wuffwaff
I'm with MikeW on this one. I put in the following:
Voltage=20V
Bias=8A
AC Gain=50%
Mosfets=4
Peak Current=8A
I don't understand why the dissipation per FET is 80W when the dissipation is 20V x 2A (ie:8A/4). Is your calculator assuming that the bias is only burdened by 2 of the four FET's in this the case above?
I'm with MikeW on this one. I put in the following:
Voltage=20V
Bias=8A
AC Gain=50%
Mosfets=4
Peak Current=8A
I don't understand why the dissipation per FET is 80W when the dissipation is 20V x 2A (ie:8A/4). Is your calculator assuming that the bias is only burdened by 2 of the four FET's in this the case above?
Hi,
the voltage is symmetrical so total voltage from plus to minus would be 40V. 40Vx8A=320 Watt is 80W/fet
After rereading your question I can also give you the following answer:
the current is not devided by 4 but by 2, so 4A is flowing in each half. It flows through the active current source into the output stage. This means 4A per fet with each fet only seing half the supply voltage: P=4x20=80 watt
William
the voltage is symmetrical so total voltage from plus to minus would be 40V. 40Vx8A=320 Watt is 80W/fet
After rereading your question I can also give you the following answer:
the current is not devided by 4 but by 2, so 4A is flowing in each half. It flows through the active current source into the output stage. This means 4A per fet with each fet only seing half the supply voltage: P=4x20=80 watt
William
wuffwaff
Thanks for your reply. I have been trying to reconcile what PSUD2 outputs to your spreadsheet so if I am to understand you correctly, the FET will dissipate double the voltage rail times the amps going through the FET, correct? I had been told that the 50% AC gain reduces the dissipation back down to the bias times the rail. I just read your edit:
By this new information, it would appear that each FET will now dissipate 160W since this new model indicates 40V times 4A. Something is wrong here. When you say that each FET only sees half the voltage, do you mean to say that is only sees +/-10V of the 20V rail or only the positive or negative side of the swing?
Thanks for your reply. I have been trying to reconcile what PSUD2 outputs to your spreadsheet so if I am to understand you correctly, the FET will dissipate double the voltage rail times the amps going through the FET, correct? I had been told that the 50% AC gain reduces the dissipation back down to the bias times the rail. I just read your edit:
By this new information, it would appear that each FET will now dissipate 160W since this new model indicates 40V times 4A. Something is wrong here. When you say that each FET only sees half the voltage, do you mean to say that is only sees +/-10V of the 20V rail or only the positive or negative side of the swing?
AudioFreak
Thanks for sounding in but your reply doesn't address how the alledged 80W are produced. Isn't the voltage swing 40V as wuffwaff suggests? If the amps in that case are 4 per FET then why isn't the dissipation 160W? How does the AC current gain factor into the expected power dissipation or isn't it a factor at all? I should also point out that many of the members whom I have interviewed that have built the Aleph-X experience junction and heatsink temperatures that suggest the dissipation is usually a litte more than half of what the wuffwaff spreadsheet and you claim in your post and I haven't even started to address the disparities between the power numbers indicated in the wuffwaff spreadsheet graph with those presented by PSUD2. Your curt and definitive reply indicates that all this must be very elementary to you but perhaps you can address my questions more completely in the future?
Thanks for sounding in but your reply doesn't address how the alledged 80W are produced. Isn't the voltage swing 40V as wuffwaff suggests? If the amps in that case are 4 per FET then why isn't the dissipation 160W? How does the AC current gain factor into the expected power dissipation or isn't it a factor at all? I should also point out that many of the members whom I have interviewed that have built the Aleph-X experience junction and heatsink temperatures that suggest the dissipation is usually a litte more than half of what the wuffwaff spreadsheet and you claim in your post and I haven't even started to address the disparities between the power numbers indicated in the wuffwaff spreadsheet graph with those presented by PSUD2. Your curt and definitive reply indicates that all this must be very elementary to you but perhaps you can address my questions more completely in the future?
40 v is the maximum potential swing, but we are talking about no signal conditions (worst case for class A amp heat dissipation)
Since you have 4 amps per side, and we'll assume that your absolute DC offset is near 0, each transistor sees 20 volts. Since 4 amps flows through each transistor, you have 80 watts per transistor.
