fender 212r

[lowell]

I'm trying to fix one of these but that's another thread. Got many Q's about this design and solid state design in general. Ok so here goes.

I'll be posting the schem later tonight as i'm on my phone right now.

First off. Can someone please explain what it means that transistors are current controlled devices?
I understand how to bias a transistor via Vbe fyi.

Q20/21 are the power devices in common collector arrangment right?
The .22 emitter resistors are ballast resistors and are there to balance the power devices better between the dual rails?
Q18/19 are the drivers?
Q16/17 are limiters? It seems as though if the signal reaches a certain point at the power devices emitters' that the limiters will turn on and via D32/33 will adjust the bias transistors Q14/15?

Let's start there. Thanks in advance for sharing knowledge.

[Enzo]

http://music-electronics-forum.com/a..._schematic.pdf

There, I lifted that link from your other thread. Hope it works.

Tubes are voltage devices. Except in certain circuits which rarely come up in guitar amps, the control grid of a tube draws no current. You vary the voltage on the grid to control the current flowing from cathode to plate. That in turn causes a voltage drop across a load resistor, which is then sampled off as a voltage and fed to the next stage. A transistor is controlled by drawing a current through its base. That then in turn controls the emitter to colector current.

Yes one transistor pushing the next makes the first one a driver. Q18 drives Q20. And if there had been a third one pushing Q18 it would have been a pre-driver.

Ballast resistors are there in part to help transistors share current with the others in parallel,, not really with those on the opposite polarity side of the amp. There are no parallel transistors in this amp, just one per polarity. But the ballast also acts like a shock absorber in my mind. The transistor has it to help limit currents. Plus it allows for easy monitoring of those current as they do in the limiter you ask about.

Yes Q16,17 are limiters. When the current through Q20 and thus R109 gets high enough, the voltage drop across that resistor allows current through the base of Q16. Q16 then conducts and tries to shunt signal away from the base of Q18. D32 is there to protect Q16 from and reverse voltage parts of the waveforms. As the signal waveform climbs more positive, once Q16 starts to turn on, any further positive increase in the waveform at base of Q18 will cause more current thrugh Q20, which will turn Q16 on harder. And that in turn will conduct harder trying to shunt the signal away from Q18 base. Q16 D32 doesn't connect directly to Q18 base, but those two small value resistors don't prevent this function.

Those limiters are not really involved with the biasinjg transistors.

[lowell]

Enzo,

A million more Q's for ya.

Jfets are voltage controlled right? Maybe I'm wrong. As I understand it a jfet is on w/ Vgs 0v. Like in a tube, the Vgs (Vgk) has to be a negative value to limit the current through it.

There's a basic electronics concept I've yet to grasp: when current flows through a resistor a voltage drop is created across said resistor. So say we have a 9v battery and connect our meter to the + and - to measure voltage. It will read 9 volts, not 0 volts. But is there current flowing? Also if you put a resistor across that battery it will read some nominal voltage, and if we reduce the resistor to a lesser value the current will increase more and the voltage will reduce more right? So that seems backwards to me. The less the resistance between a voltage and ground - the less voltage drop.

I'm not grasping the base current thing. I get that .7v has to be impressed on the base to overcome the b-e junction diode thus allowing current to flow. I just don't get how current flowing from b-e allows current to flow from c-e. Can c-e really be looked at as if being 2 diodes in series with their anodes connected at the base? If so how do 2 diodes with their anodes (or cathodes) tied together operate?

How do common collector arrangements create current gain?

Are the transistors related to the positive supply rail amplifying the entire waveform or just half as in a push-pull tube power amp?

Is C53 there to allow the Q16 limiter to be turned on more, or earlier, for higher frequencies?

Not totally clear on how D32 plays a part. Say Q16 was saturated due to a large signal and was basically shorted, would all the signal be shorted? Also, Q16 isn't "grounded" per se, its emitter is tied to the speaker out. Would not this allow the signal to still reach the speaker? Finally, what is that point in the circuit called where all the ballast resistors tie together?

[Enzo]

Jfets are voltage controlled right? Maybe I'm wrong. As I understand it a jfet is on w/ Vgs 0v. Like in a tube, the Vgs (Vgk) has to be a negative value to limit the current through it.

Yes, I should have said bipolar transistors are current devices, JFETs work a lot like tubes do. But you won;t find many of them in that FM212 power amp.

There's a basic electronics concept I've yet to grasp: when current flows through a resistor a voltage drop is created across said resistor. So say we have a 9v battery and connect our meter to the + and - to measure voltage. It will read 9 volts, not 0 volts. But is there current flowing? Also if you put a resistor across that battery it will read some nominal voltage, and if we reduce the resistor to a lesser value the current will increase more and the voltage will reduce more right? So that seems backwards to me. The less the resistance between a voltage and ground - the less voltage drop.

