Can Anyone Teach Me This Equation?

[Gary_S]

I've been studying the tech details on this link i'll provide at the end of the post. I'm still going back to relearn maths concepts and Algebra from the beginning so i'm not sure how to navigate this equation.

It's for working out the gain in a stage. I know where the numbers are from ie. plate load resistor and unbypassed cathode resistor and all that but in the top line of the equation am i adding those together or multiplying them? Then on the bottom under that we have 1 then under that we have the 1400MHOS which is the transconductance of the tube. So we have the 1 to divide into the answer from the very top line, well that'll just give the answer you get from the top line surely eg. if the top line read 1500 dividing that by 1 just gives 1500!

So i was just taking the top line and dividing it by the 1400MHOS but i'm not getting the answer that they get in the doc.

Come on you maths heads please help me learn to do this equation

http://www.ax84.com/static/p1x/p1-ex-theory.pdf  The page the equation is on is page 11. I'd write it up on the site myself but i can't write it out properly the way it appears in the document.

[kagliostro]

Do you mean what signify || ?

if so || = in parallel

're looking for this?

as to help you calculating the values, here a simply online Parallel Resistor Calculator

http://diyaudioprojects.com/Technical/Electronics/parallel-resistor-calculator.htm

Here one other

http://www.1728.org/resistrs.htm

K

[Gary_S]

Do you mean what signify || ?

if so || = in parallel

where you looking for this ?

as to help you calculating the values, here a simply online Parallel Resistor Calculator

http://diyaudioprojects.com/Technical/Electronics/parallel-resistor-calculator.htm

Hi Kagliostro,

No, what i mean is how to work the equation at the bottom of page 11? how do i do it?  The top line of the thing first; do i add those values or multiply them together? After i have those am i supposed to divide by 1 or the transconductance which is 1400mhos?

Basically i'm just asking how i go about actually doing the whole equation, to arrive at the final result, step by step.

I'm starting studying algebra again as of now but i'm a total newb and am not sure what to do with this equation!

Thanks for the resistor calculator! i'll put that to good use.

[kagliostro]

Now I understand what you mean and .......... no

I can't figure how to solve it

I think my math is much worse than your's

K

[HotBluePlates]

So it's putting example numbers into the equation presented at the top of page 11.

What Kagliostro is telling you is the parallel lines (  ll  ) is simply an electronics shorthand for "in parallel with". You have to plug those numbers into a whole other formula to get the total resistance of those 3 resistances in parallel. So it's not multiplication or addition.

The formula for parallel resistances is

Rp = 1/[(1/R1) + (1/R2) + (1/R3)]

That mean "1 divided by R1" plus "1 divided by R2" plus ...  for as many parallel resistances as you have. Once you get that sum, you divided 1 by that sum to get total parallel resistance.

A shortcut you can use for 2 resistances is "product over sum".

Rp = (R1*R2)/(R1+R2)

The shortcut is you can use this formula for any 2 of those 3 resistances you started with, then take the result and use it as your new "R1" and the last remaining resistance as your new "R2". So you do product over sum twice.

Upfront, I'll say I think the AX84 gain formula is confusing. I use something different, but we'll use their approach.

"Mho's" is a unit of conductance, and is the opposite of resistance. If resistance = volts/current (in accordance with ohm's law), than conductance = current/volts. Maybe that will come in handy for you later.

Anyway, the overall equation is "resistance stuff in parallel" divided by "reciprocal of Gm plus cathode resistance". So your approach is to get the resulting value of the parallel resistances, and use that as you new numerator. You'll then find the reciprocal of Gm, because you do multiplication/division before addition/subtraction. Add the value of Rk to that number, and you have your new denominator. Now you finish by dividing the top number by the bottom number.

So how to you do "reciprocal of .... micromhos"? That μ (micro) sign means take the number and move the decimal 6 places to the left. 1400 μmhos becomes 0.0014. Take your reciprocal (1/0.0014) and you should get about 714. Add to your "Rk" of 0Ω and you are left with 714.

However you do the parallel resistances, here's some numbers to check your steps:
For total parallel resistance (top of the equation), I get 39,810Ω (and some change). We know from above the bottom number will be 714.

39810/714 = 55.7 (and some change)

I rounded in a few places, but so did the original document. Overall, we're close enough for tubes 'n rock 'n roll.

