So I have read the articles and still want to run this by people to determine at what wattage i'm into class AB. I have a F5t V2 actually maybe 2.5. I'm using cascode on the front end. My output stage consists of two pair per channel. So 4 trannies per channel. The rails are running at 50volts +/-. I'm clipping right at 123 W/ch. Each trannie is biased to .8 amp. So from what I understand im in class AB when I reach 2X idling current. Is that based on total current per channel. 2X=3.2 amp? Or 2x=6.4A. Can someone explain please.
Freecrowder,
The best way to calculate this is to do It by example.
The F5 with its complementary output stage will remain in class A operation as long as the peak output current is no more than twice the quiescent current.
So let’s say, as an example, the amplifier has a quiescent current (Iq) set at 1A. Two times that is 2A. The peak output current is 2A.
Assuming an 8Ω load, this means that the amp will run in class A up to P = (2A)^2 * 8Ω = 32W peak. Notice I wrote peak, not RMS. It’s conventional to write Class A watts as ‘peak’ watts cause everybody is wondering at what point it transitions into Class AB. So, in effect, with a 1A quiescent bias (Iq), the amp will remain in Class A up to 32W and then transition to Class B. It stays in (A)B until it clips, which is determined by the RAIL voltage. Higher rail voltage equals higher power output.
If you haven’t already, look at this excellent article written by Pass, and calculate the peak Class A outputs for yourself which is the best way to understand it:
http://www.passdiy.com/pdf/leaving_class_a.pdf
And of course the F5 turbo article is an excellent read as well.
Best,
Anand.
P.S. Don’t worry, the amp won’t make a Klunk! sound when/if it transitions 😉
The best way to calculate this is to do It by example.
The F5 with its complementary output stage will remain in class A operation as long as the peak output current is no more than twice the quiescent current.
So let’s say, as an example, the amplifier has a quiescent current (Iq) set at 1A. Two times that is 2A. The peak output current is 2A.
Assuming an 8Ω load, this means that the amp will run in class A up to P = (2A)^2 * 8Ω = 32W peak. Notice I wrote peak, not RMS. It’s conventional to write Class A watts as ‘peak’ watts cause everybody is wondering at what point it transitions into Class AB. So, in effect, with a 1A quiescent bias (Iq), the amp will remain in Class A up to 32W and then transition to Class B. It stays in (A)B until it clips, which is determined by the RAIL voltage. Higher rail voltage equals higher power output.
If you haven’t already, look at this excellent article written by Pass, and calculate the peak Class A outputs for yourself which is the best way to understand it:
http://www.passdiy.com/pdf/leaving_class_a.pdf
And of course the F5 turbo article is an excellent read as well.
Best,
Anand.
P.S. Don’t worry, the amp won’t make a Klunk! sound when/if it transitions 😉
Again following your example. One channel has 4output transistors. 2 pair. Each transistor is idling at .8amp. .8 amp x 4=3.2A quiescent. 3.2 X 2 =6.4A. So is that when I'm leaving class A?????
Freecrowder,
Look at the ‘Leaving Class A’ article I linked above, and specifically the XA160.5. It stays in Class A for the full 160 watt power rating due to the heavy amount of 3.2A bias (which calculates to 328 watts peak). Do you see that?
So in your F5 build, do you think you’re leaving Class A?
Best,
Anand.
Look at the ‘Leaving Class A’ article I linked above, and specifically the XA160.5. It stays in Class A for the full 160 watt power rating due to the heavy amount of 3.2A bias (which calculates to 328 watts peak). Do you see that?
So in your F5 build, do you think you’re leaving Class A?
Best,
Anand.