Two things. Your meter is measuring the voltage differential between two points - the terminals of a battery for example. An ideal meter tells you that there is 9v difference betwen those points. It wouldn;t need any current because it is just observing, any more than it makes you shorter if I hold a tape measure up next to you. But in the real world, meters are not ideal, they have an internal impedance or resistance. Your typical digital meter has an impedance of over a meg, and maybe even 10 meg ohms. And that being the case, we would find that a very tiny current would be drawn by the meter itself. But for the sake of the math, let's say the meter has a 9 meg internal resistance. SInce I = V/R, that current would be 1 millionth of an amp, one thousandth of a milliamp. We can ignore that. SOme places we can't. A very common example of that is the phase inverter stages of many amps. Measuring the voltage at its grids often comes up very low dues to "meter loading."

SO with just a meter on the battery terminals, you get that tiny little current. Add a resistor and ignore the meter impedance for now. The battery will put 9v across that resistor. If we read the voltage across the resistor, it won;t be some "nominal voltage," it will be 9v.

Well, remember the ideal meter? We have an ideal battery. The battery provides a steady 9v. A real battery has a limit to its current sourcing ability. Obviously my little 9v can;t turn over the starter of my car engine, it lacks the current capability. SO if we stick a 4 ohm resistor across a real world 9v battery, it will drag down the voltage severely. But if you want to do tghe experiments with 4000 ohm resistorws, you will not have nearly the problem with that.

SO resistor gets 9v. Meter across it reads 9v. The confusion here is that you are reading the battery with your meter. The resistor is just along for the ride. Put two resistors in series across the battery and then observe one of them and all those voltage drops fall into place. If you take a resistor and get 9v across it, then reduce the resistor value, more current flows through it, yes. But the voltage across it doesn;t change because it is connected directly to the voltage source, the battery. The lower the resistance, the more current flows, the hotter the resistor gets.

In the real world, too low a resistor will cause the 9v battery to drop to lower voltage, but that is because the battery is not ideal, it has its own internal resistance. COnsider that resistance in series with the voltage.


Say you want to vary that test resistor from 1000 ohms down to zero. If you just connect only that across the battery, then you are tetsting the battery itself. But put a 1000 ohm resistor in series betwen the battery and your test resistor. Now as you vary the resistor, the voltage across it will vary just as you expect.

Is C53 there to allow the Q16 limiter to be turned on more, or earlier, for higher frequencies?

C53 is for stability. The output is smapled and Q16 uses that to control the input. The potential is there for it to oscillate, getting into continually correcting itself. C53 kills its ability to move at super high frequencies.

Are the transistors related to the positive supply rail amplifying the entire waveform or just half as in a push-pull tube power amp?

It is pretty much class B. Each half of the circuit amplifies half the waveform. A classic failure would be an open Q21 and you'd have a missing bottom half of the waveform at the output.


I have to meet the wife at dinner right this instant. I'll come back later.

[Enzo]

A bipolar transistor TESTS like it was two diodes connected together. There are a couple of semiconductor junctions there that give a rough check of the condition of things. The thing is not really two diodes, you cannot just wire two 1n4007s together and get gain. MANY texts have been written about the basic function of a bipolar transistor. I refer you to those, rather than me just parroting them. That explains how a bipolar transistor works - pulling current through the base enables the current path from emitter to collector. Besides, I am not the theory guy around here. I preach enough of it to explain the function of circuits with troubleshooting in mind, but other members are more in touch with the physics of it.


That point where the ballast resistors join goes right to the output jack. I call it the output bus. A bus is a conductor that serves many related things. The output bus here, or though I refer to them mainly as rails, you can just as well call a power rail a power bus. ANd some circuit that runs a ground wire the length of something and runs various parts to it? That wire would be a ground bus. In equipment that needs them, there are signal busses as well.


Q16's job is the control the drive to Q18. It doesn;'t have to go to ground, it just has to go across Q18. Q18 is essentially connected between the V+ rail and the output bus. The harder we tug the base of Q18 towards V+, the more it conducts - and by extension then so does Q20. As current conduction rises through Q20, so does the drop across its ballast, which in turn controls the base of Q16.

A large signal doesn't just saturate Q16, just the parts of that signal above whatever level the circuit is designed for. Think of it like a peak light. So when the signal waveform hits a certain amplitude, Q16 starts to conduct, any additional amplitude then causes Q16 to conduct even harder. it doesn;t remove all signal from Q18 base, it just removes any signal in excess of the designed amount by clipping it off. D32 is just part of the circuit path for that action. The circuit would work without D32, but it would not be happy with reverse voltage parts.

[J M Fahey]

Clap Clap Clap Clap !!!!
Standing ovation.
Excellent explanation, well beyond the call of duty.

