Well AnderwT, as I said; This gets pretty sticky dosen't it 
Well, pardon me. Wikipedia seems to say, I was suggesting something totally wrong about what you could have been suggesting. And, IMO from this description, the the F5X has 2.6A of "Bias" current.
So just what did you mean when you said "The two halves are not in parallel as far as operating like an amplifier are concerned."? Because IMO if they are not, you must not have a balanced amplifier?
I beleive this description says, in this case, it will be?
From what I've seen in the past you will not leave your interpatation open enough for any understanding but your own. If I am wrong why is it that you disagree with what I suggest other members might be thinking. You will no doubt disagree with what Wikipedia calls bias, because as I said "Others however, my take that to mean the idle current through both the positive and negative legs of the amp in total and many forget to clarify this situation." As you did. And Wikipedia does.
Biasing - Wikipedia, the free encyclopedia
But this says, I and other members might be, correct
Also for the sake of understanding I will not respond to this idea again. I will leave it to you. Personally IMO, neither of us can hold a candle to N.P. and we should both just tell them to go read his articles
I'll learn to shut up and not try to help someday
So, you could be totally correct but, I never tried to get into any imbalance in the balanced circuit. I wonder, will you agree that the Wikipedia description is correct wether you are, or not? From Wikipedia: Biasing in electronics is the method of establishing predetermined voltages and/or currents at various points of an electronic circuit to set an appropriate operating point. The operating point of a device, also known as bias point, quiescent point, or simply Q-point, is the point on the output characteristics that shows the direct current (DC), collector-emitter voltage (VCE), and the collector current (IC) with no input signal applied. The term is normally used in connection with devices such as transistors.
Well, pardon me. Wikipedia seems to say, I was suggesting something totally wrong about what you could have been suggesting. And, IMO from this description, the the F5X has 2.6A of "Bias" current.
So just what did you mean when you said "The two halves are not in parallel as far as operating like an amplifier are concerned."? Because IMO if they are not, you must not have a balanced amplifier?
Biasing - Wikipedia, the free encyclopedia
I beleive this description says, in this case, it will be?
Well, I tried to offer an opinion of what I thought the other, lets say, "less senior members" might think from your post but, I'm aparently wrong from what you are saying. I don't beleive I said "bias" and Idle" are the same but, wikipedia seems too? I also don't believe you can comprehend what some of these members might think?
From what I've seen in the past you will not leave your interpatation open enough for any understanding but your own. If I am wrong why is it that you disagree with what I suggest other members might be thinking. You will no doubt disagree with what Wikipedia calls bias, because as I said "Others however, my take that to mean the idle current through both the positive and negative legs of the amp in total and many forget to clarify this situation." As you did. And Wikipedia does.
Biasing - Wikipedia, the free encyclopedia
But this says, I and other members might be, correct
Well thanks.
As I said ealier, for the sake of understanding, I never tried to get into any imbalance in the balanced circuit. But, according to Wikipedia "That quiescent current" is bias current.
Also for the sake of understanding I will not respond to this idea again. I will leave it to you. Personally IMO, neither of us can hold a candle to N.P. and we should both just tell them to go read his articles
I'll learn to shut up and not try to help someday
I do not think so. It is another aspect of the same thing.
It shows also that the bias to take into account ( for classA) is half of the total one.
That we were trying to explain.
I think it is a very important topic you are raising.
However there are quite a few people here struggling just calculating the bias let alone determining the optimum bias for a given load.
Once people understand and agree what the bias current is then I think what you want to discuss is very important, and the next logical step.
However there are quite a few people here struggling just calculating the bias let alone determining the optimum bias for a given load.
Once people understand and agree what the bias current is then I think what you want to discuss is very important, and the next logical step.
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I was just trying to help those who want to play with balanced things and wonder about bias questions. (Sometimes, a little drawing tells more than a long speach.)
Of coarse it supposes that calculating the bias for a simple push pull is acquired, but this is nicely explained in the Nelson's F5 PDF.
Of coarse it supposes that calculating the bias for a simple push pull is acquired, but this is nicely explained in the Nelson's F5 PDF.
from Wikipedia
I will pick out one phrase.
When considering one device the bias current is the same as the quiescent current for that device only.
Now refer back to what I said originally.
