 10 uf vs. 22 uf) so I thought I would hear an enormous difference, but I heard none.

Again, it depends on what loads the buffer as much as it does what's in the buffer. The output cap carries the signal to the load. The half-power point of any single-R, single-C filter is F = 1/*2*pi*R *C).  The lowest note on a guitar in standard tuning is 82Hz, bass, 41Hz. So to make no real difference to bass response, a circuit with a 2.2uF output can could drive a load of R = 1/ (2*pi*C*F) = 1/(6.28*2.2uF*41Hz) = 1765 ohms, or anything larger. Loads larger than this will show no appreciable difference in bass response. For a 10uF, that becomes 388 ohms and for 22uF, 176 ohms.

Yep, I bet you heard no difference. The source impedance and load impedance are as much a part of the buffer circuit as any part you see in the schematic, but you have no real control of what gets hung on there.

5.  Lastly, if choosing a high gain bipolar transistor, is it true that approximate input impedance of the buffer is the HFE of the transistor multiplied times the emitter resistor?  So lets say a 2N3904 has a 500 HFE and the emitter resistor is 3.3K, the approx input impedance would be 1.6M?

See calculations above.

[Kesh]

Everyone's answered pretty much everything, but some points I thought worth mentioning.

R1 and R2 don't quite form a voltage divider when using a BJT, as there is current flowing into the base. As R1 and R2 get higher and hfe and the emitter resistor get smaller, this starts to matter, both for calculating R1 and R2 and for stability of biasing.

So you can't just pick two 4M7 resistors and have a wonderfully high impedance buffer.

You can also omit R1 and bias from the +ve supply alone.

[Gus]

IMO simple buffer needs to be designed for the load it drives in lower current designs.

Look at the input of the 3 transistor fuzz(you can find it in the "schematics" link at the top of the page) I have the emitter offset from 1/2 of 9VDC on purpose.  The resistor values were picked from parts you could buy at Radio Shack at the time, both for the offset and input resistance with higher hfe transistors

A resistor from base to + for bias is not temp stable and a bad design IMO.

Higher input resistance can be achieved with BJTs and  bootstrapping.

You can adjust the offset with one resistor in each circuit below

Sims I have posted in other threads

[R.G.]

Gus posted while I was typing. 

Everyone's answered pretty much everything, but some points I thought worth mentioning.

R1 and R2 don't quite form a voltage divider when using a BJT, as there is current flowing into the base. As R1 and R2 get higher and hfe and the emitter resistor get smaller, this starts to matter, both for calculating R1 and R2 and for stability of biasing.

So you can't just pick two 4M7 resistors and have a wonderfully high impedance buffer.

Good point. The devil is always in the details. The point was touched on by noting the input impedance of the base. But calculating rough boundaries on the current pulled by the base lets you figure out better where the emitter will end up.

To a first, crude, quick approximation, with about 4V across the emitter resistor, you get 4V/Re current flowing in the emitter, and 1/HFE (the DC HFE, note) in the base. This would be something like 1/(300) * 4V/3.3K = 4uA in the handwaved case we've been talking around. That comes out of a voltage of the divider resistors in parallel times the voltage supply, and through an equivalent resistor of the divider resistors in parallel.

For the 100K/220K/470K examples, this is 50K, 110K, and 235K and the drop from 4uA is 0.2V, 0.444V, and 0.949V. This matters if you're doing big signals again, as in the case above. The error is somewhat like the size of the base-emitter drop.

You can also omit R1 and bias from the +ve supply alone.

Yes. You can, but it's a poor choice for stabilizing the bias point in terms of different HFEs for different transistors and gains, and for gain and leakage drift. A better choice for both bias stability and for high input impedance is making the bias divider be a pair of lower value resistors, perhaps in the 10k-100k range, and a high-value resistor from this divider to the base. This has the advantage that the junction of the bias divider can be connected to the emitter by a capacitor, and effectively "bootstrapped" to a much higher impedance than is otherwise possible.

[PRR]

> at the base is exactly 1/2 the supply ..pair of 100K ...a pair of 470K resistors.

