Tech Question From Merlin's Book

[Gary_S]

Ok got Merlin's book and it's great but being a relative novice there's a concept i can't get my head around and i was hoping some of you guys could help me with it.

In the first chapter when we're talking about the amplification in the tube it says that when there's no anode (plate) current flowing in the tube then no voltage can be dropped off through the plate load resistor so the full HT reaches the anode.

Then it says imagine the tube is a short circuit and then the full HT would be dropped across the resistor and none across the tube.

So i thought that no matter what; the resistor would always lower the voltage going to the tube, depending on the value of the resistor. In the first case above how does the resistor not block a certain amount of the voltage getting to the anode? How is it possible that the full signal get's through? I thought every time voltage went through a resistor then the voltage coming out the other side would be attenuated. But in this situation it isn't.

It's obviously down to Ohm's law, which because there's no current flowing, means the resistor doesn't drop the value of the voltage but instead the whole HT get's through.

But i am trying to think of a way to get this set in my mind

[sluckey]

Well you already stated why. Maybe the best way to make yourself believe it is a simple experiment. Connect a 9 volt battery, 100Ω resistor, and switch in series. Positive of battery to the resistor. Call the negative side of the battery ground just for reference.Connect your voltmeter across the switch. What voltage do you read with the switch open? Why? What voltage do you read with the switch closed? Why?

[tubeswell]

If no current is flowing through the tube, then there isn't any plate current either, therefore there is no current flowing through the plate resistor either, so the plate side of the plate resistor would be 'resting' at the same voltage as the HT node (which the other side of the plate resistor is connected to). So the plate would also be sitting at the same (HT) voltage (because there is no voltage being dropped across the plate resistor) (because there is no current through the plate resistor). That is where you get the X-axis (horizontal axis) end of the load line.

Now on the other hand, if the tube were able to conduct at maximum current (i.e.: the point at which it couldn't possibly conduct any more current), then there would be no voltage between the plate and the cathode (because they would be at the same voltage) and (assuming - for a minute - that the cathode is at ground potential) all the current would be going through the plate resistor. So the plate resistor would be dropping the full voltage from HT to the cathode (ground potential). So that is at the '0V' end of the x-axis (where the X-axis intersects with the Y-axis). Then you simply measure the voltage drop across the plate resistor, and divide that by the plate resistor's resistance in order to get the current through the plate resistor, which you then plot on the Y-axis.

[Gary_S]

Thanks guys for the replies. Yes it makes sense but it's hard to keep track of it all when you're fairly new at it.

I just always thought that when you have your plate load resistor, say a 100K one like i have on my Marshall 4010, that sends the HT voltage to the plate and the voltage all went through the resistor and then so much was dropped on the resistor and the attenuated voltage then went to the plate. Now i'm finding out that in certain circumstances that doesn't happen as simply as that.

Well i'm learning guys. I've already learnt quite a lot just reading the first two chapters of the book. I'm glad i bought it. This is what i wanted with the amps; to learn the nuts and bolts of why everything is where it is and what it does. Other books and videos just tell you to replace this capacitor or this resistor without telling you technically why you're doing it. I find that frustrating as i like to know how things work the way they do.

It's not easy going, but then again nothing worth having is easy.

[HotBluePlates]

The problem you're having is you're trying to relate "normal operation" to the extremes Merlin is describing.

The extremes Merlin presents are useful to define the ends of a loadline when you draw it on a graph of the tube's plate curves.

It also presents a possible condition you'd see under extreme faults. If the tube plate were shorted to ground (by a piece of loose wire in the chassis perhaps), then that tube's plate voltage would be 0v, because it is connected to ground. You'd never see that in real life outside of a short-circuit, which might then draw enough current through the plate resistor to burn it up.

How the extreme can help you is if you measure a preamp tube's plate voltage, and find it is say 50vdc. If it is normally 210vdc, and you probe the B+ node supplying that tube and find it is at its normal voltage, you can assume the tube is drawing excessive current for some reason, which has dropped more than the normal amount of supply voltage across the plate load, and left relatively little at the plate.

