basic tube amplifier theory question
« on: July 19, 2005, 11:21:15 AM »
Hello everybody,

I've been lurking around for the last few weeks enjoying the posts and getting excited about learning more electronics and doing some DIY.  I've really enjoyed reading the posts on this forum.  There seems to be a minimum of b.s. and a maximum of helpful content around here, which is pretty nice.  Anyway, I found a copy of the Audio Cyclopedia, which was expensive, but awesome, and I have been digging into it.  I love it, in part, because it explains tubes.  But, of course, now I've got a question, and it's really basic.

Simply stated - In a two-stage triode amplifier, why does increasing the plate load resistor increase the voltage gain of the amplifier?

Or to quote tremaine - "To obtain a large voltage gain in a resistance coupled amplifier stage, the plate-load resistor must have as large a value as is practical."

This seems counter-intuitive to me because the higher the plate-load resistance is, the less current will flow through the plate and the greater the voltage drop across the resistor.  This means that the voltage at the plate is lowered as the resistance increases.  This implies (to me) less gain as resistance increases becuase less voltage is available at the plate.

Can somebody please explain where I went wrong with my reasoning?

For those of you that have the tremaine book (2nd edition), I'm looking at figure 12-1A.

Thanks - Andrew


Mark Burnley

basic tube amplifier theory question
« Reply #1 on: July 19, 2005, 12:01:27 PM »
Well,

The valve is acting as a voltage-to-current device effectively (i.e. transconductance). As you know a small change in Grid voltage allows a greater or lesser Anode current to flow. The Anode load resistor, across which the output signal is developed, converts this change in current to a change in voltage.

So for a given voltage change at the grid, this will result in the same anode current change. So increasing the anode load resistor will mean that this same current change will result in a larger voltage change.

Mark
O_O tape is life O_O

BYacey

    Where dogs wear thermal underwear, Alberta, Canada
  • Posts: 769
basic tube amplifier theory question
« Reply #2 on: July 19, 2005, 12:21:50 PM »
Because a tube is a relatively high impedance device, it is not capable of sinking huge amounts of current if the plate resistor was small in value. It is the modulated DC voltage drop across this resistor that develops the AC component of the signal. Therefore, as an example,  if the tube is capable of sinking a maximum current of 2 ma at 300V, you must choose a plate resistor high enough to  get the maximum voltage drop across the resistor between full conduction and cutoff of the tube.  Too low a value, and the tube will not handle the high current safely, and too high a value, the tube will not be able sink enough current. Hope this makes sense.
Bill Yacey
"Adjust R116 for least smoke"

bcarso

Re: basic tube amplifier theory question
« Reply #3 on: July 19, 2005, 01:09:14 PM »
Quote from: "cantgetnosleep"


This seems counter-intuitive to me because the higher the plate-load resistance is, the less current will flow through the plate and the greater the voltage drop across the resistor.  This means that the voltage at the plate is lowered as the resistance increases.  This implies (to me) less gain as resistance increases becuase less voltage is available at the plate.


Thanks - Andrew


What we are interested in, usually, is gain as a change around a bias voltage at an input producing a change around a bias voltage at an output.  Aside from operating the tube where it works "best", we don't really worry about what the absolute value of the plate voltage is until we are looking at our ability to swing big voltages, as outlined by BYacey.

In principle, as you make the load resistor bigger, you can always raise the voltage on the top end of the resistor to get back your desired plate voltage.  In practice you will hurt the tube if a big negative swing cuts it off and the plate arcs over to the grid, but that's another issue.

But if you increase the impedance of the plate load other ways, like putting a big inductor in place of the resistor, or using a sand state current source with high impedance, you will see higher voltage gain that finally limits out at a value called µ, "myoo", often shown as u when we're too lazy to fetch the proper greek letter.  This value is due to an intrinsic resistance called the plate resistance setting an upper limit to the circuit resistance.

Now, our voltage signal at the grid is still causing a change in plate current---as Mark describes, the transconductance ("gm") is the parameter that determines how much current change per voltage change at the grid.