I see no conflict whatsoever between what I interpret and what is quoted from Wikipedia
The bias of one output stage is 1.3A. The bias of the other output stage is 1.3A.
The sum of the bias current is 2.6A.
The bias current of one input stage is ~8mA. The bias of the other input stage is ~8mA.
The sum of all the bias currents is 1.3+1.3+0.008+0.008 = 2.616A.
The idle current of the balanced amplifier is 2.616A. The quiescent current of the no signal balanced amplifier is 2.616A.
The bias of the input stage is 8mA.
The bias of the output stage is 1.3A.
I see no part in there where interpretation can lead to different versions of the definition of bias current of the output stage.
I have read this and since some of you seem to think I am disagreeing with that (unofficial) definition of bias current I re-read it to be sure of what It says.
I will pick out one phrase.
I agree absolutely with this phrase. I agree with the whole definition as you posted it.
When considering one device the bias current is the same as the quiescent current for that device only.
Now refer back to what I said originally.
That fits exactly with the posted version of the Wikipedia definition.
I see no conflict whatsoever between what I interpret and what is quoted from Wikipedia
The bias of one output stage is 1.3A. The bias of the other output stage is 1.3A.
The sum of the bias current is 2.6A.
The bias current of one input stage is ~8mA. The bias of the other input stage is ~8mA.
The sum of all the bias currents is 1.3+1.3+0.008+0.008 = 2.616A.
The idle current of the balanced amplifier is 2.616A. The quiescent current of the no signal balanced amplifier is 2.616A.
The bias of the input stage is 8mA.
The bias of the output stage is 1.3A.
I see no part in there where interpretation can lead to different versions of the definition of bias current of the output stage.
Yes, when considering one device "with that," becomes the same thing. For one device the bias current and the quiescent current become the same thing.
You are right.
I agree with you.
But the conditional statement "for one device" must always be either written or implied.
I was quite clear I was referring to output bias current. I was referring implicitly to a series string of four off "one device". I listed them.
I clearly stated the sum of the bias currents is 2.6A and that the bias current is 1.3A. I did not imply "one device", I stated it. Interpretation did not come into it.
Thanks
So, now that that is cleared up? I think we should discuss a little complimentary push-pull for the unknowing and then move on to discussing balanced. This does seem to be the area of confussion. Of coarse there are those wrapped up with output power too!
We need to somewhat limit our statements to an F5 type of circuit or we will bring on more confusion. Having said that, the F4 is in many ways similar to the F5. They are both complimentary, push-pull circuits. That basically means there is both an N type and a P type transistor involved and operate in an opposing manor. They have a continuos quiescent current when in operation with no signal applied. This Iq current will be flowing from rail to rail through our output transistors with our output being precisely in the middle at 0 volts.
At this point we could discuss quiescent power dissapation. We know the rail voltages of both + an -, and we know the Iq. If we have +24V and -24V for the sake of this calculation we will add them together as they are in series. And, we have 1.3A Iq flowing from 1 rail to the other, then we will see 48*1.3=62.4W. (E*I=Power) That was considering 1 complimentary pair of outputs as in a stock F5.
Does this basic description sound correct to all of you? Is it understood by the individuals trying to grasp this? Now is the time to ask. Dare I say, there are no stupid questions
So, now that that is cleared up? I think we should discuss a little complimentary push-pull for the unknowing and then move on to discussing balanced. This does seem to be the area of confussion. Of coarse there are those wrapped up with output power too!
We need to somewhat limit our statements to an F5 type of circuit or we will bring on more confusion. Having said that, the F4 is in many ways similar to the F5. They are both complimentary, push-pull circuits. That basically means there is both an N type and a P type transistor involved and operate in an opposing manor. They have a continuos quiescent current when in operation with no signal applied. This Iq current will be flowing from rail to rail through our output transistors with our output being precisely in the middle at 0 volts.
At this point we could discuss quiescent power dissapation. We know the rail voltages of both + an -, and we know the Iq. If we have +24V and -24V for the sake of this calculation we will add them together as they are in series. And, we have 1.3A Iq flowing from 1 rail to the other, then we will see 48*1.3=62.4W. (E*I=Power) That was considering 1 complimentary pair of outputs as in a stock F5.