[lowell]

Ok I get it. Q16 controls the current through Q18 by tugging or not tugging on Q18's base. I was looking at it as if Q16 was stealing AC signal off of Q18's base. Or maybe it's doing both?

So D32 is there to stop reverse DC voltage or signal voltage? And stop it from reaching what parts?

[Enzo]

Q16.

The early parts of the power amp shove the output section up and down at pretty much full amplitude. The drivers and outputs amplify the current, not the voltage. SO to get the output down to -40v, the bases have to go down there too. Q16 can wind up with a collector more negatiove than the emitter.

The base of Q18 sits at about +1v at idle, that is right on the virge of conduction. As the waveform comes along, the positive half starts to climb, that turns on Q18 more, which turns Q20 on more, which allows more current through the load. And through the ballast. The higher the signal waveform goes at the base Q18 the more current the output provides to the load. Until we get to where Q16 starts to respond. Now if the input waveform TRIES to go even higher, Q16 will conduct harder to prevent it. As the waveform falls back down to more reasonable levels, Q16 relaxes its grip until it drops out. Note that without a load the amp can have the waveform all the way up to the power rail and no one cares. Without the load, no current flows through th ballast, so there will be no limiting action by Q16. The load determines the output current.

[Enzo]

A little story. I grew up on tubes, I learned my electronics in the 1950s. Later in life, I found myself working in coin op industry, and things were mechanical. I got really good at reading relay logic. Might take me a day to get used to it again, but i am sure i could get myself right back at that. PInballs, jukeboxes, etc, all chock full of relays. SOmewhere along about 1975, I got involved in video arcade serviceing logic boards, so I had to learn TTL logic, and then microprocessors happened. I learned TTL and CMOS logic and got pretty good at that. But I had skipped over transistors entirely, I had just the crudest understanding of them.

troubleshooting is troubleshooting, so I got along OK, but eventually i had to go back and learn about transistors. FIll in my blank. Over the years I suppose I have filled in those blanks reasonably well, but I admit not as well as it may appear here. Got my weak spots. But because i learned a lot of troubleshooting on relay logic and then TTL logic (ICs) I to this day look at transistors from a logic point of view. The metaphor seems to work for me so I see no reason to change it. SO that is why I tend to describe things as pulling the base towards the collector turns on the transistor. It is really just adapting logic ICs where we pull an input high and watch an output go low, or whatever. When I pull the base of a transistor up, it conducts harder which pulls its collector down. Just as making a 12AX7 conduct harder pulls its plate down. The basic transistor stage is an inverter in my mind.


SO when I grab the base Q18 and try to tug it higher, Q16 reaches up and grabs my wrist and holds me back. Q16 leaves me alone until I get to its threshold, but once it starts, then the harder I try to tug upwards, the harder Q16 pulls my wrist down and we wind up stalemated at the threshold level. I don;t in my mind visualize it as siphoning off signal or upsetting the DC applecart. The signal IS a varying DC at that point anyway, so I don'[t even think about it.

At this point it is a zen thing, and a lot of time I have to think about what I am thinking about to know what I am thinking... if you get my drift.

[lowell]

Does the positive rail side of the outputs amplify the positive half of the AC cycle? I'm gathering that that is the case. That being the case, how are the outputs on the positive side of the power amp going down to -40v? Shouldn't they go from 0v to +40v and back down to 0v and then cut off for half of the cycle?

[Steve Conner]

Yes. The positive and negative halves of the output stage are connected together to the speaker. There is only one output terminal, hence there can only be one output voltage.

On the negative half-cycle of the signal voltage, the positive half of the output stage isn't passing any current, but because it's connected to the negative half, it has to follow it down towards the negative rail.

Think of it like an engine with two cylinders. They don't fire at once, they take it in turns to generate power. But because both pistons are connected to the same crankshaft, they both move all the time.

It's also exactly the same as a push-pull tube amp, just one half of it is "turned upside down", so there's no need for a PI or an output transformer. You can't do the turning-upside-down trick with tubes, because you don't get "PNP" tubes. (I suspect they would have to be made of antimatter and handled VERY carefully.)

Drawing one of the tubes in the output stage upside-down, like those old Fender schematics, doesn't count: it only confuses things. When drawing schematics out, you should always draw everything with the more positive voltages towards the top of the page. Then you can visualize voltage as vertical position, just like on your scope screen, imagine parts of the circuit physically hopping up and down as voltages change with time, current flowing like a waterfall down the page, whatever works for you.

Enzo: Transistors behave just the same as tubes, you can neglect the differences to a first approximation. Well, except transistors work fine without the heater voltage. They are not current-controlled devices, this is a myth to pacify newbies. The Vbe is what controls the collector current, just like the grid voltage in a tube controls the plate current, but with immensely higher gain than any tube, so it only takes a tiny change of Vbe either side of 0.7V to make a big signal at the collector.

Base current is an error that transistor designers would do away with if they could.

