I'm very attracted to aleph design. They are unique and good. I'm about to built aleph power amp, but my primary limitation here is heat dissipation. So I want to built aleph based power amp just as much as I will need. I have 2 questions here, hope you elders can help me.
1. If I built aleph with +/-25 supply, how big will I get clean sinusoidal? Will I get full +/-23 clean sinusoidal in my scope? What if I built with +/-15 supply, how much clean sinusoidal trace will I get? How can I calculate the relation between supply voltage and clean output trace (below clip)?
2. What is the relation between steady current vs speaker load? I will use 3 way 4 ohm speakers, how much steady current (in second stage) will I need if I built +/-25V aleph? What is this number if I built +/-15V aleph? What is the formula for determining constant current for aleph vs load impedance for certain voltage? Is it as simple as V/R or not?
1. If I built aleph with +/-25 supply, how big will I get clean sinusoidal? Will I get full +/-23 clean sinusoidal in my scope? What if I built with +/-15 supply, how much clean sinusoidal trace will I get? How can I calculate the relation between supply voltage and clean output trace (below clip)?
2. What is the relation between steady current vs speaker load? I will use 3 way 4 ohm speakers, how much steady current (in second stage) will I need if I built +/-25V aleph? What is this number if I built +/-15V aleph? What is the formula for determining constant current for aleph vs load impedance for certain voltage? Is it as simple as V/R or not?
1. If I built aleph with +/-25 supply, how big will I get clean sinusoidal? Will I get full +/-23 clean sinusoidal in my scope?
What if I built with +/-15 supply, how much clean sinusoidal trace will I get? How can I calculate the relation between supply voltage and clean output trace (below clip)?
You will lose about 3 volts, 2 to the Mosfet and 1 to the resistor.
2. What is the relation between steady current vs speaker load?
Inversely proportional for best results.
I will use 3 way 4 ohm speakers, how much steady current (in second stage) will I need if I built +/-25V aleph?
5 amp peak = 2.5 amp idle
What is this number if I built +/-15V aleph? What is the formula for determining constant current for aleph vs load impedance for certain voltage? Is it as simple as V/R or not?
It's that simple.
What if I built with +/-15 supply, how much clean sinusoidal trace will I get? How can I calculate the relation between supply voltage and clean output trace (below clip)?
You will lose about 3 volts, 2 to the Mosfet and 1 to the resistor.
2. What is the relation between steady current vs speaker load?
Inversely proportional for best results.
I will use 3 way 4 ohm speakers, how much steady current (in second stage) will I need if I built +/-25V aleph?
5 amp peak = 2.5 amp idle
What is this number if I built +/-15V aleph? What is the formula for determining constant current for aleph vs load impedance for certain voltage? Is it as simple as V/R or not?
It's that simple.
Thank you Mr. Pass,
I am honoured to have yourself reply my questions.
Actually I have made trial pass amp on simple heatsink (but they are burnt now). They are very hot amp. I tried to make the current smaller, it results in distoreted sound, so my conclusion the steady current must be sufficient to flow into speakers. I just dont know how much is enough.
I've been studying the aleph schematic for months now. Your second stage constant current is very unique design. The output is taken from the Source from upper mosfet, but still it can gives full swing. I really do not understand this, because usually if we want to get full swing, the output is taken from the Drain, not from emitor (uses P channel in upper mosfet).
Also usually ordinary amp uses output from collector-top+collector-bottom from current mirror (instead of 390ohm drop) to be fed into second stage. This way, the output point of the first stage is "hovering", and determined by the gate voltage of VAS transistor, because the upper VAS has already determined quiscent current. But in aleph design they are unique. I've tried to adjust the 390ohm resistor to reduce DC offset, but nothing happens until some value (>450ohm) the output is high DC. I read the forums, and found out that the dc offset is not determined by the 390ohm resistor, but from matching the differential mosfets, and they work.
My conclusion is that the value of 390ohm is changeable, as long as it gives smaller voltage than the second stage VAS+resistor drop of4-5volt. Is this true, Mr pass? I observed in the inverting section of ONO preamp, you use 470ohm, not 390ohm, because the drop in second VAS is higher, due to bigger source resistor than aleph.
