Netlist
The spreadsheet for the 8A entry described above indicates that each of the 4 FETS will burden 4A each! This is the same for both the 1.0 and 1.2 version of the spreadsheet. When you say use a little more to stay in the SOA is that to mean double the amps/power dissipation? That seems peculiar to me
The spreadsheet for the 8A entry described above indicates that each of the 4 FETS will burden 4A each! This is the same for both the 1.0 and 1.2 version of the spreadsheet. When you say use a little more to stay in the SOA is that to mean double the amps/power dissipation? That seems peculiar to me
I meant:
Use a couple more (mosfets) to stay in the SOA.
Of course you'll always need multiples of four in the AlephX.
4,8,12,16 and so on.
The SOA would be something like max 30W per device so for 8A/22V supply you need 12 mosfets, for 10A/22V this would be 16.
Please do take a look at the spreadsheet. Many cells have a small red mark in the top right corner. Just moving the mouse into these cells will reveal a lot more info as well.
/Hugo
Use a couple more (mosfets) to stay in the SOA.
Of course you'll always need multiples of four in the AlephX.
4,8,12,16 and so on.
The SOA would be something like max 30W per device so for 8A/22V supply you need 12 mosfets, for 10A/22V this would be 16.
Please do take a look at the spreadsheet. Many cells have a small red mark in the top right corner. Just moving the mouse into these cells will reveal a lot more info as well.
/Hugo
Netlist
Excellent. I will be using the 1.2 version of the spreadsheet for our analysis. Earlier in this thread we agreed that cell B13 indicates the total current that must be burdened by the FETs so that an 8A entry with 4 FETs as per the original hifizen boards will result in 2A through each FET and require 250mΩ source resistors. Plug in those numbers and you will see that the source resistor value is half and the current values double!
Excellent. I will be using the 1.2 version of the spreadsheet for our analysis. Earlier in this thread we agreed that cell B13 indicates the total current that must be burdened by the FETs so that an 8A entry with 4 FETs as per the original hifizen boards will result in 2A through each FET and require 250mΩ source resistors. Plug in those numbers and you will see that the source resistor value is half and the current values double!
Coulomb
Nice of you to joing the party. The spreadsheet clearly states "total bias for one channel being one monoblock". One monoblock includes both current source and gain FETs so clearly this is an unfortunate description. It is amazing to me how many people accept what is written for gospel
Nice of you to joing the party. The spreadsheet clearly states "total bias for one channel being one monoblock". One monoblock includes both current source and gain FETs so clearly this is an unfortunate description. It is amazing to me how many people accept what is written for gospel
Netlist
I can't even reconcile his power output numbers to PSUD2 and since wuffwaff has decided to ignore me rather than address my questions directly, I'd rather not try and interpret his meaning. For the record, one channel or one monoblock make no difference to the problem I've exposed here. I admit that I don't understand what he has done with this spreadsheet and why he insists on calculating approximately twice the bias required for a given impedence load. If I try and get any further into his head I may very well go mad
I can't even reconcile his power output numbers to PSUD2 and since wuffwaff has decided to ignore me rather than address my questions directly, I'd rather not try and interpret his meaning. For the record, one channel or one monoblock make no difference to the problem I've exposed here. I admit that I don't understand what he has done with this spreadsheet and why he insists on calculating approximately twice the bias required for a given impedence load. If I try and get any further into his head I may very well go mad
This is by far not the first time this spreadsheet has been questioned. I like that because a mistake is easily made.
It's up here for more then a year and reviewed by a countless amount of people.
Wuffwaff indeed says:
"Some assumptions are made to make calculations
easier so it won’t be precise to the nth degree but error is small enough to make it useful."
You can be pretty sure there are no mistakes in the calculations.
Of course, you are free to provide us with another sheet, post it and let other users review it. This too has been done in the past.
/Hugo
It's up here for more then a year and reviewed by a countless amount of people.
Wuffwaff indeed says:
"Some assumptions are made to make calculations
easier so it won’t be precise to the nth degree but error is small enough to make it useful."
You can be pretty sure there are no mistakes in the calculations.
Of course, you are free to provide us with another sheet, post it and let other users review it. This too has been done in the past.
/Hugo
yldouright said:Netlist
I believe that requiring nearly double the bias required to stiffly drive a given impedence is not a trivial error but I guess I'm in the minority
If I understand your question correctly, the answer is that each side of the amplifier has to carry the full bias necessary to drive the load.
Look at it from a different angle. For an 8 ohm load, driven with an 8Vrms sine wave (each side of the amp 4Vrms with respect to ground, 180 degrees out of phase), each side must be able to provide the necessary peak current through the load, which is 1.41 amps (1 A rms). Without the aleph current source, this would be 2.42 amps total bias current. With an AC current gain of 50%, each side must only have a 0.705 amp bias, for 1.41 amps total bias.
In order to extend this example back to the spreadsheet, the minimum potential for a balanced amplifier to drive an 8ohm load with an 8Vrms sine wave is +/-4*SQRT(2), or +/-5.6V. Add in the 2V overhead assumed in the spreadsheet and the supply voltages must be +/-7.6V. Check the power vs. load chart and you'll see the peak power is at 8 ohms (input parameters, 7.6V supply, 1.41 amps total bias, 50% current gain) as you would expect from this calculation if the spreadsheet is working correctly.