Same for the no-current situation. This might help you in a new build, and you can't figure out why the amp is making no sound. You measure the plate voltage of all tubes, and find your first preamp stage (which should be the 210vdc mentioned above) is really at 320vdc. You measure the B+ node feeding that tube and also find 320vdc. Now you know that no voltage is being dropped across the plate load resistor, so there is no current through it, implying no current through that tube.

In the case above you ask yourself why there's no current, and determine that if the cathode resistor is missing (or has a missing wire to the tube cathode pin or maybe no solder joint) or if the tube's heater is not warmed up, then the tube will pass no current. You check the cathode resistor, find that it is present and the correct value, but you didn't run heater wires to the socket. You run heater wires to the socket and everything returns to normal, and the amp plays.

[HotBluePlates]

Ohm's Law:

When I first learned ohm's law, we were required to memorize a phrase. I was in a Navy tech course, and they actually made us write the phrase 100 times and turn it in as homework.

You might remember ohm's law the easy way (which I do most of the time) as Voltage = Current * Resistance. It's seems easier to start with this form and manipulate it to solve for whichever of the 3 you don't know.

We had to memorize it as:

"The current in a circuit is directly proportional to the voltage across the circuit and inversely proportional to the resistance of the circuit."

Which you can write as Current =  Voltage / Resistance

When trying to equate all this to the amp circuit you're looking at, it is helpful to think of the plate load resistor and the tube as 2 resistors in series, with the plate output being the junction of the two resistors. The resistor representing the tube is a variable resistor, and the input signal it receives controls whether it looks like a big resistance or a small resistance (less or more current, respectively).

So the output of a tube is really only the result of how the tube and plate load divide up the supply voltage between them on the basis of tube current.

[Ritchie200]

HBP - GREAT explanation!  Thank you!

Jim

[sluckey]

"The voltage [dropped across a resistance] is directly proportional to current in the circuit and inversely proportional to the resistance of the circuit."

Which you can write as Voltage = Current / Resistance

WRONG!

E = IR  means voltage is directly proportional to current and voltage is directly proportional to R. Increasing either I or R (while holding the other constant) will increase voltage.

[Gary_S]

Ohm's Law:

When I first learned ohm's law, we were required to memorize a phrase. I was in a Navy tech course, and they actually made us write the phrase 100 times and turn it in as homework.

You might remember ohm's law the easy way (which I do most of the time) as Voltage = Current * Resistance. It's seems easier to start with this form and manipulate it to solve for whichever of the 3 you don't know.

We had to memorize it as:

"The voltage [dropped across a resistance] is directly proportional to current in the circuit and inversely proportional to the resistance of the circuit."

Which you can write as Voltage = Current / Resistance

So if there's no current, there is no voltage dropped across that resistance.

When trying to equate all this to the amp circuit you're looking at, it is helpful to think of the plate load resistor and the tube as 2 resistors in series, with the plate output being the junction of the two resistors. The resistor representing the tube is a variable resistor, and the input signal it receives controls whether it looks like a big resistance or a small resistance (less or more current, respectively).

So the output of a tube is really only the result of how the tube and plate load divide up the supply voltage between them on the basis of tube current.

Thanks so much for this HBP this is great, and gives me a whole different view on it. I just memorised ohms law as (voltage = current . resistance) and obviously you can manipulate it the other ways as well, depending on what you're measuring for. But that is a great thing you learnt in your navy tech class; it seems to clarify what's happening more accurately.

I just had a hard time visualising how the situation of the full HT ending on the plate could happen, seeing as the HT was going through a resistor on it's way to the plate. I thought "how is it possible for a voltage to go through a resistive component and still have the same voltage on the opposite end after having gone through the resistor!"

It's pretty complicated at the start, trying to completely understand the process. But your explanation helps a lot. I'm gonna keep memorising that slogan you had to learn to imprint it on my subconscious.

Thanks a lot for all the replies here. This is what i wanted; to get into the nuts and bolts of why things are working the way they do. Merlin's book and the advice i get on here has opened up a whole new level of knowledge and ideas. Most books you read don't really explain what the circuit is doing technically and how the three things; voltage, current and resistance are changing in response to each other. Most books just tell you to replace this with this etc.

[HotBluePlates]

"The voltage [dropped across a resistance] is directly proportional to current in the circuit and inversely proportional to the resistance of the circuit."