I asked this question, because I have an idea to eliminate the lower VAS resistor (Source directly to -VCC, not via resistor). I tried this in ordinary amp (not aleph) and I think I hear better detail. What happens What if the 390ohm resistor is much smaller, like 200ohm (gives voltage drop way smaller than second stage gate voltage of 4-5volt), will aleph still works fine?
I'm not trying to challange the well time-proven design of yours, or tobe much clever than you, I'm not, I just want to try new things.
One more question. In second stage, there are various placement of 1nF stabilizer capacitor. In aleph design it places in lower G-D of mosfet, but in volksamp, it is on the top (on the constant current G-S of mosfet). Which is the right one, Mr pass?
I am honoured to have yourself reply my questions.
Actually I have made trial pass amp on simple heatsink (but they are burnt now). They are very hot amp. I tried to make the current smaller, it results in distoreted sound, so my conclusion the steady current must be sufficient to flow into speakers. I just dont know how much is enough.
I've been studying the aleph schematic for months now. Your second stage constant current is very unique design. The output is taken from the Source from upper mosfet, but still it can gives full swing. I really do not understand this, because usually if we want to get full swing, the output is taken from the Drain, not from emitor (uses P channel in upper mosfet).
Also usually ordinary amp uses output from collector-top+collector-bottom from current mirror (instead of 390ohm drop) to be fed into second stage. This way, the output point of the first stage is "hovering", and determined by the gate voltage of VAS transistor, because the upper VAS has already determined quiscent current. But in aleph design they are unique. I've tried to adjust the 390ohm resistor to reduce DC offset, but nothing happens until some value (>450ohm) the output is high DC. I read the forums, and found out that the dc offset is not determined by the 390ohm resistor, but from matching the differential mosfets, and they work.
My conclusion is that the value of 390ohm is changeable, as long as it gives smaller voltage than the second stage VAS+resistor drop of4-5volt. Is this true, Mr pass? I observed in the inverting section of ONO preamp, you use 470ohm, not 390ohm, because the drop in second VAS is higher, due to bigger source resistor than aleph.
I asked this question, because I have an idea to eliminate the lower VAS resistor (Source directly to -VCC, not via resistor). I tried this in ordinary amp (not aleph) and I think I hear better detail. What happens What if the 390ohm resistor is much smaller, like 200ohm (gives voltage drop way smaller than second stage gate voltage of 4-5volt), will aleph still works fine?
I'm not trying to challange the well time-proven design of yours, or tobe much clever than you, I'm not, I just want to try new things.
One more question. In second stage, there are various placement of 1nF stabilizer capacitor. In aleph design it places in lower G-D of mosfet, but in volksamp, it is on the top (on the constant current G-S of mosfet). Which is the right one, Mr pass?
lumanauw,
I've only got a few minutes and you've asked a lot of questions, but I'll try to get to as many as I can.
--Sufficient current for a loudspeaker isn't that hard to figure out. Really, the only variables are the impedance of the speaker and how loud you intend to play. All right, the efficiency figures in there, too, but let's leave that up to the owner. Keep in mind that most speakers have dips/peaks in the impedance curve, so you'll need to plan ahead for those. The minimums are the ones that will give you trouble. How loud you listen is something only you know. If you listen to string quartets, you're not likely to need a lot of power, even if your speakers are inefficient. If you're trying to crack the plaster in the wall, it's going to take a lot of power. Nelson's commercial designs went up to 200W/ch and it's possible to go even higher, but the heat is going to get pretty difficult to deal with in the practical sense; plan on three times as much wattage in heat as the amplifier delivers to the load.
--The voltage swing available from the Aleph current source isn't particularly hampered by the fact that it comes from the Source. Look at the output stage of Nelson's A-75...Sources on both sides. The ability of a current source to track a given signal is called compliance. The objective of the Aleph amps isn't really so much the ability to swing to within .000001V of the rail as it is to increase the theoretical efficiency of a single-ended class A amplifier. A topology that would ordinarily approach 25% effiency can now (at least on paper) approach 50%. Actual percentages are more like 30-40%, but it's still nice to have more power with less heat.