Look at the junction temperature that produces with a real heatsink. Yikes!
Since you have 4 amps per side, and we'll assume that your absolute DC offset is near 0, each transistor sees 20 volts. Since 4 amps flows through each transistor, you have 80 watts per transistor.
Look at the junction temperature that produces with a real heatsink. Yikes!
yldouright said:
It is the bias current that determines dissipation, the AC current gain simply determines the ratio between bias current and peak current for each half of the circuit (each half essentially being a standard Aleph output stage). Since we are talking about dissipation at idle, the rails are evenly shared between the 2 FETs ie. 20V across each FET.
Junction temperatures cannot be measured, since the junction is inside the case.
I used the spreadsheet to predict temperatures of my mini A by using 4 FETS, twice the real bias and half the heatsink K. figuring half the power and twice the heatsink K ought to come up with the same temperature.
Heat sink temperatures and seem pretty close to predicted once you properly factor in the resistance across the insulator. (tab temp - HS temp)/fet dissipation gives a K that is divided in two, entered in B66/67. I measured 1.2 deg/watt with pink silpads, and 1.0 deg/watt with lightly greased aluminum oxide isolators, at 17 watts dissipation per fet.
In the A75 articles, NP explains quite nicely how to calculate your heat sink requirements. Just treat thermal resistance like electrical resistance. heat sink to air is a single resistance in series with the cross isolator reistances paralleled in series with the Junction to case resistances in parallel. multiply the net resistance by the total heat load, you'll get the junction temperature rise above ambient. This is what the spreadsheet does for you, with some extra features thrown in.
hope this helps
I used the spreadsheet to predict temperatures of my mini A by using 4 FETS, twice the real bias and half the heatsink K. figuring half the power and twice the heatsink K ought to come up with the same temperature.
Heat sink temperatures and seem pretty close to predicted once you properly factor in the resistance across the insulator. (tab temp - HS temp)/fet dissipation gives a K that is divided in two, entered in B66/67. I measured 1.2 deg/watt with pink silpads, and 1.0 deg/watt with lightly greased aluminum oxide isolators, at 17 watts dissipation per fet.
In the A75 articles, NP explains quite nicely how to calculate your heat sink requirements. Just treat thermal resistance like electrical resistance. heat sink to air is a single resistance in series with the cross isolator reistances paralleled in series with the Junction to case resistances in parallel. multiply the net resistance by the total heat load, you'll get the junction temperature rise above ambient. This is what the spreadsheet does for you, with some extra features thrown in.
hope this helps
BobEllis
It seems unlikely to me that there exists no means to measure junction temperature or at least estimate it by using a passive probe but lets accept your answer for now. If you used the wuffwaff spreadsheet, why did you need to input "twice the real bias" or for that matter, half the heatsink factor if the formulas in the spreadsheet were correct in the first place? I have read that article but it does not address the questions presented here about which values are the correct ones to use in determining the expected dissipation in the Aleph-X and how those numbers are calculated.
AudioFreak
I am always delighted to read your posts. I was hoping that your reply would be less obfuscating. It sounds like you are saying that the bias current is the only determining factor of the dissipation and that voltage has nothing to do with it and that seems incorrect to me
So let's be clear, are you saying that wuffwaff was incorrect when stating that the mosfet will see the whole voltage swing?
It seems unlikely to me that there exists no means to measure junction temperature or at least estimate it by using a passive probe but lets accept your answer for now. If you used the wuffwaff spreadsheet, why did you need to input "twice the real bias" or for that matter, half the heatsink factor if the formulas in the spreadsheet were correct in the first place? I have read that article but it does not address the questions presented here about which values are the correct ones to use in determining the expected dissipation in the Aleph-X and how those numbers are calculated.
AudioFreak
I am always delighted to read your posts. I was hoping that your reply would be less obfuscating. It sounds like you are saying that the bias current is the only determining factor of the dissipation and that voltage has nothing to do with it and that seems incorrect to me
So let's be clear, are you saying that wuffwaff was incorrect when stating that the mosfet will see the whole voltage swing?