Which you can write as Voltage = Current / Resistance

WRONG!

Thanks! I must've been loopy from the cold medication!

I meant Current = Voltage / Resistance, and corrected that in my post above.

Obviously, I should've written the phrase 200 times.

[sluckey]

I knew you didn't believe that.

I had to memorize that same statement (no writing exercise though) early on in tech school. My instructor was retired Navy. This is probably the page where that statement originated...

[Gary_S]

I think i should have titled this thread as an ongoing thing like "Tech questions from Merlin's book!" rather than question.

Anyway i feel i'm learning so much from the first chapter alone! there's so much detail in this book.

I'm confused about something that's pretty simple, i think: i always thought that when the HT supply sent the DC voltage through the plate load resistor that the voltage gain of the tube would be lower with a bigger resistor.Yet with a higher value of resistor here it's increasing? I always thought that a smaller resistor in that position would let more voltage through and that would mean more gain from the stage. But it's like the inverse of this.

If i take a reading on the resistor there on the HT side before the voltage has went through the resistor would it not be a higher reading here and then it's lower on the far side after having gone through the resistor surely? But how does that equate to a higher value giving you more tube gain?

It's all to do with interpreting Ohm's law, and i must be looking at it from the wrong angle  

[HotBluePlates]

Do you know how voltage dividers work?

Get a handle on those, and how a volume control is a variable voltage divider. When you tell me you've got that idea down pat, I'll explain why tube gain appears to increase with a larger plate load resistor.

[sluckey]

It's all to do with interpreting Ohm's law, and i must be looking at it from the wrong angle  

There are no angles in Ohms's law. That comes later in AC analysis.

Now, as far as gain... Gain = Voltage out divided by Voltage in. Put 1 volt in and get 20 volts out and the gain is 20. ( Gain=E_out/E_in=20/1=20)

Hold that thought. Now consider Ohm's law and the plate load resistor. Ohm's law says E=IR, where E is voltage, I is current, and R is resistance. E is directly proportional to R. If R increases, then E must also increase. And the voltage across the plate load resistor (R) is the voltage out in the gain equation. Increasing the plate load resistor causes the voltage across it to go up which causes the gain to also go up.

IOW, gain is directly proportional to voltage out (voltage across plate resistor) and voltage across plate resistor is directly proportional to the value of the plate resistor. You're dealing with two formulas and need to know how to substitute values between the two formulas.

If I've just stirred the mud hopefully someone else will filter out what I meant to say and it will become clear to you.

[PRR]

> Do you know how voltage dividers work?

+1.

[tubeswell]

I think i should have titled this thread as an ongoing thing like "Tech questions from Merlin's book!" rather than question.

Anyway i feel i'm learning so much from the first chapter alone! there's so much detail in this book.

I'm confused about something that's pretty simple, i think: i always thought that when the HT supply sent the DC voltage through the plate load resistor that the voltage gain of the tube would be lower with a bigger resistor.Yet with a higher value of resistor here it's increasing? I always thought that a smaller resistor in that position would let more voltage through and that would mean more gain from the stage. But it's like the inverse of this.

If i take a reading on the resistor there on the HT side before the voltage has went through the resistor would it not be a higher reading here and then it's lower on the far side after having gone through the resistor surely? But how does that equate to a higher value giving you more tube gain?

It's all to do with interpreting Ohm's law, and i must be looking at it from the wrong angle  

The voltage drop across the plate resistor (i.e.: the voltage drop from HT to plate) increases if you increase the size of the resistor (all other things being equal). This gives the plate voltage more range within which to swing up and down (up to a point, because if the plate resistor is made too large, plate current will stop flowing altogether)

[catnine]

Yeah OK.

 Yes the plate load resister drops voltage but only if there is current draw. And the preamp tube has it's own resistence. I f there is no path through the tube for current to flow ( no cathode resister ) then I would expect the voltage after the plate load resister to rise in reference to ground. If the tube has all the proper connections then a low value plate load resister will increase the plates voltage. Don't you aim for a center biased tube ie in a working amp if you have the same value plate voltage and you lower the cathode resister value the tube is biased hotter sooner breakup. If you rised the plate load resister value you would have to lower the cathode resister value how much you go in either one depends on the tubes internal resistence. A vol pot voltage divider , divides the voltage off the plate between plate + and ground -. It's never a dead short because it always provides resistence before ground.