--The 392 ohm load resistor in the front end does double duty. It serves to generate a signal from the varying current coming from the 9610, and it also sets the bias point for the output MOSFET (i.e. the bottom one) by generating a DC offset against the Gate of the output--enough to make it "turn on." Trying to adjust DC offset at the output by changing the value of the resistor is going to be more trouble than it's worth. If you still want to try it, bear in mind that you'll only want a few ohms adjustment either way. Something like the adjustable triplet that I put into the Aleph-X current source would be the way to go, although you'd need to recalculate the actual values to center around 392 ohms instead of 221. Another possibility would be to adjust the current going through the front end, in which case you could simply plug in the Aleph-X front end current source. If you change the front end bias, the DC drop across the 392 ohm resistor will change too. In a 'regular' Aleph it's best just to use matching MOSFETs in the front end and let the feedback loop handle the rest. The only real exception is if you're using something other than IRF MOSFETs in the back end that might have a radically different Vgs. At that point, you'll have some fiddling to do to make all the pieces of the puzzle fall into place. In the real world, anything under 100mV of offset at the output is good enough.
--Using a 200 ohm resistor as the load in the front end will give you a number of problems. Assuming that you're using IRF MOSFETs in the output, they won't bias. Being enhancement devices, they need the Gate pushed a certain number of volts positive relative to the Source before they will conduct. Another problem will be that the gain of the amp will drop. Actually, this may be a benefit from your point of view, but the Alephs are fairly low-gain amps to begin with. Dropping the gain from 20dB down into the mid-teens is going to require more voltage swing from your preamp. More subtle things will include changes in the bandwidth, etc. In principle, you could increase the front end bias current in order to use a 200 ohm resistor, but be prepared to work with it a while in order to get the bugs out. Watch the heat dissipation on the front end MOSFETs if you increase the bias.
--The cap you mention isn't even in some of the designs. The stability of the amps will vary depending on the number of output devices (cumulative Gate capacitance), board layout, lead length, and a number of other factors. Just bear in mind that you may need it (or may not) depending on how your final amp is configured. Look at the output on an oscilliscope. Are high frequency square waves showing any evidence of ringing? Is the amp oscillating? Does it react poorly to reactive loads? If so, add the cap. If not, then you may not need the cap at all.
--As a final note, trying to analyze the Aleph circuit as though it was a normal amp will give you headaches. It's best just to approach it as a stand-alone design and understand it on its own terms. The front end is a completely ordinary differential. The only thing to watch out for is that the load resistance has to generate the right DC offset to bias the output. The output device (the one on the bottom) is a plain vanilla grounded Source amplifier. Nothing fancy. The current source is the unusual part. Left on its own, it produces a constant current. If a signal is present (read through the current sensing resistors at the output), it will vary its output somewhat. Any time I phrase it that way, Nelson comes along and notes that the current averages to a constant value over time, so I'll beat him to the punch and say it myself.
There's no way around it--you're going to need a lot of heatsinking for the outputs. Unless you just happen to like the little puffs of smoke as the MOSFETs burn up.
Grey
I've only got a few minutes and you've asked a lot of questions, but I'll try to get to as many as I can.
--Sufficient current for a loudspeaker isn't that hard to figure out. Really, the only variables are the impedance of the speaker and how loud you intend to play. All right, the efficiency figures in there, too, but let's leave that up to the owner. Keep in mind that most speakers have dips/peaks in the impedance curve, so you'll need to plan ahead for those. The minimums are the ones that will give you trouble. How loud you listen is something only you know. If you listen to string quartets, you're not likely to need a lot of power, even if your speakers are inefficient. If you're trying to crack the plaster in the wall, it's going to take a lot of power. Nelson's commercial designs went up to 200W/ch and it's possible to go even higher, but the heat is going to get pretty difficult to deal with in the practical sense; plan on three times as much wattage in heat as the amplifier delivers to the load.
--The voltage swing available from the Aleph current source isn't particularly hampered by the fact that it comes from the Source. Look at the output stage of Nelson's A-75...Sources on both sides. The ability of a current source to track a given signal is called compliance. The objective of the Aleph amps isn't really so much the ability to swing to within .000001V of the rail as it is to increase the theoretical efficiency of a single-ended class A amplifier. A topology that would ordinarily approach 25% effiency can now (at least on paper) approach 50%. Actual percentages are more like 30-40%, but it's still nice to have more power with less heat.