Then this current change develops a voltage change based on the plate resistance Rp, and you have a voltage gain from grid to plate of µ.  That is, µ = gm*Rp.  Since the plate voltage goes negative for a positive change in grid voltage the actual voltage gain is -gm*Rp.

When you capacitor-couple that signal to the next stage you lose a bit because of the grid bias resistor as well.

basic tube amplifier theory question
« Reply #4 on: July 19, 2005, 01:38:43 PM »
Wow. You guys are fast.  Thank you everybody for the quick replies.  I was indeed making a very simple mistake in my reasoning that taught me I need to spend more time looking at the equations and less time staring at the pretty schematics when I'm trying to figure things out.  However, thanks to your help I think I've got a better grasp now.

Increasing the plate resistor increases the amplification because, for a given change in current, the subsequent change in voltage is directly proportional to the resistance.  Increase the resistance and the given change in current creates a larger change in voltage.  However, this only holds true so long as you operate within the range of currents that the given tube can sink.  

I kept thinking (mistakenly) that the larger resistor would just choke off the current and decrease the overall voltage (fuzzy thinking, sorry).

Thanks for the help - Andrew

bcarso

basic tube amplifier theory question
« Reply #5 on: July 19, 2005, 01:58:54 PM »
Between pondering this stuff for the first time AND being in Iraq, I can see why you can't get sleep  :grin:

Welcome to the forum in any event.

BYacey

    Where dogs wear thermal underwear, Alberta, Canada
  • Posts: 769
basic tube amplifier theory question
« Reply #6 on: July 19, 2005, 03:04:49 PM »
Quote from: "cantgetnosleep"
 I was indeed making a very simple mistake in my reasoning that taught me I need to spend more time looking at the equations and less time staring at the pretty schematics when I'm trying to figure things out.  

I wouldn't spend too much time with the formulae; it's more important to have an understanding what is really taking place  - abstract thinking within the schematic. This way when you look at a formula you can have an idea of what to expect for an answer when you do the math.
Bill Yacey
"Adjust R116 for least smoke"

CJ

basic tube amplifier theory question
« Reply #7 on: July 19, 2005, 04:50:31 PM »
You will never remember this stuff. At least I don't. So I make up hypothetical examples at two extremes to get a grip.

Imagine a fixed plate resistor. 100K is popular.
Next, imagine if the tube can swing 10 ohms. Predict the output of the "voltage divider" (tube circuit).
Next, Imagine if the tube could swing 100 k ohms. Predict the output.

Then reverse the tube and the plate resistor and you can re-invent the wheel where ever you go.

cj
If I can't fix it, I can fix it so nobody else can!
Frank's Tube Page: www.mif.pg.gda.pl/homepages/frank/vs.html
Guitar Amps: http://bmamps.com/Tech_sch.html

PRR

basic tube amplifier theory question
« Reply #8 on: July 19, 2005, 11:21:42 PM »
> This implies (to me) less gain as resistance increases because less voltage is available at the plate.

Gain and voltage are not quite the same. "Gain" can be measured with microscopic voltages. Yes, a 12AX7 fed with 250V from a huge plate resistor may bias-up at +1V on the plate, and overload at less than a volt, but we can put in 1 milliVolt and get out 80 milliVolts.

In the question before, it mentions staying on the linear portion of the characteristic. That means when you tamper the plate resistor, you have to make complementary changes to the grid-cathode bias or cathode resistor. As a rough starting place, put the plate at about half the supply voltage. For highest voltage gain, you would go lower; for maximum output swing you would go a little higher, but half is usually the starting place for a volt-amp.

12AX7, 200V supply, cathode resistor bypassed, cathode resistor adjusted to keep plate near 100V. If we start with a 100K plate resistor, we need to adjust for 1mA plate current. If we have a 400K plate resistor, we need to adjust for 0.25mA plate current, to hold that ~100V plate voltage.

If we do this with a transistor, voltage gain is unchanged. Gm (transconductance) is proportional to emitter current, so the 0.25mA situation has 4 times lower Gm which balances the 4 times higher load impedance.

But in a tube, Gm does not vary so much with current. Different tubes are different, but as a general approximation Gm goes down as square-root of cathode current.