And, you'll never guess what TTL logic chips were made of inside.

[Enzo]

Actually I don;t have to guess what is inside TTL chips. Point was that learning about them as logical elements instead of analog things provides a different way to look at circuits. Most guys don;t look at a 12AX7 triode and thing in terms iof pullups and pull downs.


I recommend against looking at transistors as tube analogs because the natural tendency becomes looking for the exact parallel, element by element on the part of the novice. However many parallels there are, then the differences that DO exist get ignored. And also especially in the power amp like we discuss here, we see the output driven by the driver, driven by the predriver, driven by maybe even a pre-pre-driver to get current through the speaker. In the tube amp we just have to ramp up the signal voltage to get the power tube to work, then the transformer "transforms" the signal to high current.

For example, that 1.5k grid stopper resistor on some 6L6 tube does a fundamentally different job from a 1.5k resistor in series with the base of a transistor I am using to run an LED or a relay or whatever.


I am looking at it from a training perspective. Once you understand the things, then you can marvel at how alike they are under the skin.

If I may... I grew up on the imperial mesurments 1/4" 3/8" etc. If I want to learn about metric hardware, I think it better to just learn, these are 7mm and these are 9mm, rather than oh, these are the ones pretty close to 1/4" and these are the ones pretty close to 3/8". I recognize and reach for a 10mm 0r 11mm nutdriver to tighten up pots without reverting to "this is the one that my 1/2 doesn't quite fit." SO I look to get someone thinking in terms of transistors rather than thinking in terns of tubes that look funny.

[lowell]

Ok so I think I grasp that the output bus moves up and down with the AC, otherwise how could we get sound? So then the bases of the power amp transistors must move WITH the output bus in the same direction right? My thinking is that if the bases don't move down the SAME amount of voltage that the output bus moves then the bias will be changing with the signal as Vbe changes. It's ONLY when the top half of the power amp signal goes positive that the top half of the power amp's bias changes, and the same when the bottom half of the power amp gets a negative going cycle. Correct?

Is D28 there to stop the negative going cylce of the AC voltage on Q18's base from pulling down Q9's collector?

[Enzo]

I don;t like drawing tube parallels, but think of how a cathode follower works. Not look at the output transistor as a cathode follower, except here it is an emitter follower. Yes, the output bus follows the bases.

Hopefully the bias never changes. This is why I don;t like tube-think applied to transistors. Tube bias is totally different from transistor bias as applied to a circuit. Forget bias a moment, we just tie the V+ driver bases to the V- driver bases. Normally they sit at zero - centered between V+ and V-. The output sits there too. The output follows the bases. SO grab that point where the bases are tied together. SHoe it upwards, the output goes upwards, and shove it downeards and the output goes downwards... MUSIC>

Problem with that is there would be crossover distortion. A little segment at the middle would be snipped out.

To get an output transistor conducting, we have to get its base about a half volt more positive than its emitter. Same thing for a driver. So since the driver drives the output transistor, and its base needs to be a half volt over its emitter, that means when they are stacked in the amp, that the base of the driver has to be about 1 volt higher than the output bus. That means that with +1v on the V+ side driver base, we have the V+ side output transistor just on the edge of conduction. Any signal pushes the base higher than that +1v means the output goes as well. SO now with that 1v of bias, the output doesn;t wait to follow, it is right there right now. Without that bias, the signal would have to climb from zero up to 1v before the outputs start to follow. DO you follow?

The same thing is going on in the V- side, except negative.

Now we want the output to follow the signal, and we need the driver base to be 1v more pos than the output bus it sits above. And the negative also a volt away. SO what we need is a way to make sure the two opposing bases STAY 2v apart - 1v pos plus 1v neg. That is the bias, that 2v apart. At any given instant, the voltage with respect to ground from those bases will be whatever the signal tells it to be. But the V+ driver will always have its base 2v more positive than the V- driver base. They move in unison. Now when the V+ driver base is more negative than its emitter, it is shut off and not conducting, same in reverse down below. When properly biased, the two sides seamlessly hand off back and forth without creating crossover distortion. So the bases do indeed follow the signal all the way up and down, just when they go reverse on the transistor it simply shuts it off.

The bias is a voltage difference betwen the two drivers, NOT to ground. Basically the amp takes the signal voltage and yanks the bias stack up and down with it.

[lowell]

Yes! Makes sense. I can visualize the entire power amp (excluding the VAS) now fluctuating up and down with the signal. The bases of the drivers being 2v apart, I get. The output bus and bases following the AC signal up and down... providing no voltage gain but plenty of current gain. I can visualize it and understand it far more than before. I'm sure there's more aspects of it to grasp, but I'll learn those as I continue on.

Ok so how about:

Is D28 there to stop the negative going cylce of the AC voltage on Q18's base from pulling down Q9's collector?