[jojokeo]

You simply need to understand "orders of operation" meaning which calulation to do first. And then I think where you're getting stuck is the lower part, the conversion. uMHOS needs to called and thought of as micro MHOS. Therefore, 1400uMHOS is 1400 to the minus 6th power (in spelling/writing it out for you). This means move the decimal point to the left 6 places to get a standard measurement in MHOS. This then turns 1400 uMHOS = .0014 MHOS
Whenever you do a calculation for resistance, current, voltage, power, etc. you have to convert the number to the actual unit of measure (not use mA, uA, nA, etc...). So for ex if you're figuring out current in A you divide Voltage by Resistance. Your result is always is Amperage. If the number is small like .000153 Amps, then change it to 0.153mA or go another three decimal places to the right and the same result can be called 153 uA. So, 0.000153 A = 0.153 mA = 153 uA. You will learn about significant figures in basic algebra also which is how many actual numbers to use or not to get the desired result you're looking for. Meaning too many numbers becomes insignificant, redundant, cinfusing and/or useless therefore rounding up or down is used to the amount of necessary numbers wanted or needed.

So do the top three values in parallel, you get 39.810K
1/.0014 OR the recriprocal of 1400 uMHOS = 714.3 MHOS, now adding "0" to anything is still zero right?! Lower value still is 714.3 MHOS
39,810 R / 714.3 MHOS = 55.7 is your gain

[jojokeo]

Damn HBP, if I'd known you were answering I could've saved some time, lol. 

[kagliostro]

Many Thanks HBP

I was completely missing this

That μ (micro) sign means take the number and move the decimal 6 places to the left. 1400 μmhos becomes 0.0014. Take your reciprocal (1/0.0014) and you should get about 714.

K

p.s.: Damn Jojokeo, if I'd known you were answering I could've thanks also you   

[Gary_S]

You guys rock! you're the best on here! I was putting this question up on another amp site a few minutes ago, in case i didn't get any luck on EL34, i then came back and you've broken it down and answered the question for me!!

Thanks very much guys, it's great when experienced folks take the time to help a novice like myself come to terms with new stuff, because this stuff can get confusing!!

HBP, Jojokeo, Kagliostro   

[HotBluePlates]

You're welcome!

We all started as inexperienced novices, too. 

[PRR]

If Rk is zero, as seen here, then there is a MUCh simpler way. (total plate load) TIMES gM.

39,000 * 0.0014 = 55.7

Engineering calculator or a plain +-*/ 4-banger?

With a TI-type engineering calculator, 1,400uMho is 1400, EE, 6, -.

Otherwise, as well-said, 1400uMho is 0.001,4.

When you have fractions within fractions, it often helps to work-out the bottom first. Especially if you have a 1/x key. Do the bottom, 1/x, then do the top.

But there's other ways to do it.

That form assumes all the input (cathode) stuff is series (+) and all the output stuff is parallel (||). Then the gain is simply the ratio.

While transconductance is given in uMhos, it is VERY helpful to think of 1/Mho as a cathode resistance (which it is). Then the gain is all the stuff in the output (parallel) /OVER/ all the stuff in the cathode (1/gM + any Rk).

[Colas LeGrippa]

Gary S,

one simple trick to begin with when estimating the total load of paralleled resistances is to take the value of the smallest resistance value. Then, the total load is smaller than this value, everytime. So the total resistance of  those in parallel :  1 ohm, 1 meg, 470K, 68k and 10 ohm  is less than 1 ohm, which is the smallest value.

Colas

[Gary_S]

If Rk is zero, as seen here, then there is a MUCh simpler way. (total plate load) TIMES gM.

39,000 * 0.0014 = 55.7

Engineering calculator or a plain +-*/ 4-banger?

With a TI-type engineering calculator, 1,400uMho is 1400, EE, 6, -.

Otherwise, as well-said, 1400uMho is 0.001,4.

When you have fractions within fractions, it often helps to work-out the bottom first. Especially if you have a 1/x key. Do the bottom, 1/x, then do the top.

But there's other ways to do it.

That form assumes all the input (cathode) stuff is series (+) and all the output stuff is parallel (||). Then the gain is simply the ratio.

While transconductance is given in uMhos, it is VERY helpful to think of 1/Mho as a cathode resistance (which it is). Then the gain is all the stuff in the output (parallel) /OVER/ all the stuff in the cathode (1/gM + any Rk).

Thanks PRR, more ways to look at the subject there. Man you guys know your stuff!!!!! This is heady material when you start getting into this realm!  

See this is the difference between someone who just knows a little about amps, ie. me, and someone who knows the maths and nuts and bolts behind it all.

I hope to get up there eventually, but Rome wasn't built in a day, and this is a complicated subject, no doubt about it.

It's great to be able to ask questions from folks who have real knowledge!  

[HotBluePlates]

If Rk is zero, as seen here, then there is a MUCh simpler way. (total plate load) TIMES gM.

39,000 * 0.0014 = 55.7

And assuming a bypassed cathode resistor, this is always how gain of a pentode is figured (triode gain can be figured using a different method that does not use Gm).