--The 392 ohm load resistor in the front end does double duty. It serves to generate a signal from the varying current coming from the 9610, and it also sets the bias point for the output MOSFET (i.e. the bottom one) by generating a DC offset against the Gate of the output--enough to make it "turn on." Trying to adjust DC offset at the output by changing the value of the resistor is going to be more trouble than it's worth. If you still want to try it, bear in mind that you'll only want a few ohms adjustment either way. Something like the adjustable triplet that I put into the Aleph-X current source would be the way to go, although you'd need to recalculate the actual values to center around 392 ohms instead of 221. Another possibility would be to adjust the current going through the front end, in which case you could simply plug in the Aleph-X front end current source. If you change the front end bias, the DC drop across the 392 ohm resistor will change too. In a 'regular' Aleph it's best just to use matching MOSFETs in the front end and let the feedback loop handle the rest. The only real exception is if you're using something other than IRF MOSFETs in the back end that might have a radically different Vgs. At that point, you'll have some fiddling to do to make all the pieces of the puzzle fall into place. In the real world, anything under 100mV of offset at the output is good enough.
--Using a 200 ohm resistor as the load in the front end will give you a number of problems. Assuming that you're using IRF MOSFETs in the output, they won't bias. Being enhancement devices, they need the Gate pushed a certain number of volts positive relative to the Source before they will conduct. Another problem will be that the gain of the amp will drop. Actually, this may be a benefit from your point of view, but the Alephs are fairly low-gain amps to begin with. Dropping the gain from 20dB down into the mid-teens is going to require more voltage swing from your preamp. More subtle things will include changes in the bandwidth, etc. In principle, you could increase the front end bias current in order to use a 200 ohm resistor, but be prepared to work with it a while in order to get the bugs out. Watch the heat dissipation on the front end MOSFETs if you increase the bias.
--The cap you mention isn't even in some of the designs. The stability of the amps will vary depending on the number of output devices (cumulative Gate capacitance), board layout, lead length, and a number of other factors. Just bear in mind that you may need it (or may not) depending on how your final amp is configured. Look at the output on an oscilliscope. Are high frequency square waves showing any evidence of ringing? Is the amp oscillating? Does it react poorly to reactive loads? If so, add the cap. If not, then you may not need the cap at all.
--As a final note, trying to analyze the Aleph circuit as though it was a normal amp will give you headaches. It's best just to approach it as a stand-alone design and understand it on its own terms. The front end is a completely ordinary differential. The only thing to watch out for is that the load resistance has to generate the right DC offset to bias the output. The output device (the one on the bottom) is a plain vanilla grounded Source amplifier. Nothing fancy. The current source is the unusual part. Left on its own, it produces a constant current. If a signal is present (read through the current sensing resistors at the output), it will vary its output somewhat. Any time I phrase it that way, Nelson comes along and notes that the current averages to a constant value over time, so I'll beat him to the punch and say it myself.
There's no way around it--you're going to need a lot of heatsinking for the outputs. Unless you just happen to like the little puffs of smoke as the MOSFETs burn up.
Grey
Look at it this way. Suppose you've got a 10mA AC signal going into a 1k resistor. You'd see a 10V AC waveform across the resistor. Okay, same current (remember that the Aleph front end is driven by a current source--one way or another that current is going to flow), but now use a 511 ohm load resistor. Presto! 5.11V signal, or roughly a 6dB loss of gain.
Going from 392 ohms to 200 ohms will result in a similar loss. About 5.8dB. If you're using a CD player and a preamp with plenty of gain, that 6dB loss may not be a problem.
You could, I suppose, reduce the feedback to compensate, but distortion will rise, damping factor will fall, and bandwidth will narrow somewhat. However, the sun will still rise in the east and the amp would still play music. Depending on how you feel about distortion and other sundry factors you might even feel that the amp sounded better that way.
Grey
Going from 392 ohms to 200 ohms will result in a similar loss. About 5.8dB. If you're using a CD player and a preamp with plenty of gain, that 6dB loss may not be a problem.
You could, I suppose, reduce the feedback to compensate, but distortion will rise, damping factor will fall, and bandwidth will narrow somewhat. However, the sun will still rise in the east and the amp would still play music. Depending on how you feel about distortion and other sundry factors you might even feel that the amp sounded better that way.
