A purely theoretical question here.
I was a bit confused by something I read in 'Valve Amplifiers' earlier this morning.
I understand classification on a stage, but something the Author mentioned about efficiency left me thinking, mainly because he didn't make any attempt to explain the statement he'd made.
He was discussing the maximum theoretical efficiency of different stage classifications. He was measuring this in terms of sine wave at the output.
He mentioned 50% for class A, and 78.5% I think for class B pp.
At first I thought this would be, efficiency of sine wave reproduction at the output for a given sine wave input.
But that's not right. Because otherwise the class B stage would have to be 50% per valve. And the maximum efficiency would depend greatly on how much distortion the stage was producing anyway.
Then I thought that perhaps he was discussing it as a process relative measurement. For example, a valve only conducts in one direction. So one that is expected to produce a negative form as well as a positive form is against the nature of the processor.
If that bassis was right, a class B valve would be 100% efficient, so that can't be right either.
So I'm still kind of confused about the bassis of his efficiency measurement and, in consequence, what the measurements are relative to.
I've also begun wondering how pp amplifiers got the name push PULL, since neither valve does any pulling, they just push against each other less. Is there an output stage arrangement which genuinely has one valve pushing current whilst the other pulls?
I was a bit confused by something I read in 'Valve Amplifiers' earlier this morning.
I understand classification on a stage, but something the Author mentioned about efficiency left me thinking, mainly because he didn't make any attempt to explain the statement he'd made.
He was discussing the maximum theoretical efficiency of different stage classifications. He was measuring this in terms of sine wave at the output.
He mentioned 50% for class A, and 78.5% I think for class B pp.
At first I thought this would be, efficiency of sine wave reproduction at the output for a given sine wave input.
But that's not right. Because otherwise the class B stage would have to be 50% per valve. And the maximum efficiency would depend greatly on how much distortion the stage was producing anyway.
Then I thought that perhaps he was discussing it as a process relative measurement. For example, a valve only conducts in one direction. So one that is expected to produce a negative form as well as a positive form is against the nature of the processor.
If that bassis was right, a class B valve would be 100% efficient, so that can't be right either.
So I'm still kind of confused about the bassis of his efficiency measurement and, in consequence, what the measurements are relative to.
I've also begun wondering how pp amplifiers got the name push PULL, since neither valve does any pulling, they just push against each other less. Is there an output stage arrangement which genuinely has one valve pushing current whilst the other pulls?
Efficiency of a tube just means what percentage is of its DC input is converted into useful output power, the rest burned off as heat. So, a 50% efficient tube just means if let's say 250V @ 100mA is sent to the plate (25 watts), 12.5 watts is turned into useful output, weather this is audio signal, RF, or whatever and the other 12.5 watts is dissipated by the anode and (if applicable) screen grid.
I've got that, but he defined these as specific limits on the efficiency of a particular classification. Rather than general limits. He's refering to some sort of process limitation as opposed to real world limitation.
Also, if he was refering to the efficiency of reproducing the stage's input, a class B stage would always be 50%. Any more and it would be an AB stage. Any less, and it'd be moving into C
Instead, he says that the maximum efficiency of a class A stage is 50%, and a B stage 78.5%. So he can't mean the efficiency of input reproduction.
This's what annoys me. He made the statement and then made no effort to explain where it was coming from.
You write a 600 pages plus book on electronics and you better expect the guys reading it to be the kind who won't willing accept such statements without an explaination!
Also, if he was refering to the efficiency of reproducing the stage's input, a class B stage would always be 50%. Any more and it would be an AB stage. Any less, and it'd be moving into C
Instead, he says that the maximum efficiency of a class A stage is 50%, and a B stage 78.5%. So he can't mean the efficiency of input reproduction.
This's what annoys me. He made the statement and then made no effort to explain where it was coming from.
You write a 600 pages plus book on electronics and you better expect the guys reading it to be the kind who won't willing accept such statements without an explaination!
eeka chu said:
I think Morgan has to. The book is thick enough already, and if he explained everything in detail, many people would not buy it (because it looks 'too hard'), or they would get lost reading it.
You're referring to page 384 right?
I don't know how to arrive at the 78.5% figure myself (it's actually pi/4 or something), but you might find this useful http://sound.westhost.com/efficiency.htm#classb
The efficiency is a percentage, independent of actual parameters.
So, the idea is simply this: when you go from class A --> AB --> B the relative quiescent power goes *down* while the relative output power goes *up*. Thus the apparent increase in % efficiency.
Efficiency here is taken as the ratio of all power being dissipated vs. the actual output power.
The 78.3% figure is best case optimistic, theoretical max.
It's simply the heat vs output ratio in practice. Class A is high heat. Class B, low heat.
_-_-bear
PS. someone got the efficiencies backwards on PP vs. SE class A, PP is higher than SE.
So, the idea is simply this: when you go from class A --> AB --> B the relative quiescent power goes *down* while the relative output power goes *up*. Thus the apparent increase in % efficiency.
Efficiency here is taken as the ratio of all power being dissipated vs. the actual output power.
The 78.3% figure is best case optimistic, theoretical max.