Re-check that observation. What I figure: with a hi-gain 2N3904 the bias junction will sag 0.7V with 2X470K, and less with lower resistors. You should see a half-volt difference.

The 10uFd is plenty-ample for signal. Now run your cable over big power lines, or around fluorescent lamps, humm/buzzz sources. With 10uFd, the emitter can suck-up some of that hash so you won't hear it in the amp. With 0.02uFd (ample for most cord-stuff signal purposes) the hash-suck effect is much less.

Also 10uFd electro is often the cheapest cap available. Pot of tarnished alloy instead of careful layers of hyper-fine sandwich-wrap.

The half-supply bias: Yes, if the load were infinite then for MAXimum possible output you want the emitter at half-supply. Of course 0.7V "off" in a 9V system is only 1.5dB less.... if the one plan isn't enough, the other isn't so much better as to ensure total happiness.

Typical 100K-1Meg loads are "infinite enough" compared to the 3.3K DC load.

But consider if you had to drive a 3K load. Going up, the 100mA transistor can easily supply 4 or 6mA with near-zero drop. But going down, the limit is the 3.3K DC load fighting the 3K external load. Now the optimum is more like 2/3rd of supply. Bias at 6V, transistor can pull-up 3V, the other way the 3.3K and 3K split the 6V to give (about) 3V load swing. It's symmetrical, and (nearly) as good as you can get without higher idle current (battery dies sooner).

That's with BJT transistors working at small fractions of their clog-up/melt-down current. Go with tubes, and instead of ~~~10 ohms inside the slammed transistor you find the 2K-40K of electrons in empty space. Since this is comparable to the 5K-100K DC loads we would use, we have to compute both pull-up and pull-down. (There's a simple optimum but it has a square-root.)

In DIY: assume half-supply is a good starting point. Beat it. If it never clips, that's fine. If it clips, asymmetrically (a 'scope is handy), then re-diddle it (you can also use ears).

[fuzzy645]

Thank you all for your detailed responses. I have learned quite a bit by reading this.

Or - approximately half the supply is good, not mandatory, unless you are buffing big signals.

That makes sense, thanks.

Details matter. As noted in the earlier answers, the two resistors are effectively in parallel as regards input impedance. Two 100ks give an equivalent 50K, two 220ks give 110k, and two 470ks give 235K. Each of these appears in parallel with the base input impedance of the transistor, which for the emitter follower connection can be approximated by current gain times the emitter resistor and any AC loading put on it. Assuming that the loading will be no worse than the emitter resistor itself and a current gain of 300, the input impedance of the transistor is 300*16.5K, or 4.95M. Big. A transistor gain of 100 gives about 1.65M.
So the input impedance is dominated by the bias resistors until they get up to about 3M each.

2 follow ups about this.   

1).  Wouldn't that provide an argument to always use high resistor values (lets say 2M each) for the bias resistors.  2M in parallel will yield 1M.    I'm asking this because I'm sure there must be an argument against it since I don't think I have ever seen a buffer circuit posted with a pair of 2M resistors.   

2).  Is the 16.5K you mentioned above is assuming the emitter resistors is 16.5K?

Also, it matters A LOT whether it's some other electronics driving it or a guitar pickup. Pickups have a huge inductive component, and that is what makes them have bigger SOURCE impedance at high frequencies. The source impedance of a guitar pickup may be up in the 100K range at treble frequencies. To avoid losing those high frequencies, you need input impedances in buffers **for guitar pickups as a signal source** of significantly larger than 100K, and "significantly larger" usually means 10:1 bigger. Hence the 1M nominal input in most guitar input circuits.

That makes perfect sense, thanks.

The output cap carries the signal to the load. The half-power point of any single-R, single-C filter is F = 1/*2*pi*R *C).  The lowest note on a guitar in standard tuning is 82Hz, bass, 41Hz. So to make no real difference to bass response, a circuit with a 2.2uF output can could drive a load of R = 1/ (2*pi*C*F) = 1/(6.28*2.2uF*41Hz) = 1765 ohms, or anything larger. Loads larger than this will show no appreciable difference in bass response. For a 10uF, that becomes 388 ohms and for 22uF, 176 ohms.

