Help Me out with some calcs please!

[Gary_S]

Hi guys, been studying the 5F6A Bassman schematic that i'll put on as an attachment.

I've been trying to anaylse the voltage drops across some of the parts of the circuit and i can get the figures using the voltahes fender have put on the schematic but am finding something a bit confusing regarding parallel resistors.

The part i'm referring to is the output from the PI. Now if i'm reading this correctly the voltage from the filter cap of 385v is being impressed on the junction of the two resistors, the 82k and the 100k one. Ok so if i take the voltage of 235/82k i should get the current through that resistor. Same with the other side if i take the 230/100k i should get the current through that side. From what i've read aren't those two resistors in parallel though? which would mean 45k of total resistance there? If my calculation was right!

Anyway my real question is what's the best way of calculating that voltage drop on the 82k and the 100k if the 235 and 230v were not written on the chart? Voltage drop is I*R but when i do the math with the 385 as the voltage i'm not getting the right answer.

It's obviously ohm's law but i can't nail it down exactly.

Schematic attached, ignore the circuit diagram of the board layout along with it, unless you want to look at that as well. Quite educational following the schematic and studying the board layout as well for me though to see all the connections.

[Ed_Chambley]

I think it would be a voltage divider.  The ability to determine the volts in the circuit would depend on knowing exactly how much loading you would have and how the voltages are reduced.  I am sure there is a way, but you would have to know much more about transformers and how loading of that specific transformer reacts to load.

Not sure if you could get an exact number, but don't take my word for it.  I am sure someone will come by and let us know.

[Gary_S]

Thanks Ed appreciate it. Yeah i'm just studying the circuit and practicing up on applying Ohm's law to work out things voltage wise, just for practice more than anything.

Using the voltages at the plates of the PI written on the schematic then i can work it out ok but i was just wondering if there was a way to do it without that being stamped on the schematic? I was wondering if there was a way to work it out with the 385v and the two resistor values? 

I know about Ohm's law and it's variants but i'm pretty new to analysing a circuit with it with all it's twists and turns.   

[tubeswell]

if i take the voltage of 235/82k i should get the current through that resistor. Same with the other side if i take the 230/100k i should get the current through that side.

You are dealing with current that is flowing through a triode stage in each case. So at idle with those voltages, you have (385V-235V)/82,000R = 1.83mA through the LTP's input (inverting) triode, and (385V-230V)/100,000R = 1.55mA through the LTP's 'follower' (non-inverting) triode.

From what i've read aren't those two resistors in parallel though? which would mean 45k of total resistance there?

No that's wrong. Each plate resistor is part of a different gain stage. The plate resistors are only connected at one end and they are each conducting separate amounts of current.

Anyway my real question is what's the best way of calculating that voltage drop on the 82k and the 100k if the 235 and 230v were not written on the chart? Voltage drop is I*R but when i do the math with the 385 as the voltage i'm not getting the right answer.

It's obviously ohm's law but i can't nail it down exactly.

Do what I did above. V/R=I (Voltage divided by resistance = current). You measure the voltage drop across each resistor and then divide that by the resistance of each resistor.

[Gary_S]

Thanks tubeswell that helps me out and explains things.

[tubeswell]

I re-edited my calcs just after I posted. (Its 5:47 am here and I only just woke up before going to work)

[RicharD]

The 82k and 100k are the plate resistors for the 2 halves of the PI.  Those 2 circuits are in parallel, not the resistors themselves so they don't really affect each other for this calculation.  Starting at the top (B+), the plate resistor is in series with the internal plate resistance of the 12AX7 & the cathode resistor.  We can ballpark the plate resistance of a 12AX7 at 62500 ohms.  This will change depending upon the biasing of the tube but we're estimating here.  So... 82k + 62.5k + the cathode resistance.  There's 2 tubes in parallel hanging off the cathode so in this case, double RK and call it 21K.  We get 165.5k  Divide B+ by this and we get about 2mA.  Multiply 82k * 2mA and you get 165volts across that resistor.  The schematic shows 150 volts.  Pretty close for using a ball park figure for Rp.  You could take the voltages from the drawing, figure out expected current, and derive the equation to finger out what Rp actually is.