In fact, triode gain goes up as current goes down, but very slowly. In fact I would not say that gain increases with large plate resistors: it doesn't increase much for practical values. But it will go down quite badly with small resistors and low resistor voltage drop.

You will usually be doing fine if the plate resistor is about twice the tube's internal plate resistance at the proposed operating point. For 12AX7, typically 100K-330K.(*) If that does not give enough gain, you probably want a different tube (or an added tube), not fine-fiddling with part values.

basic tube amplifier theory question
« Reply #9 on: July 20, 2005, 02:32:56 AM »
OK.  I think I'm getting a clearer picture.  Let me see if I'm getting this right.

The plate and cathode resistors form a voltage divider which sets the plate current and voltage.

Increasing the plate-load resistor, along with a subsequent increase in the cathode resistor to keep the plate voltage bias around half of the supply voltage, decreases the current through the tube.

In transistor, trasnconductance is directly proportionate to emitter current such that biasing does not affect amplification.  In tubes, transconductance is not as closely linked to plate current.  As the total current decreases, a given change of the grid voltage will cause a proportionally larger change in current through the plate.  This causes a proportionally larger change in the voltage of the voltage divider formed by the biasing resistors and the tube.

Right or wrong?

For example (I realize these number are probably way off, they are just for illustration), imagine a given tube swings 10mA through its linear region.  If the current through the plate is 25mA, then the linear region is swinging through 40% of the available current.  If we increase the current through the tube to 250mA, the 10mA linear region only encompases 4% of the total current and we get a 10x loss in voltage change at the plate.

Does that sound right?

Here's another question.  Actually its really more my current thought process, please correct me if I'm wrong.

A tube (or transistor) in cutoff is essentially like an open circuit.  A tube (or transistor) in saturation is essentially like a closed circuit as long as you don't exceede the maximum current rating of the device.  This implies that, as the tube or transistor approaches cutoff, the voltage at the plate (or collector) approaches the supply voltage.  As the tube or transistor approaches saturation, the voltage at the plate (or collector) approaches the bias value defined by the voltage divider - R2/(R1+R2).

Correct?

Thanks for the help.

bcarso - The name actually comes from my days when I was working (and partying) as a DJ a lot.  I didn't get a whole lot of sleep back then.  Iraq's not actually that bad, at least on the military bases.  I wouldn't want to hang around in town, but luckily I don't have to go into town much, except for the ocassional convoy.  It's just long hours and all work.  I've only been to the bunker once in the last couple weeks (because of mortars or rockets).  We used to go on a daily basis.  The insurgents seem to be targeting the iraqis a lot more than the bases these days.


analag

basic tube amplifier theory question
« Reply #10 on: July 22, 2005, 08:25:44 PM »
Grounded Cathode Amplifier
Read it, then read it again.
Audio engineering suffers from misinformation, disinformation, and downright lying more than most fields of endeavour.

PRR

basic tube amplifier theory question
« Reply #11 on: July 22, 2005, 10:44:46 PM »
> A tube (or transistor) in saturation is essentially like a closed circuit as long as you don't exceede the maximum current rating of the device.

There is a Saturation Resistance.

On a BJT, this is so low, like 100Ω, that you will blow-up the transistor before you notice the small collector-emitter voltage in a practical circuit.

On a small vacuum tube, the saturation resistance is more like 10K ohms. You can't think of it as a short circuit.

In a typical case: 12AU7 with a 100K plate resistor and 100V supply, when we take the grid-cathode voltage to zero (we never take the grid positive in audio: it is too much work), the plate only pulls down to about 10V. In many happy circuits, it won't even pull down that far.

The concept of Amplification Factor is critical. Plate(-cathode) voltage affects current, but grid(-cathode) voltage affects current much more.

> Grounded Cathode Amplifier

Good.

leigh

basic tube amplifier theory question
« Reply #12 on: January 21, 2006, 01:18:57 PM »
Quote from: "analag"
Grounded Cathode Amplifier
Read it, then read it again.


Thanks for this link, that's a well-written explanation for a novice like me!