Now on some pedal schematics (such as some overdrive circuits) I have observed a typical value for the output capacitor to be  .01 uf.  Using the reactance formula above R = 1/ (2*pi*C*F) then a circuit with a .01 output cap could drive a load of 388.18K or higher without showing a difference in bass response.  Based on this, it would seem to me that most often it would be use a much larger output cap along the lines of 10 uf.

Yes. You can, but it's a poor choice for stabilizing the bias point in terms of different HFEs for different transistors and gains, and for gain and leakage drift. A better choice for both bias stability and for high input impedance is making the bias divider be a pair of lower value resistors, perhaps in the 10k-100k range, and a high-value resistor from this divider to the base.

Would that imply the pair of 100K's (still in parallel) yield 50K, but that is technically in series with the 1M which yields a higher iinput impedance?

This has the advantage that the junction of the bias divider can be connected to the emitter by a capacitor, and effectively "bootstrapped" to a much higher impedance than is otherwise possible.

3 follow-up questions on this one:

1. When you say this capacitor is connected to the emitter are you referring to the fact that the emitter is grounded, and this additional capacitor is also grounded? 
2. How does this capacitor affect the input impedance?
3. I don't understand what you mean by "bootstrapped" here.

> at the base is exactly 1/2 the supply ..pair of 100K ...a pair of 470K resistors.

Re-check that observation. What I figure: with a hi-gain 2N3904 the bias junction will sag 0.7V with 2X470K, and less with lower resistors. You should see a half-volt difference.

I double checked, and you are right!! 

The 10uFd is plenty-ample for signal. Now run your cable over big power lines, or around fluorescent lamps, humm/buzzz sources. With 10uFd, the emitter can suck-up some of that hash so you won't hear it in the amp. With 0.02uFd (ample for most cord-stuff signal purposes) the hash-suck effect is much less. 

So a larger cap can actually minimize some of the extraneous noise/buzz.  That is very cool.

Everyone's answered pretty much everything, but some points I thought worth mentioning.
R1 and R2 don't quite form a voltage divider when using a BJT, as there is current flowing into the base. As R1 and R2 get higher and hfe and the emitter resistor get smaller, this starts to matter, both for calculating R1 and R2 and for stability of biasing.

I guess what I am confused about is it looks like every example I have seen posted of voltage dividers. 

[R.G.]

Details matter. As noted in the earlier answers, the two resistors are effectively in parallel as regards input impedance. Two 100ks give an equivalent 50K, two 220ks give 110k, and two 470ks give 235K. Each of these appears in parallel with the base input impedance of the transistor, which for the emitter follower connection can be approximated by current gain times the emitter resistor and any AC loading put on it. Assuming that the loading will be no worse than the emitter resistor itself and a current gain of 300, the input impedance of the transistor is 300*16.5K, or 4.95M. Big. A transistor gain of 100 gives about 1.65M.
So the input impedance is dominated by the bias resistors until they get up to about 3M each.

2 follow ups about this.   

1).  Wouldn't that provide an argument to always use high resistor values (lets say 2M each) for the bias resistors.  2M in parallel will yield 1M.    I'm asking this because I'm sure there must be an argument against it since I don't think I have ever seen a buffer circuit posted with a pair of 2M resistors.   

It would, if other issues did not arise. The bias resistors contribute noise proportional to their resistances, so high resistance also leads to higher noise. In addition, the unavoidable capacitances across and around the resistors cause some treble loss, depending on the relative values. The three-resistor setup I mentioned, with two lower valued resistors making a voltage divider and a higher value resistor in series from there to the base is one attempt to get around some of these issues. There are other ways, too, but they all share the issue of using more parts. In this, like many circuit design questions, there are many conflicting requirements to compromise among.

2).  Is the 16.5K you mentioned above is assuming the emitter resistors is 16.5K?

Oops. Sorry, I slipped a digit. That should have been 1.65K, not 16.5K. The emitter resistor was 3.3K, I assumed an external load resistor after the output cap of no less than the same 3.3K, or 3.3K parallel with 3.3K. Then I got the decimal point wrong. S'wat I get for doing math in my head.