Take the total bias current, multiply it by 2 and that is the minimum figure for leaving class A. Example: two parallel transistors at .8 amp bias is 1.6 amps bias for a push-pull output stage with 4 devices.. Minimum current leaving class A will be 3.2 amps, so peak class A into 8 Ohms is 3.2 X 3.2 X 8 or roughly 80 watts peak, 40 watts rms. This the minimum, as square law for Fets or exponential for Bipolars could give you higher figures depending on the degeneration (resistance in series with Source or Emitter pins . For example, undegenerated power fets routinely give class A envelopes twice those numbers, or even better.
sum Iq is twice 0.8A
as current is flowing from one rail to second, you have two vertical paths ( from N to P mosfet)
writing that just to correct your 0A8 multiplying by 4
everything else as already written
I have a related question: Suppose one wants to measure the class A/AB transition point. Can it be done this way?
(Assume we have source resistors.)
Use a load (for example 8 ohms) on the output, and apply a variable signal to the input, say a nice clean1k sine wave.
Observe the AC waveform across any source resistor with a scope and increase the input level until the AC waveform
shows the onset of clipping/flattening. The amp output at this signal level should then be the class A/AB transition
point for this load.
Is this correct?
Thanks,
Dennis
(Assume we have source resistors.)
Use a load (for example 8 ohms) on the output, and apply a variable signal to the input, say a nice clean1k sine wave.
Observe the AC waveform across any source resistor with a scope and increase the input level until the AC waveform
shows the onset of clipping/flattening. The amp output at this signal level should then be the class A/AB transition
point for this load.
Is this correct?
Thanks,
Dennis
yup
with large signal present, when you see 0Vac across one source resistor, that's !KlunK! point
with large signal present, when you see 0Vac across one source resistor, that's !KlunK! point
Thanks ZM.
And '!KlunK! point' sounds much nicer than "class A/AB transition point'. 🙂
Cheers,
Dennis
And '!KlunK! point' sounds much nicer than "class A/AB transition point'. 🙂
Cheers,
Dennis
ZM & Papa Pass,
Thanks for the correction. I wasn’t sure what the “SUM” Iq was in the Freecrowder’s F5 build but I am hoping my equations were ‘ok.’ Which is why I was illustrating only by ‘example.’
Best,
Anand.
Thanks for the correction. I wasn’t sure what the “SUM” Iq was in the Freecrowder’s F5 build but I am hoping my equations were ‘ok.’ Which is why I was illustrating only by ‘example.’
Best,
Anand.
Thanks for your response, Hui. I was wondering about doing that exact same thing as far as measuring it directly.
I'm a visual learner, so this would be fascinating for me. I am trying to up my measurement game. 😀
For a complimentary output stage, could I -
Measure input signal,
Measure source resistor on N side,
Measure source resistor on P side,
Measure output signal?
I have a 4-channel scope, and I think all the voltages would be fine and within the specs. Currently, I've only used two of the channels, but it could be super fun to see the onset of clipping variance between the N and P sides. I also don't think I've ever driven any of my amps to the Klunk! point yet even on the bench, so it could be a very interesting bit of learning. If my load resistors don't get too toasty, I may even see if I can drive the output to clipping.
Might be wise to do it on a headphone amplifier or something of lower power first...
Thoughts?
Hi Patrick,
I think practically speaking you don't need to measure the input signal. Most likely you will just crank it up until you see the 'clipping' behaviour and you note the output voltage.
Actually one thing I would first do is find the input voltage that will give clipping at the amp output. Note that you may not see any AC waveform clipping at the source resistors (until the amp itself clips) if the amp is fully biased for class A into the test load impedance. (For example, a standard F5 is good for 25W into 8 ohms class A so you shouldn't see anything with an 8 ohm load. On the other hand, you should see a Klunk! transition if you drive it hard into 4 ohms.) So it's good to know how high an input signal you can use before the amp clips for your test.
It would be neat to see if there's any noticeable difference in the waveform measurement between the N and P.
Cheers,
Dennis
I think practically speaking you don't need to measure the input signal. Most likely you will just crank it up until you see the 'clipping' behaviour and you note the output voltage.