I didn't ask the author (Wuffwaff) but as small co-author I would like to see the AXE-1 adapted for OpenOffice.
Only the graph would have to be redone.
So if anyone with some experience in that matter could help us?
I know the sheet is write protected but that can be arranged privately with possible candidates.
/Hugo
Only the graph would have to be redone.
So if anyone with some experience in that matter could help us?
I know the sheet is write protected but that can be arranged privately with possible candidates.
/Hugo
Hi Hugo,
that is no problem but I would wait a few days until my AXE-1.2 is ready. It is the same sheet with some added features, a few more digits in some places and a mistake removed pointed out to me by Kropf (will change the max power dissipation possible at a given junction temperature, here the wrong cell is refered to giving too high values but of no further consequence as it is only an info point))
William
that is no problem but I would wait a few days until my AXE-1.2 is ready. It is the same sheet with some added features, a few more digits in some places and a mistake removed pointed out to me by Kropf (will change the max power dissipation possible at a given junction temperature, here the wrong cell is refered to giving too high values but of no further consequence as it is only an info point))
William
yldouright said:
The junction we are referring to here is the semiconductor junction on the chip inside the device. Sure, you measure junction temperature by drilling a hole in the case without destroying the chip. Failing that you can measure the tab temperature and add to that the device's thermal resistance (from datasheet) x power dissipated to estimate junction temperature as WuffWaff does in the spreadsheet.
As far as output device heatsinking is concerned, an aleph is half of an AX. So if I want to calculate the temperature rise of an Aleph using the AX spreadsheet without rewriting it, I can simulate my results as a pair of alephs on a single power supply. So double the number of output devices (draws twice the current) and double the heatsink (halves the thermal K). If you cannot accept this, just plug the values into the formulas in the articles mentioned.
I am NOT saying that WuffWaff was wrong, each device will see a voltage swing coming close to twice the rail voltage. IIRC, the sheet uses a max swing within 2 volts of the rails to determine power output. NP has said that you're more likely to only get to within 6 volts of the rails due to the Vgs of the output devices. So, voltage limited output estimated by the spreadsheet may be a little high.
However, the maximum voltage swing the output devices see is irrelevant to determining heat sink requirements. In a class A amplifier, the worst case is idle. Since the transistors never cut off, any current delivered to the load means that less is being sourced or sunk by the output stage. this lowers the dissipation of the stage.
To simplify we'll use 20 volt rails and an aleph biased at four amps and a non reactive load. Take a "snapshot" in time when the output swings to the positve rail (+20). The source FET(s) carry all of the current, but there is no voltage across it. The 8 ohm load carries 2.5 amps. The Gain Fet(s) carry 1.5 amps, the rest of the bias. The gain FET sees 40 volts at 1.5 amps, for 60 watts dissipation in the stage. In reality, you won't get 20 volts peak, and the current sources will always dissipate some heat. Pick any output voltage you want, as long as the bias supports class A operation you'll have less total dissipation than at idle.
At idle both current source and gain device see 20 volts and carry 4 amps for 80 watts dissipation each. That's 160 total. A bit worse than the maximum voltage swing case.
So, for class A operation we size our heat sinks for idle conditions.
kapish?
BobEllis said:
Hi Bob,
the nice thing about the bootstrapping cap is that you can even get above power supply voltage at the gates
What Nelson did say is that it is probably better to stay 3V or so away from the output voltage wich is easy to say if you have voltage to spare
I can change it to 3V or make a new input although I think it is better to keep it simple as it is difficult enough already.
William
just measured my Aleph-X with 22V / 8A.
It gives around 28.5 volts or 40.3V peak wich is more or less 2V away from the rail.
For 4 Ohms peak voltage was 38V wich is 4 volts from the rail.
At 2 Ohms it doubled again to a bit over 8 volts (not current limited!).
I will change this in the next AXE-1 sheet to get a better result at lower impedances.
William
It gives around 28.5 volts or 40.3V peak wich is more or less 2V away from the rail.
For 4 Ohms peak voltage was 38V wich is 4 volts from the rail.
At 2 Ohms it doubled again to a bit over 8 volts (not current limited!).
I will change this in the next AXE-1 sheet to get a better result at lower impedances.
William
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