Now on some pedal schematics (such as some overdrive circuits) I have observed a typical value for the output capacitor to be  .01 uf.  Using the reactance formula above R = 1/ (2*pi*C*F) then a circuit with a .01 output cap could drive a load of 388.18K or higher without showing a difference in bass response.  Based on this, it would seem to me that most often it would be use a much larger output cap along the lines of 10 uf.

Depends on what you're driving. It's that unseen and unspecified load impedance again. If the person selecting cap values wants full range bass, they have to pick the cap based on what they think someone *might* connect to it. A 1M guitar amplifier input would be one choice one could make, and that would give full range bass. Or, the person picking caps may have *wanted* to drop out some bass. Or they may have had no clue what C = 1/(2*pi*F*R) does for you.

If you plop in 10uF everywhere, you can get way, way subaudio response. This is not bad for the audio, necessarily, because all the bass will be there, but it may also cause things to take a long time to come on, charging up that 10uF through a high load impedance. One really does have to know and/or guess at what will be connected to the output.

Would that imply the pair of 100K's (still in parallel) yield 50K, but that is technically in series with the 1M which yields a higher iinput impedance?

The way this is intended to be done is to use a couple of resistors of some value that doesn't eat up too much current from the power supply - pick a value - to make the bias voltage. A pair of 100K's is on the high side, but probably OK. This voltage divider output is then decoupled to ground with a capacitor, which makes it a quite-low impedance to AC, and also shunts the noise from the bias resistors to ground. Then a large resistor, like a 1M if that's what you want to use, goes from the bias point to the base. The AC impedance is the series biasing resistor in parallel with the base input impedance to a first approximation.

1. When you say this capacitor is connected to the emitter are you referring to the fact that the emitter is grounded, and this additional capacitor is also grounded? 

No. This is a different variation of the "noiseless biasing" from above. The emitter is still going through a resistor to ground, still being an emitter follower and putting out 0.99 or so of the input voltage. The capacitor connects the emitter to the junction of the bias voltage divider, which has no effect on its DC voltage, but driving the bias voltage with an AC voltage only slightly smaller than the input voltage.

What this does is reduce the AC voltage across the 1M bias (in this case) bias resistor. It's got the input signal, 1.00 times the input voltage, on the base end, and the emitter (AC) voltage, about 0.99 of the input voltage on the other. Ohm's law still applies so the resistor eats an AC current of only 1.0 times the input signal from the base end minus 0.99 times the input signal at the other end, and so the current through it is (1.0-0.99) = 0.01 as much. In effect, bootstrapping makes the bias resistor look 100 times (in this made-up example) as large as it is to AC signals. So you could use a 100K bias resistor instead of 1M, but the bootstrapping from the emitter would make this perform like a 10M input resistor in terms of loading on the input signal.

The setups are similar. Noiseless biasing decouples the bias voltage divider to ground. Bootstrapping connects it to the output, in a form of positive feedback which is always guaranteed to be less than unity, so it does not cause oscillation. Resistors same connection, capacitor connection different.

2. How does this capacitor affect the input impedance?

AS noted, it multiplies the effective AC value of the series bias resistor by the inverse of the ratio of the output voltage to the input voltage. A bipolar follower generally has an output which is 0.95 to 0.99 times the input voltage, so the factor is generally 20 to 100 times larger. This is a very broad brush. To get the real numbers, one has to do the math.

3. I don't understand what you mean by "bootstrapped" here.

The output voltage, which is nearly as big as the input voltage, and has a low impedance, is used to raise and lower the input bias voltage in synchronism with the signal. That is, the low impedance output is partly used to buffer its own input - the term comes from the phrase to pull ones self up by ones own bootstraps.

[Gus]

Here is a thread with two circuit sims left side is a one pickup single coil guitar/cable sim.  It has a X10 gain stage with a switchable input biasing setup one way is bootstrapped and the other is textbook biasing like in a LPB type circuit.  I think this might help you understand biasing if you study what I did.  I worked out circuit  resistor values for a transistor like 2N5088/5089 so I could switch biasing without much change in the circuit voltages and currents at the collectors, emitters and bases.

Anyone build it yet?

http://www.diystompboxes.com/smfforum/index.php?topic=99139.0

There have been other threads in the past at this site about buffers and offsetting the emitter voltages from 1/2 for the load it drives.