Does this basic description sound correct to all of you? Is it understood by the individuals trying to grasp this? Now is the time to ask. Dare I say, there are no stupid questions
yep i'm trying those Rifas, but may also play with mundorfs for a tweak, nice caps, quite sexy to look at and small for the rating, which is a big bonus. i wonder who makes them for mundorf under contract; as mundorf doesnt seem to make anything themselves
My last post was mostly about the quiescent or DC mode of a stock F5 with no signal. Lets try a little AC.
The F5 is Single ended, meaning there is one output node and it is refrenced to ground. You need to connect your speaker to the output and ground. With no signal the output is at zero. When a signal comes along and drives the output transistors, one FET will be turning on harder and he other FET will begin to shut off. Remember there is 1.3A of quiescent current flowing through both at no signal. This would be about the same for the F4.
Lets get right to the meat, or is it the bone? If you drive this amp with enough signal to get maximum ClassA output, one FET goes from flowing 1.3A down to 0A while the other goes from 1.3A up to 2.6A. Basically 1 FET turns off and the other tries to turn on and connect the output to the rail. As soon as this current shift happens, the voltage at the output swings + or -. It will actually swing in the direction of the FET turning on's rail. If your speaker is at the output, it will have this voltage applied to it and the current will now flow through the speaker and the FET that is on. If this sytem were perfectly linear, ideal etc. Whatever it takes to change the current in 1 FET, will cause the equal but opposite effect on the other FET and 2.6A of current will be flowing through the speaker and the 1 turned on FET. As the signal swings the other direction, the opposite will begin to happen. The FET that is turned on starts to turn off and the FET that is off starts to turn back on. The voltage at the output will go back down to 0 and increase in the opposite direction.
You probably have heard the statement, push-pull can swing twice it's bias current in classA. That was an example. Yes, I left out a few details but, basically that is ClassA operation in a stock F5. Unless you have a high impeadance speaker, like 16 ohms for example your amp should do this. For speakers greater than 8 ohms the F5 output voltage(20V max) is not high enough to leave ClassA operation. That would be, drive more than 2.6A to the load. In fact, a 16 ohm speaker may only reach a peak current of about 1.25A. 20V/16ohms=1.25A. If there were no questions before, there must be now?
The F5 is Single ended, meaning there is one output node and it is refrenced to ground. You need to connect your speaker to the output and ground. With no signal the output is at zero. When a signal comes along and drives the output transistors, one FET will be turning on harder and he other FET will begin to shut off. Remember there is 1.3A of quiescent current flowing through both at no signal. This would be about the same for the F4.
Lets get right to the meat, or is it the bone? If you drive this amp with enough signal to get maximum ClassA output, one FET goes from flowing 1.3A down to 0A while the other goes from 1.3A up to 2.6A. Basically 1 FET turns off and the other tries to turn on and connect the output to the rail. As soon as this current shift happens, the voltage at the output swings + or -. It will actually swing in the direction of the FET turning on's rail. If your speaker is at the output, it will have this voltage applied to it and the current will now flow through the speaker and the FET that is on. If this sytem were perfectly linear, ideal etc. Whatever it takes to change the current in 1 FET, will cause the equal but opposite effect on the other FET and 2.6A of current will be flowing through the speaker and the 1 turned on FET. As the signal swings the other direction, the opposite will begin to happen. The FET that is turned on starts to turn off and the FET that is off starts to turn back on. The voltage at the output will go back down to 0 and increase in the opposite direction.
You probably have heard the statement, push-pull can swing twice it's bias current in classA. That was an example. Yes, I left out a few details but, basically that is ClassA operation in a stock F5. Unless you have a high impeadance speaker, like 16 ohms for example your amp should do this. For speakers greater than 8 ohms the F5 output voltage(20V max) is not high enough to leave ClassA operation. That would be, drive more than 2.6A to the load. In fact, a 16 ohm speaker may only reach a peak current of about 1.25A. 20V/16ohms=1.25A. If there were no questions before, there must be now?
flg and the others in this conversation have made some bulbs go off here in my head. Thanks for the explanations, I am starting to get a different understanding, but still learning.
Is there a simple explanation for what is happening to current sharing between FET's using less than the signal required for maximum output? Some of the Iq simply remains but is unused? In other words, there is always 2.6A going somewhere?
Is there a simple explanation for what is happening to current sharing between FET's using less than the signal required for maximum output? Some of the Iq simply remains but is unused? In other words, there is always 2.6A going somewhere?