Leigh

Kev

basic tube amplifier theory question
« Reply #13 on: January 21, 2006, 04:20:21 PM »
really, really good thread
 :green:
with answers like those above
I don't miss Tech Talk so much
now all you need is a few of those manufacturers to visit more often

a  cheerio  and a  thankyou  to all, involved in this thread
 :thumb:
Kev
DIY Factory

skipwave

basic tube amplifier theory question
« Reply #14 on: January 24, 2006, 12:57:52 PM »
This is a really informative link, though I prefer the HTML version because I hate PDFs :evil:

Grounded Cathode Amplifier (HTML Version) :thumb:
Quote from: PRR
Now, maybe you don't, or shouldn't, grab the ribbon for far-harpsichord, nor the hot condenser for snare-kissing... yet often we do.

caps

Re: basic tube amplifier theory question
« Reply #15 on: January 25, 2006, 10:51:52 PM »
Quote from: "bcarso"
Quote from: "cantgetnosleep"


This seems counter-intuitive to me because the higher the plate-load resistance is, the less current will flow through the plate and the greater the voltage drop across the resistor.  This means that the voltage at the plate is lowered as the resistance increases.  This implies (to me) less gain as resistance increases becuase less voltage is available at the plate.


Thanks - Andrew


What we are interested in, usually, is gain as a change around a bias voltage at an input producing a change around a bias voltage at an output.  Aside from operating the tube where it works "best", we don't really worry about what the absolute value of the plate voltage is until we are looking at our ability to swing big voltages, as outlined by BYacey.

In principle, as you make the load resistor bigger, you can always raise the voltage on the top end of the resistor to get back your desired plate voltage.  In practice you will hurt the tube if a big negative swing cuts it off and the plate arcs over to the grid, but that's another issue.

But if you increase the impedance of the plate load other ways, like putting a big inductor in place of the resistor, or using a sand state current source with high impedance, you will see higher voltage gain that finally limits out at a value called µ, "myoo", often shown as u when we're too lazy to fetch the proper greek letter.  This value is due to an intrinsic resistance called the plate resistance setting an upper limit to the circuit resistance.

Now, our voltage signal at the grid is still causing a change in plate current---as Mark describes, the transconductance ("gm") is the parameter that determines how much current change per voltage change at the grid.

Then this current change develops a voltage change based on the plate resistance Rp, and you have a voltage gain from grid to plate of µ.  That is, µ = gm*Rp.  Since the plate voltage goes negative for a positive change in grid voltage the actual voltage gain is -gm*Rp.

When you capacitor-couple that signal to the next stage you lose a bit because of the grid bias resistor as well.



This brings up a point of interest for me I was going to make a seperate thread. However, I have probably made too many of them recently. :grin:

Id like everyones opinions on Constant Current Sources. hmmmm, gain approaching u, less distortion etc....

Any drawbacks to CCS's that one should know/learn about? For all its potential worth, I havent seen alot of them in use. I also havent been able to find alot of info on the web about them, having extensively searched. A little, but nothing that goes into the nuts and bolts of how a CCS circuit actually provides constant current etc...

The sort of information I was looking for, is what devices can be used as CCS'S? How exactly do these devices provide constant current? How would one judge the suitability of a particular device for a CCS? etc. etc...

So, any links/info you may have that go into a bit more depth on CCS's would greatly be appreciated by me at least, and im sure a few others here. Thanks

caps

basic tube amplifier theory question
« Reply #16 on: January 28, 2006, 09:14:49 AM »
anyone?

caps

basic tube amplifier theory question
« Reply #17 on: February 01, 2006, 12:28:05 AM »
no one can enlightne us further on the ins and outs of CC's?

burdij

basic tube amplifier theory question
« Reply #18 on: February 01, 2006, 01:59:19 PM »
Here is an article that explains the CCS concept fairly well. The popular current source circuits use FETs. The general theory is, I believe, that the bias point relationship no longer depends on the "voltage divider" effect of the plate and load resisitances. You can optimize the bias point and then dial in the current value that will give you the correct plate voltage for maximum output swing. As long as the current source has sufficient drop across itself for correct operation, the plate can approach the B+ voltage quite closely.

http://home.pacifier.com/~gpimm/Active_loads_and_signal_current_control.html


 

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