Actually one thing I would first do is find the input voltage that will give clipping at the amp output. Note that you may not see any AC waveform clipping at the source resistors (until the amp itself clips) if the amp is fully biased for class A into the test load impedance. (For example, a standard F5 is good for 25W into 8 ohms class A so you shouldn't see anything with an 8 ohm load. On the other hand, you should see a Klunk! transition if you drive it hard into 4 ohms.) So it's good to know how high an input signal you can use before the amp clips for your test.
It would be neat to see if there's any noticeable difference in the waveform measurement between the N and P.
Cheers,
Dennis
Thank you, Dennis. Extremely helpful as always.
I've SWAGed the input sensitivity for all my amps into 4 and 8 ohms, but I agree. It would be an excellent exercise to measure it. I am going to need to get some beefier load resistors / heatsinks to measure a few amps. I have been a major fraidy-cat. However, if they have all that oooooomph, I may as well see how they perform on the bench at estimated rated power output into 4 and 8 ohms.
One of the reasons I was thinking of tossing up the input signal is to make an attempt to "overlay" all the signals on the screen. Not sure if it would work all that well, so I appreciate the insight. Sometimes I can't tell if a waveform is "perfectly rounded" / sinusoidal / symmetrical just by looking at it on the screen. I was wondering if I could see small deviations from the "shape" as distortion begins to occur if I had the input fundamental (set to a different scale obviously) on the screen at the same time. I wanted to include the output also b/c it would be fascinating to watch the amp 'Klunk' while the output signal stays "clean".
I'll try to get some parts on order with another couple sets of not-too-expensive probes and see what I can get done without invoking the lyrics to a famous Talking Heads song. 🙂
I've SWAGed the input sensitivity for all my amps into 4 and 8 ohms, but I agree. It would be an excellent exercise to measure it. I am going to need to get some beefier load resistors / heatsinks to measure a few amps. I have been a major fraidy-cat. However, if they have all that oooooomph, I may as well see how they perform on the bench at estimated rated power output into 4 and 8 ohms.
One of the reasons I was thinking of tossing up the input signal is to make an attempt to "overlay" all the signals on the screen. Not sure if it would work all that well, so I appreciate the insight. Sometimes I can't tell if a waveform is "perfectly rounded" / sinusoidal / symmetrical just by looking at it on the screen. I was wondering if I could see small deviations from the "shape" as distortion begins to occur if I had the input fundamental (set to a different scale obviously) on the screen at the same time. I wanted to include the output also b/c it would be fascinating to watch the amp 'Klunk' while the output signal stays "clean".
I'll try to get some parts on order with another couple sets of not-too-expensive probes and see what I can get done without invoking the lyrics to a famous Talking Heads song. 🙂
Yeah, it can be tricky seeing the deformation in the waveform. But another thing is to note that the Klunk occurs when one side switches off so the current is zero and hence the measured voltage across the source resistor is also zero. So when the tip of the sine wave you're measuring across a source resistor touches zero, you've reached the Klunk point. 🙂
^ Exactly. The neat part for me using a scope (again, I'm a visual guy) will be to observe the output signal staying "clean" / non-deformed during the "Klunk" and even after the transition. Then, once I start to reach max power at output, I can see the output signal begin to clip. This is probably standard measurement for those "in the know", but I've never actually thought about it until you posted it.
I know it happens... I can sorta visualize it happening... I even think I understand why/how it happens; but I love the idea of watching it. It took ZM showing me a simple circuit simulation with the waveforms for current and voltage to understand the relationships in a push-pull output stage. I struggled with it mightily until I saw that one little animation. Then... KLUNK! I got it.
I know it happens... I can sorta visualize it happening... I even think I understand why/how it happens; but I love the idea of watching it. It took ZM showing me a simple circuit simulation with the waveforms for current and voltage to understand the relationships in a push-pull output stage. I struggled with it mightily until I saw that one little animation. Then... KLUNK! I got it.