Grey
Yes, that would be true of an amp with no feedback- running open loop.
But (and I could very well be wrong here) I thought the gain of amp with a feedback loop (which the Alephs have) is determined by the feedback resistor divided by the input resistor (this may be a crude approximation). It should remain the same, as long as the open loop gain is sufficient. By reducing the open loop gain, you would reduce the amount of feeback being applied, but as long as there is sufficient open loop gain, the voltage gain should stay the same. So, then the question is whether the Alephs have enough open loop gain with less gain from the front end to support the closed loop gain of the circuit (which, if I remember right, is around 26db?).
Thank you GR Rollins, for detailed explenation. So aleph will still works fine if I use 200ohm instead 390ohm in differential? How about eliminating the source resistor in the bottom mosfet in the second stage (source directly to -vcc), is this also do-able? Or it will gives bad effect?
I observed that the quality of sound is somehow connected to how hot the devices are. Since the voltage I will be using is quite low, I will be able to get >20ma of steady current into TO-220 mosfets to get about 1W in open air.
But since my background on electronics begin with bipolar transistors, I always imagine transistors as "water Valve". If you put 1ma current into base, with hfe of 50, you will get 50ma in emitor. The bigger current you get in the emitor, you will need bigger current in base.
I know that mosfets and tubes are transconductance devices, they do not need current fed into their base. But it still worries me, for example : If I make constant current for aleph differential up to 50mA (instead of 20mA that aleph standard use), will it still have the same sensitivity to small signals? Or if I raise steady current into mosfets it will becomes less sensitive to small signals? This will result in lack of detail in music, in my imagination. From my experiment bigger quiscent current gives "stiffer" sound. But I dont have any equipment to measure distortion or frequency response, so what will be the effect if I raise quiscent current? Will it still have the same sensitivity to small signal and plays same detail in music?
I observed that the quality of sound is somehow connected to how hot the devices are. Since the voltage I will be using is quite low, I will be able to get >20ma of steady current into TO-220 mosfets to get about 1W in open air.
But since my background on electronics begin with bipolar transistors, I always imagine transistors as "water Valve". If you put 1ma current into base, with hfe of 50, you will get 50ma in emitor. The bigger current you get in the emitor, you will need bigger current in base.
I know that mosfets and tubes are transconductance devices, they do not need current fed into their base. But it still worries me, for example : If I make constant current for aleph differential up to 50mA (instead of 20mA that aleph standard use), will it still have the same sensitivity to small signals? Or if I raise steady current into mosfets it will becomes less sensitive to small signals? This will result in lack of detail in music, in my imagination. From my experiment bigger quiscent current gives "stiffer" sound. But I dont have any equipment to measure distortion or frequency response, so what will be the effect if I raise quiscent current? Will it still have the same sensitivity to small signal and plays same detail in music?
Yes, but no...
An ideal opamp has infinite gain. For better or worse (actually for better), the Aleph circuit isn't a pre-packaged opamp with a gazillion dB of open loop gain. There's only about 40dB, split roughly 50/50 between the front end and the output. (Yes, the output has gain. Weird, huh?) Anyway, at that point it becomes a question of simple subtraction: 40dB (open loop gain)- 20dB (negative feedback)= 20dB (closed loop gain). Anything that subtracts from the available open loop gain is going to show up as a reduction in the closed loop gain of the final circuit.
So the Aleph fails as an "ideal" opamp--who cares? Actually most amps do, including pretty much all tube designs. My killer tube amps here only have about 36 dB open loop. I run them with 10dB feedback, so the closed loop gain is around 26 dB, which is about average for a power amplifier. If you go back to the '70s when amps routinely used crushing amounts of feedback in pursuit of distortion specs with multiple zeros to the right of the decimal point, the open loop gain was astronomical. The Aleph wasn't designed with that in mind. There are several ways you could boost the gain, but then you're going to lose the sound quality.
Ugh. I'd rather not.
