Instead of my posts being buried in cyberspace to never be found again, except by my laborious and repeated typing... ya know I wouldn't mind going on to beat in the point of newbies coming in not knowing a damn and wanting it all told to their face, but... I understand this fact of life, so for posterity I shall list some of my longer and interestinger posts here. Enjoy.
In response to:
What makes class d amps more efficient than class a?
How do class a amps work that makes them so inefficient?
What's the difference between class a and b amps?
I know that class d amps switch the output transistors at high frequencies, but I don't see how that will make them more efficient.
Well first of all let's go over the classes: Class A = always conducting.
Class B = push-pull only, one side drives the positive half of the wave, the other side drives only the other. Hence each is on for half the waveform.
Class AB = a cross between A and B where, within a certain (small) signal range, both devices are conducting simultaneously as a class A amplifier.
Class C = both are off for less than 1/2 the wave. For small signals, rather than being treated well (as with class AB), they are cut out entirely because neither side (positive or negative) is conducting, hence unable to amplify it.
Class D = always in two states: ON or OFF.
Class A must burn a distinct amount of power constantly, because it has to be "ready" to accept a signal. This alone makes it atrocious because in this condition it is 100% inefficient, consuming power for no output whatsoever. At maximum output it is maybe 33% efficient(?), even with perfect components (which transistors approach pretty well).
Class B gains a significant bit of efficiency due to the devices being off at rest. Sorta. Pure B is a mathematical curiosity, unattainable in practice because devices always conduct a little bit, even in the cutoff region. So technically it's either class AB or C, never exactly B. For this reason of course, there is a practical definition, but I don't know what it is. Efficiency is maybe 65% max.
Class C produces tremendous amounts of distortion (known as crossover distortion) because the devices are off for a certain percentage of the waveform. No good for audio. It does, however, find use in radio because the distortion products are filtered out by the tuned circuits. Efficiency is up around 75% for some transmitting tubes.
Now, class D would seem utterly useless for audio because you don't listen to ones and zeroes! (If you have a CD player that's old enough to play data CD's as audio, you'll know how horrible this sound is.) However, there's this magical technique known as pulse width modulation where you take a square wave and control how much time it spends ON vs. OFF, thereby causing an average shift in the output, after integrating out the carrier wave that is. Transistors make awesome switches and as such, efficiency is easily above 90%.
To sum up, class A amplifiers are inefficient because the varying current, which produces the very signal itself that you hear, must always be flowing, and controlled by the device. Class D has this current flowing from a filter instead, which is then supplied by pulses of HF current from the PWM amplifier.
on Today at 3:31pm, T3sl@ wrote: [that's me, by the way]
I would expect your hum to be either 50Hz spikes (one up, one down; period is faster than 10kHz and quickly damped so it bounces just a little), 50Hz short peaks (one up, one down, corresponding to charging current - typical of a grounded power cord type external ground loop) or an even 50 or 100Hz noise.
I'll elaborate on this a bit.
The spikes correspond to diode switching, dunno if a tube diode will cause them. Probably not. Often transmitted through PT capacitance to the heaters, grounding them or bypassing with a few uF usually cures it. Sometimes a hum balance control might be necessary to remove every last bit; I think this is what's wrong with Revision 3 (my 6L6 SE amp).
The short bumps corresponding to charging pulses may be both positive-going, negative-going or alternating. Most common on grounded equipment where an induced voltage in the safety ground wiring appears at the chassis of each piece of equipment; when several are connected together (for instance, the amp, a scope and your computer) this puts a small voltage across the ground in the patch cables between them, but not in the signal wires. As a result you get hum pickup. This will produce up and down bumps (like a sine wave with a hell of a lot of crossover distortion) because most everything uses full wave rectification, at least once traced back to the line. (A FWCT circuit may not appear FW at first glance but transformer actions fix that.)
If the bumps go the same direction, it's something local, an internal ground loop or PSU hum. Although PSU hum, as I'm sure you know, is usually ramped or sinusoidal in form.
But...you say it's like a rounded squarewave. And you're sure it's at 50 or 100Hz? I've never seen a squared out hum signal. Square anything basically means something is switching on and off, that is, distorting an otherwise clean signal. Or it's creating its own signal. If it's just a small signal then it can't be an oscillation (which would grow in amplitude 'til it maxes out the amplifier), and it can't be distortion (because there is ample clean room above and below, I presume).
So unless your signal generator was tuned to 50 or 100Hz and some was leaking through and you didn't realize it (don't feel bad, sometimes I track down stray 1kHz signals only to realize it's the not-quite-off volume control on the input, leaking signal in! ), I'm stump-diddly-umped.
How does the [graphing] process differ when designing a CF stage? Since the currents are essentially the same do you use the same process?