It's simply the heat vs output ratio in practice. Class A is high heat. Class B, low heat.
_-_-bear
PS. someone got the efficiencies backwards on PP vs. SE class A, PP is higher than SE.
bear said:
oops....that would be me ...
...so I'll just muddy the waters -
according to Mr. Broskie :
"The point that needs clearing up is the part about the relative efficiencies of output stage topologies. The efficiency difference between a single-ended and a push-pull Class-A amplifier is zero, both are 50% efficient if inductively loaded. (If constant current source load, both are 25% efficient; resistive loaded, 12.5%.) In other words, 8 output tubes in a totem-pole push-pull Class-A amplifier equal 8 output tubes in a single-ended Class-A amplifier. In the push-pull version, the output tubes must see an idle current equal to half the peak output current; for example, if the peak output current swing is 2 amperes, then the idle current must equal 1 ampere and as totem-pole amplifier has four output tubes in parallel per bank, each tube must draw 250 mA at idle.
On the other hand, in the single-ended amplifier the output tubes must see an idle current equal to the peak output current; for example, if the peak output current swing is 2 amperes, then the idle current must equal 2 amperes, with eight output tubes in parallel, each tube must draw 250 mA at idle. "
I see, literally how much the amplifier is going to cost you to run then.
But why would a single ended Class A stage need to draw it's peak current at idle? Surely he just means a higher idle current right?
If your valve is biased to it's peak current at idle, it could only swing in one direction, restricting current. Since it can't have a peak peak current.
But why would a single ended Class A stage need to draw it's peak current at idle? Surely he just means a higher idle current right?
If your valve is biased to it's peak current at idle, it could only swing in one direction, restricting current. Since it can't have a peak peak current.
First you have to refer to loadlines to understand.
First, let us define a perfect amplifier device. It has a flat resistance characteristic (like a pentode or transistor), zero saturation voltage (sweeps achieve 30-50V while transistors do 1-3V) and to simplify things, a linear characteristic. We'll use a transconductance device (MOSFET or pentode) because it takes no power to drive (capacitances aside), but this can apply to BJTs (and class 2 triodes, such as most transmitting types) as well. Input voltage is proportional to drawn output terminal current.
Now that we have defined our device, we can graph. Take a load resistance of say, 100 ohms and a supply of 50V. As the amplifier's current draw is increased, voltage drops and ultimately reaches a certain limit, in particular, 50V/100ohm = 0.5A. Thus, saturation comes at 0.5A and 0V. Class A's definition is that the amplifier is constantly conducting. For it to always conduct while symetrically amplifying the input signal, it must be biased exactly halfway between the limits (cutoff and saturation). (Realistic devices are nonlinear and usually bias a little higher, due to softened cutoff regions for instance.) Thus our situation finds 25V 0.25A as the bias point. A certain amount of drive will start clipping in an exact way (as our device cuts off at exactly 0A and saturates at exactly 0V), and just below this is the maximum output. This situation gets exactly 50Vp-p at clipping, which is 17.68V RMS, across a 100 ohm resistor is 3.13W. Total dissipation is 50V * 0.25A = 12.5W or 33.35% for this resistive load, or if it were inductively supplied (as with an output transformer or choke load), from 25V, the dissipation is cut in half (it eliminates the load resistor's idle dissipation), 50%.
You can similarly calculate class B and C efficiencies, remembering to include the opposing device in an inverse manner (which reflects its being on the other side of the transformer).
Tim
First, let us define a perfect amplifier device. It has a flat resistance characteristic (like a pentode or transistor), zero saturation voltage (sweeps achieve 30-50V while transistors do 1-3V) and to simplify things, a linear characteristic. We'll use a transconductance device (MOSFET or pentode) because it takes no power to drive (capacitances aside), but this can apply to BJTs (and class 2 triodes, such as most transmitting types) as well. Input voltage is proportional to drawn output terminal current.
Now that we have defined our device, we can graph. Take a load resistance of say, 100 ohms and a supply of 50V. As the amplifier's current draw is increased, voltage drops and ultimately reaches a certain limit, in particular, 50V/100ohm = 0.5A. Thus, saturation comes at 0.5A and 0V. Class A's definition is that the amplifier is constantly conducting. For it to always conduct while symetrically amplifying the input signal, it must be biased exactly halfway between the limits (cutoff and saturation). (Realistic devices are nonlinear and usually bias a little higher, due to softened cutoff regions for instance.) Thus our situation finds 25V 0.25A as the bias point. A certain amount of drive will start clipping in an exact way (as our device cuts off at exactly 0A and saturates at exactly 0V), and just below this is the maximum output. This situation gets exactly 50Vp-p at clipping, which is 17.68V RMS, across a 100 ohm resistor is 3.13W. Total dissipation is 50V * 0.25A = 12.5W or 33.35% for this resistive load, or if it were inductively supplied (as with an output transformer or choke load), from 25V, the dissipation is cut in half (it eliminates the load resistor's idle dissipation), 50%.
You can similarly calculate class B and C efficiencies, remembering to include the opposing device in an inverse manner (which reflects its being on the other side of the transformer).
Tim
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