[HotBluePlates]

... Don't you aim for ...

Gary's first question about ohm's law and tube voltages deals with d.c. tube conditions. However, Gary's new question really asks about the tube's a.c. conditions and should not be confused with idle conditions.

Gary:

For a moment, forget everything you know about a triode. Imagine that you replace a preamp triode with a perfect voltage generator. The output of the generator will be equal to the applied grid voltage (vg1) times the triode's amplification factor, mu. Refer to the figure below.

now imagine that the output of this perfect generator is applied to a voltage divider, akin to a volume control. The upper leg of the volume control is the internal plate resistance of the tube, rp1. The lower leg of the voltage divider is the actual plate load resistor, RL1.

If the plate load resistor is made bigger, or the internal plate resistance of the tube is somehow made smaller (or both), this is akin to "turning up the volume" on our voltage divider.

The external plate load resistor works against the internal plate resistance of the tube to define the resulting voltage gain of the tube, which is generally less than the published amplification factor of the tube. That's because to have measured voltage gain equal to mu, the plate load resistor would have to be infinitely large, but that also require infinitely-large supply voltage.

In practice, we shoot for a plate load resistor 2-5 times bigger than the internal plate resistance of the tube at the chosen operating point. More than this results in diminishing returns unless you resort to trickery like using a solid-state device as an active plate load.

Carrying this further, pretend the coupling cap (Cc2 in the figure) is replaced with a piece of wire. Now the plate load resistor RL1 is in parallel with the grid resistor of the following stage, Rg2, which reduces the effective value of the plate load and reduces the measured voltage gain. We generally pick Rg2 to be 2-5 times bigger than the previous stage's plate load resistor to minimize its effect. There is a limit given by manufacturers for how big a grid resistor can be before leading to problems due to grid current.

A design procedure might then be pick the biggest allowable grid resistor Rg2 for the following stage. Divide that value by 4-5 and use this value for your plate load resistor. Draw a loadline with the available supply voltage for the stage, then select a first-guess likely good idle point. Calculate the internal plate resistance at this operating point, then check against the value of selected plate load resistor. Ensure the plate load is several times bigger than internal plate resistance, then use the formula to predict the gain of the stage. If everything checks out, use the chosen bias point's grid voltage and plate current to calculate the needed cathode resistor. Use a bypass cap which fully bypasses the cathode resistance and parallel tube resistance at the lowest frequency of interest to ensure you get the full gain predicted by the formula.

[Gary_S]

Thanks guys for the replies. I've been reading them all. It was just a bit confusing because i'm reading (or should i say re-reading because i've been poring over it a lot) the fundamentals of amplification bit the first chapter and studying the load line diagrams.

HBP: that helped me a lot with the diagram.

Thanks folks. It's pretty complicated, the amplification process going on, with AC signal voltages getting superimposed on to DC voltages and then working out how the tube is processing it all. It's not an easy study, it's hard work.   But it's very satisfying and enjoyable to learn.

[HotBluePlates]

... It's pretty complicated, the amplification process going on, with AC signal voltages getting superimposed on to DC voltages ...

The ocean is at sea level, which is interpreted as one specific elevation, all the time. (d.c.) We might describe the heights of mountains as so many feet above sea level.

But the sea itself has waves, sometime small, sometimes big. (a.c.)

If sea level is some steady median value as though the waves didn't exist, you might imagine the waves were fluctuations in water height above and below the median sea level. (a.c. superimposed on d.c.)

When thinking about tube audio circuits, it's usually helpful to simplify and think only in terms of sea level, and at other times only in terms of waves, as though the other didn't exist. That helps you understand each individually, and makes the process of seeing how they work together seem much less complicated.

[Gary_S]

Cheers HBP. A nice way to think of it and makes sense. I tell you reading Merlin's book has already opened my mind a lot to what's going on in those triodes. Before i started reading and studying it i didn't think it was possible to have AC and DC going to the same spot and being superimposed on one another!!!!

Think it's best to do what you say and think more in terms of DC, as thinking of the two and trying to keep track of them both is confusing at first!!!!