Grey
An ideal opamp has infinite gain. For better or worse (actually for better), the Aleph circuit isn't a pre-packaged opamp with a gazillion dB of open loop gain. There's only about 40dB, split roughly 50/50 between the front end and the output. (Yes, the output has gain. Weird, huh?) Anyway, at that point it becomes a question of simple subtraction: 40dB (open loop gain)- 20dB (negative feedback)= 20dB (closed loop gain). Anything that subtracts from the available open loop gain is going to show up as a reduction in the closed loop gain of the final circuit.
So the Aleph fails as an "ideal" opamp--who cares? Actually most amps do, including pretty much all tube designs. My killer tube amps here only have about 36 dB open loop. I run them with 10dB feedback, so the closed loop gain is around 26 dB, which is about average for a power amplifier. If you go back to the '70s when amps routinely used crushing amounts of feedback in pursuit of distortion specs with multiple zeros to the right of the decimal point, the open loop gain was astronomical. The Aleph wasn't designed with that in mind. There are several ways you could boost the gain, but then you're going to lose the sound quality.
Ugh. I'd rather not.
Grey
I never said it was. I said that it will try to keep the voltage gain the same, as long as there is enough open loop gain. Since the output is operating common source, it would make sense that it has gain.
I disagree... there will be some loss of closed loop gain, as the Aleph isn't a perfect opamp (as you correctly noted). It will not be as simple as subtracting the reduction in open loop gain from closed loop gain. The amp is still trying to achieve the gain setup by the feedback loop. Being that it isn't a perfect opamp, it will not succeed, but it will still be close. Well, I simulated two Aleph circuits. They were different only in the load resistor on the front end, and the amount of bias current through the front end (to get the output stage to bias up right, as you noted before). The first had a load resistor of 500 ohms. The second had a load resistor of 100 ohms. The gain of the first circuit (closed loop) was approximately 27 db. The gain of the second circuit was approximately 26.3 db- a change of .7db in closed loop gain. This would be negligible when considering what kind of voltage the preamp would have to put out, which was what we were talking about anyway.
Why was it so similar? Because, even though the open loop gain was reduced (in the front end), the overall open loop gain was high enough to support that level of feedback. What happened as a result, is that there is less overall feedback being applied globally. This had little effect on the overall voltage gain (closed loop) of the amp. Even though it isn't a perfect opamp (there was a .7db difference in gain) it was good enough that there will be really no need to adjust what you drive the amp with, which was my point in the first place.
Now, the practical application would be that Nelson obviously set the amp up for a certain amount of feeback, and it is VERY likely that this an optimized value for the design. Deviating from that gain/feeback setup may not prove to be beneficial (or maybe it will- we all like different things).
Steve,
Ah, but you changed two parameters, not just one. By increasing the bias current through the front end, you increased the gain. This offset the loss due to lower load resistance. Now, given the way Nelson set the amp up, you pretty much have to do something of that nature in order to bias the output stage, but it's equivalent to reaching inside an opamp and fiddling the internals.
Ah, the glory of DIY...you can twiddle to your heart's content. How can the chip guys stand not being able to control things?
Lumanauw,
I come from a tube background, so I find FETs more intuitive, being depletion devices, etc. Bipolars drive me crazy, requiring current into the base, thermal runaway, and such. Positively unnatural, they are.
Assuming that you're using IRF MOSFETs in the output, I'd stick with 392 ohms unless you're prepared to adjust the current source as Steve did. On the other hand, if you're wanting to go to 50mA, then you'll almost be forced to use a lower load resistance. (There are other options--you could cap couple to the back end and set bias separately, for instance.) Higher current is a good thing. Higher voltage is a good thing. The high heat that results has to be watched carefully. Just don't burn up your inputs. The gain will change, depending on what current and load resistance you choose. Don't lose sight of the various parameters you're juggling: front end gain, bandwidth, distortion, heat dissipation, and back end bias. It's not that it can't be done, it's that you have to stay alert. And, no, all other things being equal, it won't hurt the sound quality to raise the front end bias current.
Grey
Ah, but you changed two parameters, not just one. By increasing the bias current through the front end, you increased the gain. This offset the loss due to lower load resistance. Now, given the way Nelson set the amp up, you pretty much have to do something of that nature in order to bias the output stage, but it's equivalent to reaching inside an opamp and fiddling the internals.
Ah, the glory of DIY...you can twiddle to your heart's content. How can the chip guys stand not being able to control things?