[Kesh]

Everyone's answered pretty much everything, but some points I thought worth mentioning.
R1 and R2 don't quite form a voltage divider when using a BJT, as there is current flowing into the base. As R1 and R2 get higher and hfe and the emitter resistor get smaller, this starts to matter, both for calculating R1 and R2 and for stability of biasing.

I guess what I am confused about is it looks like every example I have seen posted of voltage dividers.  

I guess every voltage divider you have seen (as they usually do) assumes the load on the dividing point can be discounted, as it draws so little current/has such high resistance.

a pic is worth a thousand words


In this thread, R_L in the picture above is the transistor's "resistance" from base to ground. for DC biasing purposes it is approximately the emitter resistor x hfe

[R.G.]

Correct, Kesh!

This is one reason there is a rule of thumb that the current in the resistor divider chain needs to be 10x or more the current going out of the node; I went through some of this in the article on Geofex on picking vbias resistors: http://geofex.com/circuits/Biasnet.htm

If you can't make the load current negligible compared to the divider string current, then you have to take the loading into account.

There is another way to calculate it, as well, using Thevenin/Norton equivalent circuits. If you have your V1, R1, and R2, you can replace those three components with two, a voltage source and one resistor. The voltage is V1*R2/(R1+R2), and the resistor is the parallel combination of R1 and R2. The output voltage and current will be identical for all loading to the three-part network, and the effect is usually easier to work with. The math reduces to exactly the equations you quote.

But the nice thing is that you only need to do that parallel resistor calculation once, and the math afterwards can largely be done in your head.

[amptramp]

Details matter. As noted in the earlier answers, the two resistors are effectively in parallel as regards input impedance. Two 100ks give an equivalent 50K, two 220ks give 110k, and two 470ks give 235K. Each of these appears in parallel with the base input impedance of the transistor, which for the emitter follower connection can be approximated by current gain times the emitter resistor and any AC loading put on it. Assuming that the loading will be no worse than the emitter resistor itself and a current gain of 300, the input impedance of the transistor is 300*16.5K, or 4.95M. Big. A transistor gain of 100 gives about 1.65M.
So the input impedance is dominated by the bias resistors until they get up to about 3M each.

2 follow ups about this.   1).  Wouldn't that provide an argument to always use high resistor values (lets say 2M each) for the bias resistors.  2M in parallel will yield 1M.    I'm asking this because I'm sure there must be an argument against it since I don't think I have ever seen a buffer circuit posted with a pair of 2M resistors.   

It would, if other issues did not arise. The bias resistors contribute noise proportional to their resistances, so high resistance also leads to higher noise. In addition, the unavoidable capacitances across and around the resistors cause some treble loss, depending on the relative values. The three-resistor setup I mentioned, with two lower valued resistors making a voltage divider and a higher value resistor in series from there to the base is one attempt to get around some of these issues. There are other ways, too, but they all share the issue of using more parts.

The resistors in the voltage divider are in parallel with the input source resistance for the purpose of noise calculations.  If you have a 7000 ohm pickup driving the input, it will not matter what the divider resistors are, you cannot exceed 7000 ohms as long as the coupling capacitor passes the frequencies of interest.  There will be more noise below the input coupling turnover frequency, but this should be well below the audio range.

[R.G.]

The resistors in the voltage divider are in parallel with the input source resistance for the purpose of noise calculations.  If you have a 7000 ohm pickup driving the input, it will not matter what the divider resistors are, you cannot exceed 7000 ohms as long as the coupling capacitor passes the frequencies of interest.  There will be more noise below the input coupling turnover frequency, but this should be well below the audio range.

A guitar pickup is not a resistive source - the vast majority is inductive, and that inductive impedance rises with frequency. This inductance keeps the low wire resistance from any shunting of the thermal noise of the bias string. 

At frequencies where the inductive impedance is larger than the wire resistance, the wire resistance increasingly stops mattering.