[sluckey]

From what i've read aren't those two resistors in parallel though? which would mean 45k of total resistance there?

They are not in parallel.

Anyway my real question is what's the best way of calculating that voltage drop on the 82k and the 100k if the 235 and 230v were not written on the chart? Voltage drop is I*R

As you say E=IR. You must know the current through that resistor to calculate the voltage across it.

There is a way to approximate the current flowing thru those two resistors. If you look closely you'll see there is 1.5V dropped across the 470Ω cathode resistor. That means the current flowing thru that 470Ω resistor is 32ma. That 32ma splits and flows thru each triode and then thru the plate load resistors. Assuming the PI is perfectly matched would mean half of the 32ma flows thru each triode plate. IOW, 16ma flows thru each plate resistor. 16ma is not far from the calculated current values using Fender's posted voltages. Now you can calculate the voltage drop across each resistor.

[PRR]

> 1.5V dropped across the 470Ω cathode resistor. That means the current flowing thru that 470Ω resistor is 32ma.

So hot and sweaty today your decimal points are sliding off.

3 point 2. (Likewise 1.6 below.)

[PRR]

> We can ballpark the plate resistance of a 12AX7 at 62500 ohms.  This will change depending upon the biasing of the tube but we're estimating here

60K is nearly the lowest resistance of 12AX7. Audio swings both ways. So we usually have to (as you imply) bias for a higher resistance, so as we can swing the resistance up and down from there.

A possible first crack guess is double the show-off resistance. 120K.

In this case we find 200V across each triode, 1.6mA through, 125K effective DC resistance.

[sluckey]

your decimal points are sliding off

The decimal points were fine. My eyeballs need adjusting! Thanks for keeping me honest.

[Gary_S]

Thanks guys for the replies. I appreciate all your help.

Anyone know why the two filter caps on the far right don't have any voltages marked on those?

[Gary_S]

Here i am back again with another question about calculation figures from Merlin's books.

Ok this one is from Merlin's other book "Designing Valve Preamps"  On page 24 when talking about the slope of the DC AC load lines he has the calculation -1/Ra for the DC line and -1/RaRl in parallel (two resistors in the second calculation in parallel, just to be clear)

Anyway my question is where's the -1 coming from in the calculations? I've looked and read and re-read and can't see it!

  Gary

[PRR]

The tube graph is drawn with Tube Voltage.

We want Resistor Voltage.

Assuming B+ is fixed, then resistor is (B+)-(tube). That's where the neg sign came from.

Why a 1/ in the slope? I'd have to look in a geo/trig book.

[Gary_S]

Thanks PRR, appreciate the reply mate.

[Gary_S]

Here's another query i have regarding voltage both positive and negative. I have a hard time envisaging +v as compared to -v. I mean ok if we measure a positive voltage then it's in relation to ground and vice versa a negative voltage. But electrons that constitute the flow of current are always a negative charge you don't get positive electrons!

This came into my mind recently and i was thinking about current and voltage and what constitutes them when i was studying about semi conductors. Now the flow of current in a semi conductor is different; you have electrons with a negative charge and then you have holes that constitute a positive charge. After reading that description that actually made a lot of sense to me; the positive charge in a semi conductor is due to a preponderance of positively charged holes rather than negatively charged electrons.

But this flow of positively charged holes only applies, as far as i can tell, in the case of semi conductors.

So what actually physically constitutes a positive voltage of say 300v? Voltage is I*R and the electrons are all negatively charged so you'd think that it'd be a negative voltage.

Sorry for the long winded explanation but + and - voltage always kinda confuses me at times. As i say above i find it hard to get my head around a +v resulting from the flow of - electrons.