In this case you will dissipate 8W in the load and 10W in the amplifier for 18W total power.
Harkening back to an earlier post, if you have 8A total bias and 4 mosfets, the bias current running through each transistor will be 4A; two FETs on each side are in series so the total bias gets split only two ways and not four.
Jeremy
kropf
I understand what you are presenting and let's say for the moment I accept it, the power figures derived from the calculator don't appear to reflect the reality. Say I have speaker with a worst case impedence dip of 2Ω and I build my amp with 20V rails and 8A of total bias (now described as 1/2 of the channel). According to my numbers and those presented by PSUD2 that amp will make around 120W into that load. At 4Ω it'll deliver 90W and at 8Ω it'll make 60W. IOW, the amp won't sag power into the lower load. Now plug those numbers into the calculator and see what happens. Much more pessimistic isn't it? Now, to someone relying on this spreadsheet and building by its specs to meet a minimum impedence of 2Ω, they will require twice the total bias! Now do you understand the problem? The spreadsheet is flawed because it forces you to sacrifice voltage swing for current delivery you'll never use. I also have problems with the AC current gain calculator but let's kill the elephants before we go chasing the mice.
I understand what you are presenting and let's say for the moment I accept it, the power figures derived from the calculator don't appear to reflect the reality. Say I have speaker with a worst case impedence dip of 2Ω and I build my amp with 20V rails and 8A of total bias (now described as 1/2 of the channel). According to my numbers and those presented by PSUD2 that amp will make around 120W into that load. At 4Ω it'll deliver 90W and at 8Ω it'll make 60W. IOW, the amp won't sag power into the lower load. Now plug those numbers into the calculator and see what happens. Much more pessimistic isn't it? Now, to someone relying on this spreadsheet and building by its specs to meet a minimum impedence of 2Ω, they will require twice the total bias! Now do you understand the problem? The spreadsheet is flawed because it forces you to sacrifice voltage swing for current delivery you'll never use. I also have problems with the AC current gain calculator but let's kill the elephants before we go chasing the mice.
Although I haven't looked at any of the spreadsheets here on the site, the basic concept of current vs. voltage swing is easy enough.
Start with a regular Aleph. Make it a simple one like the Mini-A, so that what's going on is really obvious. The output stage--in this case a single output/CCS pair--swings all of the output voltage and half of the current. It sees all of the load impedance, all the time. A bridged amp, such as the Aleph-X--and again let's keep it simple with a single output pair per side--doesn't see all the load. Each side of the circuit sees half of the load. This means that if you're running a nominal 8 ohm load each side of the amp sees only 4 ohms. This will require twice the bias current per side. The flip-side of this is that each side only has to swing half the voltage The laws of physics aren't broken, just restated, trading the terms voltage and current. The overall efficiency of the circuit is the same as an equivalent Aleph. It will dissipate roughly three times the RMS output as heat.
Before getting all wrung in the withers over current gain calculations, note that the actual resistor value will change depending on the environment of the output devices due to the fact that MOSFETs react somewhat to heat. Don't assume that an arbitrarily calculated resistor value is going to be the right one. The easiest way to determine the correct resistor value is to use a pot, set it, then measure the pot to see what value to use. That way you'll know.
Ditto for bias current.
I wrote up a spreadsheet after calculating the stuff by hand a number of times. Although I threw a bunch of things into the spreadsheet, the only thing I ever really used it for was to estimate device dissipations for different numbers of devices. Given that you want, say, 50 watts, do you want one pair per side, two, three? It got tedious so I started letting the computer handle the repetitive stuff, even though the calculations themselves were trivial.
Grey
Start with a regular Aleph. Make it a simple one like the Mini-A, so that what's going on is really obvious. The output stage--in this case a single output/CCS pair--swings all of the output voltage and half of the current. It sees all of the load impedance, all the time. A bridged amp, such as the Aleph-X--and again let's keep it simple with a single output pair per side--doesn't see all the load. Each side of the circuit sees half of the load. This means that if you're running a nominal 8 ohm load each side of the amp sees only 4 ohms. This will require twice the bias current per side. The flip-side of this is that each side only has to swing half the voltage The laws of physics aren't broken, just restated, trading the terms voltage and current. The overall efficiency of the circuit is the same as an equivalent Aleph. It will dissipate roughly three times the RMS output as heat.
Before getting all wrung in the withers over current gain calculations, note that the actual resistor value will change depending on the environment of the output devices due to the fact that MOSFETs react somewhat to heat. Don't assume that an arbitrarily calculated resistor value is going to be the right one. The easiest way to determine the correct resistor value is to use a pot, set it, then measure the pot to see what value to use. That way you'll know.
Ditto for bias current.
I wrote up a spreadsheet after calculating the stuff by hand a number of times. Although I threw a bunch of things into the spreadsheet, the only thing I ever really used it for was to estimate device dissipations for different numbers of devices. Given that you want, say, 50 watts, do you want one pair per side, two, three? It got tedious so I started letting the computer handle the repetitive stuff, even though the calculations themselves were trivial.
Grey
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.