[amptramp]

The resistors in the voltage divider are in parallel with the input source resistance for the purpose of noise calculations.  If you have a 7000 ohm pickup driving the input, it will not matter what the divider resistors are, you cannot exceed 7000 ohms as long as the coupling capacitor passes the frequencies of interest.  There will be more noise below the input coupling turnover frequency, but this should be well below the audio range.

A guitar pickup is not a resistive source - the vast majority is inductive, and that inductive impedance rises with frequency. This inductance keeps the low wire resistance from any shunting of the thermal noise of the bias string. 

At frequencies where the inductive impedance is larger than the wire resistance, the wire resistance increasingly stops mattering.

Correct, I forgot about that and the possibility of a pot at the output of the guitar.  The inductance of the pickup has no noise of its own, but it does uncouple the pickup winding resistance.

[R.G.]

Yeah, the big inductance of those pickups keeps finding ways to make guitar electronics tough! 

[PRR]

> it looks like every example I have seen posted of voltage dividers.

EXAMPLE voltage-dividers have no load. If you mean the kind in text-books.

EVERYTHING useful is a voltage-divider. But (as seen in Kesh's helpful pictures) you must include any and all loads. 1Meg+1Meg, measured with a 10Meg meter, is not 0.5 but 0.4762. You account for this by combining the lower 1Meg with the 10Meg meter to get 0.9090909Meg. Or omit the meter and apply a transistor which happens to suck a lot like a 1Meg resistor: divider ratio is now 0.3333.

> an argument to always use high resistor values ...for the bias resistors.

See above.

Or get extreme. 1Meg is good, try 1,000Meg. (Mouser has 'em.) Effectively parallel to audio input, 500Meg, yowsa! Also parallel to transistor base current. If the bias stayed at 4.5V, so emitter was (rounded) 4V, that's 4V/3K or 1.3mA emitter current. Base current is hFE less. Say hFE is 400, 0.003mA or 3uA/ 3uA in 500Meg is 150 Volts. Hummm, can't be right, obviously the divider and base voltage will sag all to heck.

A hasty approximation is to figure the base looks-like the emitter resistor (3.3K) times hFE (400?), or 1.3Meg. Now you see that 1,000Meg+1,000Meg won't do what we want with 1.3Meg sucking on it. Even 1Meg+1Meg (500K) will sag very noticably.

OK, change the divider to maybe 1Meg+2Meg.

However, hFE is not known exactly. A given part-number may be 100-400. That gives 0.33Meg to 1.3Meg. Yes, YOU can spend time trimming resistors to the ONE transistor in your hand/board. Now you get a contract to deliver 100,000 by Friday, you can't hand-trim, you need to slap-and-ship. OR you publish the plan on the interwebs where any fool can copy it and complain when it does not work.

Jack shows 220K+220K, equivalent 110K, because with 3.3K under and hFE 100-300, 330K-1,320K, bias-point voltage sags little with 330K and less with 1320K. Base voltage could be 3.3V to 4.15V--- if we have a large-Volt for swing and 0.6V for base-emitter, it will work, this gives over 2V of swing worst-case. Real slap-and-play always-works design.

> ...some overdrive circuits... output capacitor to be  .01 uf.

That's too small for "good" sound. As you say, 0.01u is good for 20Hz with 1Meg load and 41hz with your 400K(*) load, but many guitar-cord inputs are lower. (* There's no "388.18K" when figuring first-order filters. The roll-off is so gentle that 2-digit answers are plenty good.)

Overdrive is different. Put in 82Hz, it spews lots of 164Hz 246Hz 328Hz etc. The primary products are always _higher_. We value them, and will sometimes cut some of the pure sound (especially strong bass) to bring-out these new sounds.



> vast majority is inductive

AND _capacitive_.

For naked pickup, the result is impedance low everywhere eXcept an octave-wide ring in the 1KHz-5KHz zone where the ear is very sensitive. The ringy hiss may be worse than broadband hiss.

In most guitars, there is a 250K pot. Worst-case impedance is 250K, and under 100K for reasonable pickup values and not-maxxed pot position.

> the math afterwards can largely be done in your head.
> I got the decimal point wrong. S'wat I get for doing math in my head.

Should be using the slide-rule. (If only mine had the USB "Paste" option so I did not have to re-type the decimal points in the wrong place.)