[HotBluePlates]

Well, there are several points in the circuit where response at/below 100Hz is rolled off. So it doesn't seem worthwhile to extend the response downward.

Given it's a small amp which the user is likely to crank for output tube, and maybe preamp, distortion, it seems like reducing the treble repsonse to 10kHz would be handy to help roll off high end harshness.

But the specified OT is a 15w Hammond SE transformer; use of output tubes smaller than 6L6/EL34 will probably result in less power throughput and somewhat wider bandwidth than spec.

[PRR]

> realized there were no equal signs in the equation much less the article.  Both the subtraction sign and equal signs looked the same.   Ouch!!

??? Clear here. Dunno if it could be your glasses or your PDF reader?

Page 11 is cited. There's no subtraction here. The equals sign in the top equation is clear to me.

The page before has subtraction and they do not look the same as the equals.

Here's GIF screen-grab:

[PRR]

> For those of you familar with fortran equations.
top =1/(1/(70*1000)+1/(100*1000)+1/(1.2*1000000))
bottom = 1/(1400*(1/1000000)+0.0
gain =top/bottom

Thanks. That's good even without a FORTRAN background. You compute "top", "bottom", then divide for "gain".

I can do it on the sci-calc, but when I tried to set-down the keystrokes I got lost. And indeed I would often work similar flippy-fractions as top, then bottom, then top/bottom.

[frus]

I usually do bottom, then 1/x, then top. it's somehow harder to make a mistake that way 

[DummyLoad]

this may help?

Please Excuse My Dear Aunt Sally.

Parenthesis
Exponents
Mult.
Div.
Add.
Subtr.

--DL

[Willabe]

Please Excuse My Dear Aunt Sally.

Huh? 

Oh! I just got it.


                  Brad     

[Gary_S]

this may help?

Please Excuse My Dear Aunt Sally.

Parenthesis
Exponents
Mult.
Div.
Add.
Subtr.

--DL

Thanks, that's a cool way of remembering. Great stuff guys in this thread!  some day i hope to have some of this kind of knowledge. I suspect that most of you guys have been studying for a long time!

Do you know the British version?....  BODMAS which stands for:

[HotBluePlates]

2.  Range of o/t, low Hz important to me, and where I play music.  I play bass, both upright, and electric, so having the fundamentals at 40Hz, is important.  If I had a 5 string bass with a B0, then 30 Hz.  I referenced the cut sheet for the 100 Hz cut off.  I regularly use a mixing board (SS) that allows a 100Hz filter to be used on each channel.  Guitars sound tinny, when the filter is selected.  Might as well play a ukulele.

Your question was different:

... the schematic for this project indicates the O/T range is 100 to 15K Hz, Would the gurus prefer an 0/T with a wider range for a guitar amp?

There's no one-size fits all. This amp has intentional low-end roll-offs to keep heavy distortion from turning into a muddy mess.

As I said in another thread, I just built one amp that has a 25w output stage, but an 8lb OT that's rated for 30Hz-30kHz at 60w of output; it conceivably has even wider bandwidth at 25w. But that amp doesn't have the same sonic goals as the referenced project.

It's also worth noting that what sounds good for a solo guitar sound (especially in recording) may be mud in an ensemble setting. So again, no one size fits all.

[HotBluePlates]

I go back to the comment regarding 100Hz to 15Hz.   The low E on the Guitar is about 80Hz.  So, go back to my question,  how much compromising is made when you give up the bottom end of a guitar.

It's a hill, not a brick wall. There is still response at 80Hz. You even get output at 20Hz. However, each will be a little or a lot below mid-band level.

Play some music through an EQ (even the eq on your computer's audio, if you also listen with decent headphones). Reduce the 100 or 150Hz EQ band about 2-3dB (which ain't much). Since a single RC will roll off at 6dB per octave, Set the 50 or 75Hz at -9dB (better if you can land on 50Hz and 100Hz, as those are the ones that apply to the guitar amp example).

The -3dB at 100Hz will clear things up but not result in a gutless amp.

An amp that produces heavy distortion has to strip out a lot of low end to avoid a nasty rolling, muddy distortion due to intermodulation. I very much dislike a JCM800 clean and don't like Celestion Vintage 30's (because both sound harsh and bright when played cleanly), but both sound great for high-gain metal. The reduced low end keeps the distortion tight and clears sonic space for the bass player.

There's no one size fits all.

I purposely avoided comments regarding speakers, which is a totally different subject.

The question was about bandwidth. Speaker bandwidth and OT bandwidth may share some aspects in common, namely the reasons for choosing a speaker or OT that imparts a sonic signature. Whether that signature is desirable to a player depends on the context. There's more than one right answer.

[Willabe]

Yeah but that's with a SE OT, wouldn't be as bad with a PP OT.


            Brad