Lumanauw,
I come from a tube background, so I find FETs more intuitive, being depletion devices, etc. Bipolars drive me crazy, requiring current into the base, thermal runaway, and such. Positively unnatural, they are.
Assuming that you're using IRF MOSFETs in the output, I'd stick with 392 ohms unless you're prepared to adjust the current source as Steve did. On the other hand, if you're wanting to go to 50mA, then you'll almost be forced to use a lower load resistance. (There are other options--you could cap couple to the back end and set bias separately, for instance.) Higher current is a good thing. Higher voltage is a good thing. The high heat that results has to be watched carefully. Just don't burn up your inputs. The gain will change, depending on what current and load resistance you choose. Don't lose sight of the various parameters you're juggling: front end gain, bandwidth, distortion, heat dissipation, and back end bias. It's not that it can't be done, it's that you have to stay alert. And, no, all other things being equal, it won't hurt the sound quality to raise the front end bias current.
Grey
"I come from a tube background, so I find EFTs more intuitive, being depletion devices, etc."
Ummm.......... The IR. MOSFETs under discussion here are enhancement mode devices.
The second way in which a MOSFET can operate is called enhancement mode. When there is no voltage on the gate, there is in effect no channel, and the device does not conduct. A channel is produced by the application of a voltage to the gate. The greater the gate voltage, the better the device conducts.
http://whatis.techtarget.com/definition/0,,sid9_gci214511,00.html
"Steve,
Ah, but you changed two parameters, not just one. By increasing the bias current through the front end, you increased the gain. This offset the loss due to lower load resistance."
Not quite.... If you half the load resistance and double the bias current in the front end you will only wind up with the same gain
if the transconductance of the fets doubles at twice the bias current. I believe with the IRF9610s half the load resistance and double the bias current losses a little under 3dB in gain.
"Anyway, at that point it becomes a question of simple subtraction: 40dB (open loop gain)- 20dB (negative feedback)= 20dB(closed loop gain). Anything that subtracts from the available open loop gain is going to show up as a reduction in the closed loop gain of the final circuit."
Actually more precisely the closed loop gain is determined largely by the feedback resistors, so the amount of negative feedback is the open loop gain minus the closed gain. For a fairly large amount of open loop gain, changes in the amount of negative feedback make small differences in the closed loop gain.
"Ah, the glory of DIY...you can twiddle to your heart's content.'
Or you can learn a little theory, and make circuit changes with some insight of the influence on performance. It could save you a whole lot of time over trying everything, by having some basic ideas of what you are trying to achieve. Most successful trips start with a map.
Ummm.......... The IR. MOSFETs under discussion here are enhancement mode devices.
The second way in which a MOSFET can operate is called enhancement mode. When there is no voltage on the gate, there is in effect no channel, and the device does not conduct. A channel is produced by the application of a voltage to the gate. The greater the gate voltage, the better the device conducts.
http://whatis.techtarget.com/definition/0,,sid9_gci214511,00.html
"Steve,
Ah, but you changed two parameters, not just one. By increasing the bias current through the front end, you increased the gain. This offset the loss due to lower load resistance."
Not quite.... If you half the load resistance and double the bias current in the front end you will only wind up with the same gain
if the transconductance of the fets doubles at twice the bias current. I believe with the IRF9610s half the load resistance and double the bias current losses a little under 3dB in gain.
"Anyway, at that point it becomes a question of simple subtraction: 40dB (open loop gain)- 20dB (negative feedback)= 20dB(closed loop gain). Anything that subtracts from the available open loop gain is going to show up as a reduction in the closed loop gain of the final circuit."
Actually more precisely the closed loop gain is determined largely by the feedback resistors, so the amount of negative feedback is the open loop gain minus the closed gain. For a fairly large amount of open loop gain, changes in the amount of negative feedback make small differences in the closed loop gain.
"Ah, the glory of DIY...you can twiddle to your heart's content.'
Or you can learn a little theory, and make circuit changes with some insight of the influence on performance. It could save you a whole lot of time over trying everything, by having some basic ideas of what you are trying to achieve. Most successful trips start with a map.
Huh!?!