The F5 output is capable of driving 8 ohm speakers to approximately 25Wrms. Basically this is due to the output voltage limit. I beleive the spec says 20V max. If you can get 4V into the F5 and it has a gain of about 5, you can get 20V to your speaker. 20Volts on an 8 ohm speaker is 2.5A max. 20V*2.5A=50Watts Max. maybe I should be using the term Peak? Instead of max? To calculate RMS power you need to use a different formula or use a different measurement.
20Volts on an 8 ohm speaker is 2.5A Peak. 2.5A Peak is 1.7675Arms (2.5A*.707=1.7675ARMS). O.K. now using RMS Amps, 20V*1.7675Arms=25Wrms.
you can also use the E squared/R method to compute this figure but you must be using rms Volts to get rms power. Speakers do not have rms ohms.
So the next question likely to come up is; What ahppens with a 4 ohm speaker?
I beleive the deffinition of ClassA includes the fact that neither output transistor ever totally turns off. If the F5 tries to drive such a load that more than 2.6A are required, we will say that it is leaving ClassA. The F5 has the ability to continue driving a load higher than 2.6A with only one transistor operating. Distortion will get worse, but it will became a ClassAB amp and get the job done. When driving 20V to a 4 ohm speaker, you will need 5A. You will be into Class AB operation but you will be doing 50Wrms output power. Questions? Comments?
20Volts on an 8 ohm speaker is 2.5A Peak. 2.5A Peak is 1.7675Arms (2.5A*.707=1.7675ARMS). O.K. now using RMS Amps, 20V*1.7675Arms=25Wrms.
you can also use the E squared/R method to compute this figure but you must be using rms Volts to get rms power. Speakers do not have rms ohms.
So the next question likely to come up is; What ahppens with a 4 ohm speaker?
I beleive the deffinition of ClassA includes the fact that neither output transistor ever totally turns off. If the F5 tries to drive such a load that more than 2.6A are required, we will say that it is leaving ClassA. The F5 has the ability to continue driving a load higher than 2.6A with only one transistor operating. Distortion will get worse, but it will became a ClassAB amp and get the job done. When driving 20V to a 4 ohm speaker, you will need 5A. You will be into Class AB operation but you will be doing 50Wrms output power. Questions? Comments?
About Class AB operation. Say you running two F5's (built per Nelson's original schematic), and then you bridge them. So you get effectively twice the power (watts) into double the impedance. The bridged F5 will effectively be optimized (more or less) for a 16 ohm load. Now the way I understand it, is your total bias is now effectively 5.2A which gives you a RMS output of around 52W.
Say you are trying to run a 8ohm load with this bridged F5 setup. The bridge amp does not have a high enough bias current to reach the 52W in Class A, so it will switch over to Class AB. Again this is my understanding and please let me know if I am wrong.
So the question is, does the amp automatically switch over to Class AB on its own because of the relationship of the bias current a the voltage? Or does one have to take additional steps to enable the amp to make this switch? I guess I'm a little be foggy on how these amps switch from A to AB. Everyone on this forum strives to stay Class A all the time, so it seems like a topic that is usually not discussed too often.
Say you are trying to run a 8ohm load with this bridged F5 setup. The bridge amp does not have a high enough bias current to reach the 52W in Class A, so it will switch over to Class AB. Again this is my understanding and please let me know if I am wrong.
So the question is, does the amp automatically switch over to Class AB on its own because of the relationship of the bias current a the voltage? Or does one have to take additional steps to enable the amp to make this switch? I guess I'm a little be foggy on how these amps switch from A to AB. Everyone on this forum strives to stay Class A all the time, so it seems like a topic that is usually not discussed too often.
Maybe thats a good call Zen Mod. I read that article a while back, and maybe its time for me to go through it again.
Flg made a good job of explaining the way a ClassA push pull operates and the way it (automatically) transitions to ClassAB if the current demand from the speaker exceeds the ClassA current limit.
The F5 has a bias current of 1.3A. Two F5 have a sum of their bias currents equal to 2.6A. Nowhere did Flg say you quadruple the bias current when you bridge.
My argument is that the bias current of each F5 remains at 1.3A, irrespective of whether it is single F5 or a pair of F5s, or a bridged pair of F5s, or a balanced F5x The recommended voltage drop across the emitter resistor does not change for an F5, the recommended bias current does not change.
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