In the long run it doesn't really make a difference but i'd like to just get it clearer in my mind.

In a way it reminds me of the confusion (for me) of conventional current flow v's real electron flow. Electrons are negatively charged and flow to a positive anode for instance yet proponents of the conventional current idea say it flows from anode to cathode, but it doesn't! As i said above about the semi conductor idea; that is easier to visualize because a + charge is a hole and it can flow to a - terminal influencing a change in charge but in other components i don't think that applies.

Most people read and just accept it is what it is, but i always like to be able to fully understand why it is that way and why it works!

Edit: Went back to revising the material to do with the atom in the NEETS material. I think this has answered my question; a positive voltage is caused by a preponderance of positive ions requiring an electron/s, a negative voltage would be a majority of negative ions with excess electrons.

Great material the NEETS stuff! a lot of cool knowledge in there.   

[PRR]

+ or - is just arbitrary bookkeeping.

You have a checking account. You probably keep track this way:

+$90 pay
-$30 gas
-$20 food
-$50 rent

In effect you are tracking "wealth". Or "money in". And (as usual) you end up with negative wealth (-$10).

You could instead label the checkbook "Debt", "money out":

-$90 pay
+$30 gas
+$20 food
+$50 rent

Now you have +positive+ $10 of debt. And in some real way, this *is* positive: you are using someone else's money.

Or from a different view: your rent is money-out for you, money-in for your landlord. Same dollar, same absolute direction, different sign because different relative direction (from you, to landlord).

So it's just bookkeeping. You have equal but opposite effects, you use +/- symbols to keep it straight.

[Gary_S]

"I know a few guys that play with or have played with positive electrons", I even saw the cloud patterns of a few when I was in college. they work at Los Alamos, and Sandia National Labs.  all physicists.  We know those positive electrons as positrons. 

^^  Thanks guys appreciate the help.

[Gary_S]

This is a thing i see coming up again and again the more i study amp circuits and electronics in general. It can be very confusing, especially to someone who's beginning.

Conventional current is an ok way to look at it, from that + to - slant. But vacuum tubes seem harder to picture that on, as we're thinking more of the passing of electrons from cathode to anode ie. negative charged electrons going to the positive anode.

To picture it the other way around seems strange although in certain areas it's probably more convenient to think in terms of conventional current. 

[PRR]

> the passing of electrons from cathode to anode

Have you ever SEEN an electron? Passing cathode to anode?

Tube current flows from B+ to plate to cathode to B-. That's the way it was drawn by old-timers. Actual electrons are beneath our notice, a physicist's fantasy.

Does water flow down from St Louis to New Orleans? Or does dryness flow up from NO to SL? I was out there one year when crops were drowning and the flow of dryness was a major concern.

If you keep your + and - straight, normal pos/neg-number arithmetic gives you the right answers.

[Gary_S]

> the passing of electrons from cathode to anode

Have you ever SEEN an electron? Passing cathode to anode?

Tube current flows from B+ to plate to cathode to B-. That's the way it was drawn by old-timers. Actual electrons are beneath our notice, a physicist's fantasy.

No, it doesn't. Tube current is electrons boiling off the cathode and going to the anode/plate of the tube. Is it not pretty well established that the old timers were wrong in their assumption of current direction? It's the opposite way.

This is what i mean about people being given conflicting/opposing opinions about conventional current/electron flow.

In an amp particularly when we're dealing with what's going on inside a tube it makes no sense current flowing from anode to cathode. Electrons are negative charges and won't flow to the cathode unless we need to redefine Physics and say that like charges attract one another

[PRR]

> Tube current is electrons boiling off the cathode

So "they" say, based on evidence like...

> those fuzzy pictures of tungsten atoms first hand

I don't accept or deny UFOs or BigFoot looking at the fuzzy pictures. Maybe the "tungsten atoms" shot is an unfocused snap of a plate of peas?

> the old timers were wrong in their assumption of current direction?