Since plate current equals cathode current in class 1 (no grid current) amplifiers, you can place the load in the plate circuit for your analytical example. Then you have to connect output voltage in series with input voltage, in such a way that plate remains constant and cathode (and B+ as it would be in a normal plate loaded stage) move instead... but that's complicated so we'll ignore that for now. And we can, since it doesn't have anything to do with how the tube sees things.
Say you have a 12AU7 triode running from +300V and the grid is biased at +150V from an external source (perhaps the plate of a preceeding stage) and there is a 10k cathode resistor. The first assumption you can make is that cathode voltage will be near grid voltage, and you can eyeball that there will be about 15mA flowing (150V/10,000ohms).
(If the tube were cutoff, which is impossible because we know current is flowing, there might be 20V between grid and cathode. That places the cathode at 150 + 20 = 170V, which means there are 17mA unaccounted for. So the tube cannot be cut off. Of course, cathode isn't equal to grid voltage, as the tube would then be saturated, fully on; but that requires a load line to find what voltages and currents would be present because a triode has an indefinite saturation point.)
For further analysis (and an assumption like this is often sufficient!), we'll have to refer to the plate curves. Operating point is around 300-150 = 150V plate to cathode and 15mA. You can mark this point if you want.
Now, say current drops to zero: cathode voltage will disappear. 300 - 0 = 300V will be across the tube, so mark a point at 300V, 0mA. If it were possible for the tube to turn fully on and become a short, Vp-k = 0V and so 300V are across the load resistor. 300/10,000 = 30mA, so mark 0V, 30mA. Connect these two dots with a straight line. This is the load line.
If you marked the earlier point, you'll see it falls exactly on the line. That's because we used ohm's law to find the current, so of course it's going to be on the line! It's not the exact operating point, but it's close enough.
Now, you can see it's pretty darn close to the Vg=0V curve. If you move over to the intersection, you'll see you can only get +15V swing before grid current sets in. Compared to the available -20Vg swing, which would produce almost -150V output, you can see this is a pretty trashy operating point. Because of the high current, it'll drive a line pretty well, but you don't have the voltage output available for more than line level output (so no good if you are trying to say, drive output tubes into class 2 (grid current)).
Anyway, that point at 150V on the loadline is around -1.5Vg. So if grid is 150V, cathode is 151.5V and actual current is 15.15mA. See how little difference that makes to the assumtion? If you must, you could bump the point over a pixel or two to represent 151.5V but gee, it's not worth it, and your tube is less accurate (to the data) than that!
Within the available swing of +/-1.5Vg, the extremes of 0Vg and -3Vg land at approx. 132V and 170V, for +18 and -20V. Average is 18+20 / 2 = 19V (peak) for 1.5V peak input, so gain is 13. I don't remember the equation for distortion but it'll probably be like, 3% 2nd harmonic.
Now, at last, we can apply NFB. Output voltage appears 'under' the signal, hence Vin = Vo + Vtube, where Vtube is the grid voltage. We said a swing of 19V for 1.5V in, so Vin = 19 + 1.5 = 20.5V. This is for the same output voltage of 19V, so what happened? NFB happened, and it reduced our gain (to less than unity!). But it pulled distortion and Zo with it. How much? By the change in input voltage: 20.5 / 1.5 = a factor of 13.7, reducing gain and distortions. If 2nd H was 3%, it is now .22%.
There ya have it, that's pretty much it. Really should edit this monster of a post though, it's far longer than necessary... oh well!
First you need a scope. Hook up an unrectified B winding (say 300VAC) to the tube's plate. Whatever voltage you want to swing up to. A variac or rheostat (NO LAMP DIMMERS!) will come in handy to change voltage if you don't have a suitable transformer on hand, or want it continuously variable. Make sure the rheostat can handle the tube current!
The tube will rectify on its own. You can put in a diode to protect it (or two or four if you want to be fancy-shmancy and add a negative load so there is little DC in the transformer) if you care. Hook up a C- supply (variable grid bias with voltmeter) and A supply (heater; 6.3V filament transformer or whatever) and you're almost there.
Connect a small (10 ohm or so) resistor in series with the cathode, or transformer return (current that flows in the plate flows throughout the supply winding and back to ground, equal at all points because it's a series circuit). This senses current, connect to the VERTICAL (Y) scale on your scope. It goes 10mV/mA so 10mV/div reads 1mA/div, and so forth. You'll have to invert the display if you use the ground return method. Connect the HORZONTAL (X) scale to the plate, or if voltage swing is greater than you can read, add a voltage divider. 100k to 1k will give 1V/100V (not exactly, but we don't care about the 1% in the divider, let alone 5% in the resistors or ~3% in the scope, depending on what you have), which is 500V full screen (10 div) at 500mV/div.