" 'Anyway, at that point it becomes a question of simple subtraction: 40dB (open loop gain)- 20dB (negative feedback)= 20dB(closed loop gain). Anything that subtracts from the available open loop gain is going to show up as a reduction in the closed loop gain of the final circuit.'
Actually more precisely the closed loop gain is determined largely by the feedback resistors, so the amount of negative feedback is the open loop gain minus the closed gain. For a fairly large amount of open loop gain, changes in the amount of negative feedback make small differences in the closed loop gain."
Did I catch that right ...
Grey says: "open loop gain - negative feedback = closed loop gain"
Fred says: "negative feedback = open loop gain minus closed loop gain"
Beyond a superficial rearrangement of a simple arithmatic equation, how is this different?
"For a fairly large amount of open loop gain, changes in the amount of negative feedback make small differences in the closed loop gain."
That seems at first glance to be a contradiction.
OK Fred, I know your going to take this as an attack but please trust me, it is not! You're trying to make a point, I want to understand it, I'm just not getting it. Could you elaborate?
Metalman
" 'Anyway, at that point it becomes a question of simple subtraction: 40dB (open loop gain)- 20dB (negative feedback)= 20dB(closed loop gain). Anything that subtracts from the available open loop gain is going to show up as a reduction in the closed loop gain of the final circuit.'
Actually more precisely the closed loop gain is determined largely by the feedback resistors, so the amount of negative feedback is the open loop gain minus the closed gain. For a fairly large amount of open loop gain, changes in the amount of negative feedback make small differences in the closed loop gain."
Did I catch that right ...
Grey says: "open loop gain - negative feedback = closed loop gain"
Fred says: "negative feedback = open loop gain minus closed loop gain"
Beyond a superficial rearrangement of a simple arithmatic equation, how is this different?
"For a fairly large amount of open loop gain, changes in the amount of negative feedback make small differences in the closed loop gain."
That seems at first glance to be a contradiction.
OK Fred, I know your going to take this as an attack but please trust me, it is not! You're trying to make a point, I want to understand it, I'm just not getting it. Could you elaborate?
Metalman
Actually I was trying to make a point....
"Actually more precisely the closed loop gain is determined largely by the feedback resistors, so the amount of negative feedback is the open loop gain minus the closed gain. For a fairly large amount of open loop gain, changes in the amount of negative feedback make small differences in the closed loop gain"
I really wasn't to be cute but trying to make a bit clearer. The close loop gain is the target gain of what you are trying shoot for with the amp design. The amount of open loop gain you design in how determines much negative feedback you wind up with. With a pretty good amount of open loop gain increasing the open loop by 6dB will make a fairly small change in the closed loop gain for the same value of feedback resistors. It will half the output impedance though. I wasn't trying to contradict or confuse but just to look at it in the way useful to me during an amp design, that is changing the amount open loop gain to get a given amount negative feedback for the desired closed loop gain one starts with. The other description looked a bit like the tail wagging the dog to me and i thought might be a bit confusing to the novice. It could be be just a matter of perpective and the impotant thing is knowing what closed loop gain and open loop gain are and that the amount negative feedback is the relation between the two. subtraction of the closed loop gain from open loop gain is only usefull when looking at the values in dB voltage gain of course.
"Actually more precisely the closed loop gain is determined largely by the feedback resistors, so the amount of negative feedback is the open loop gain minus the closed gain. For a fairly large amount of open loop gain, changes in the amount of negative feedback make small differences in the closed loop gain"
I really wasn't to be cute but trying to make a bit clearer. The close loop gain is the target gain of what you are trying shoot for with the amp design. The amount of open loop gain you design in how determines much negative feedback you wind up with. With a pretty good amount of open loop gain increasing the open loop by 6dB will make a fairly small change in the closed loop gain for the same value of feedback resistors. It will half the output impedance though. I wasn't trying to contradict or confuse but just to look at it in the way useful to me during an amp design, that is changing the amount open loop gain to get a given amount negative feedback for the desired closed loop gain one starts with. The other description looked a bit like the tail wagging the dog to me and i thought might be a bit confusing to the novice. It could be be just a matter of perpective and the impotant thing is knowing what closed loop gain and open loop gain are and that the amount negative feedback is the relation between the two. subtraction of the closed loop gain from open loop gain is only usefull when looking at the values in dB voltage gain of course.
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