No. Current is a mathematical quantity. No common-experience experiment can say what the charges are or which way they go. When electrons were discovered/invented, it was quickly realized that they are NEGATIVE charges. If negative charges flow "up" then "current" flows "down". Since there ARE positive charges that flow, there is no contradiction.

> inside a tube it makes no sense current flowing from anode to cathode.

Until I can see the electrons, it makes no sense to watch the electrons, or worry about what kind of charges do my work. If I keep my signs straight, the sums come out right.

BTW, I have *done* the Millikan experiment which "weighs an electron". Long afternoon. The only reason I believe I was "weighing" some negative charge is by correctly accounting for the polarity of the electric field I applied and the polarity of gravity. (Either or both could be reversed in an anti-matter universe... there is a real question if it is either OR both.)

[Gary_S]

PRR Current may be a mathematical concept but that's of no use to the vast majority of people on a forum such as this. We're guitar players primarily or amp builders we need to see things as they actually are in the most convenient easy to understand concept as possible.

Current, being electrons, does not flow from positive to negative, that's impossible. If you're referring to current as meaning something else than electron travel then define what you mean more clearly as there seem to be a lot of assumptions in your post. In the real world current is a flow of electrons and electrons do not flow from positive to negative. As i said before if they did we'd be better redefining the whole realm of Physics. Current can flow the other way if you're viewing it from a semi conductor standpoint where holes with a positive charge can actually move.

What positive charges flow in a vacuum tube?

[Gary_S]

Hi guys another question from Merlin's preamp book: on page 48 the drawing at the bottom of the page of the pentode gain stage. How do you work out the voltage drop on the two resistors Rg2 and Ra?

So it looks like 255v going into a parallel connection with Rg2 1M and Ra 220k. Each branch of a parallel circuit see's the same voltage so that'd be 255v going to each resistor initially. I tried to get the current going through each resistor by using I= E/R which would have been 255/1M and 255/220k in both cases.

Anyway this doesn't give me the voltage dropped on each resistor. According to the book; on the far side of Rg2 there's 99v and on the far side of Ra there's 81v. So subtract those from the 255 total and you get the voltage drop on each.

[sluckey]

How do you work out the voltage drop on the two resistors Rg2 and Ra?

The voltage on each side of the resistors is given. Just subtract the voltage on each side of the resistors to get the voltage drop.

I tried to get the current going through each resistor by using I= E/R which would have been 255/1M and 255/220k in both cases.

You're using the wrong value for the voltage. You must use the voltage drop ACROSS a resistor to calculate the current thru a resistor. Just knowing the voltage on one side of the resistor is not good enough. The voltage drop across those resistors is 255-99=156 for Rg2 and 255-81=174 for Ra.

[Gary_S]

You're using the wrong value for the voltage. You must use the voltage drop ACROSS a resistor to calculate the current thru a resistor. Just knowing the voltage on one side of the resistor is not good enough. The voltage drop across those resistors is 255-99=156 for Rg2 and 255-81=174 for Ra.

Thanks Sluckey. So without having that other voltage on the far side of the resistor there's no other way to work that out with just the other values that are given? I thought that must be the case but wanted to check with you guys on here to clarify.

The only thing that confused me about that was; i study the NEETS electronic stuff and there's a circuit in there where there's a power supply giving out 60v then a resistor in series with that, then that go's to two resistors in parallel then back to the other side of the power supply. Now this was an example circuit and they worked out how many volts were dropped on the two resistors in parallel being given nothing but the power supply voltage and the resistor values. That's what confused me about this. I said "why can't i work out the voltage drop here when i'm basically doing the same thing?"

[sluckey]

I said "why can't i work out the voltage drop here when i'm basically doing the same thing?"

It's a similar problem but not the same. In your example from Merlin you are provided with the voltage on each side of both resistors and you ask how to calculate the voltage drop across each resistor. That's all the info you need to calculate the voltage drop across each resistor. Just subtract. No need to calculate the current.

[Gary_S]

Thanks Sluckey appreciate the advice and help. 

[PRR]

> Each branch of a parallel circuit see's the same voltage

IF they go to the same places, both ends.

I can't see the picture, but one goes to Plate and the other goes to G2, right? Then there is some stuff (electrons?) happening inside, possibly *different* for Plate and G2 (since you *can* see they are different sizes and distances from the cathode where "the current comes out").

Merlin probably covers this better than I care to.

But the first crack is to find the tube's data-sheet and estimate the P/G2 current ratio. For most small pentodes, for most voltages, G2 current is about 1/5th of Plate current. (We would like G2 to draw no current, but it does, but the Tube Designer makes it modest.) Power tubes, even 1/5th is a bunch, more like 1/10th. A few high-gain tubes, G2 is 1/3rd of Plate. Every designer had different pressures from his boss and customers. So you gotta look at typical data.

That and the resistors will say the approximate *relative* drops. But this does not tell you the TOTAL cathode current. You need G1 voltage *and* G2 voltage for that. (Plate voltage has very little effect on cathode current.)

You want another tip? Don't try to design, even understand, Pentode biasing. It can be done, it WAS done, for you, 50+ years ago. Find the RC amplifier tables, your load (use 200K unless you know better), use those values, play a lot, THEN tamper values to get different clipping action.

[Gary_S]

You want another tip? Don't try to design, even understand, Pentode biasing. It can be done, it WAS done, for you, 50+ years ago. Find the RC amplifier tables, your load (use 200K unless you know better), use those values, play a lot, THEN tamper values to get different clipping action.

Thanks PRR i appreciate it. Yeah sometimes trying to really understand what's goin' on there in the small signal pentodes is damn hard work. 

I'll never understand the nuclear physics that are goin' on at the deepest most complicated level i just like to understand the basics of it so i have a handle on it.

I think to totally understand everything to the nth degree you'd have to have a serious background in Physics.

But maybe that's not totally necessary in guitar amps. Get a baseline and then work from there.

A lot of folks use an already existing amp circuit eg. 5F6A Bassman but tweak the values around to change it to what they want.

[Gary_S]

I have another question. This is regarding voltage gain of say a triode for instance; Now a resistor restricts the flow of current so i'm having a hard time reconciling that with the fact that a larger plate load resistor will give you more voltage gain. I'd have thought it was the other way around and a smaller resistor would give you more gain because it isn't restricting the flow of current as much.

I know a larger value gives you more possible output voltage swing and that's how it happens, but it's kinda weird trying to envisage it.

[sluckey]

What's the question?

Ohm's Law says E = I * R, so...

    1mA of current flowing thru a 100K plate resistor causes a 100v drop across that resistor.
    1mA of current flowing thru a 220K plate resistor causes a 220v drop across that resistor.

[Gary_S]

What's the question?

Ohm's Law says E = I * R, so...

    1mA of current flowing thru a 100K plate resistor causes a 100v drop across that resistor.
    1mA of current flowing thru a 220K plate resistor causes a 220v drop across that resistor.

Ok so that's a voltage drop across the resistor as in 100v or 220v getting eaten up by the resistor. But how does that work with the triode having more gain? i just can't picture that. I know it's correct and everything but it's hard to picture in the minds eye.

In order for the triode to have more gain i thought one way was to increase the voltage on the anode/plate but can't see how you're doing that with the bigger resistor because you're dropping more volts than with a smaller one.

It's kinda hard to describe what i'm meaning.

[sluckey]

Gain = V_out / V_in

Suppose the voltage on the grid is 10V (that's V_in). And that voltage caused 1mA to flow thru the tube. That 1mA will cause a 100V drop across the 100K plate resistor (that's V_out. Therefore gain is 100/10=10.

Now repeat the same exercise above, but using a 220K plate resistor. Now gain is 220/10=22.

[Gary_S]

Thanks Sluckey it makes sense now, appreciate the help man.