Understanding output stage SOA??
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 21st November 2007, 02:21 PM #1 Zero Cool   diyAudio Member     Join Date: Sep 2004 Location: MN Understanding output stage SOA?? I am working on a high power amplifier at the moment and i am studying the output stage. I am trying to understand how a designer chooses devices and the number of them for a particular design. for example. the amp i am working on now has 16 output devices per channel MJ21194/93 devices 250v 16 amp rated devices. Studying the spec sheet i was looking at the SOA curve and noticed that at 100V the devices is rated for about 2 amp's for 1 second and probably 1 amp for long periods. Stay with me now i am covering a lot of ground. So if each device can handle 1 amp safely at or near the voltage that is going to be used. How to you calculate the number of devices needed for a particular power output? Working the problem backwards This amp is rated for 365w/PC @ 8 ohms 650 watts @ 4 ohms and 1000 watts @ 2 ohms. the math on this equates to 6.7 amps for 8 ohms, 12.75 amps for 4 ohms and just over 22 amps at 2 ohms. So if we figured 1 amp per output device. thats 8 amps per polarity. and 16 amps total. So question #1 do you figure for only one output stage polarity? as only one bank will be conducting at a time. If that is the case then the amp crosses the SOA at 4 ohms. 8 devices x 1 amp = 8 amps If you use all output devices, in this case 16, 16 x 1 amp = 16 amps. then reaching the current rating at 4 ohms is within the SOA but at 2 ohms. we come dangerously close to the max SOA of the devices. If we pushed the devices right up to the edge of there SOA at 1.5 amps per device, we get 16x1.5= 24 amps and we are within limits again. BUT, that amp had better have some good cooling or POOF! SO.....If my reasoning here is correct. If i was calculating an amplifier with a target output power of 100 watts @ 8 ohms, 200 watts @ 4 ohms per channel. doing the math I come up with 5 amps @ 4 ohms. to be safe i would need at least 6 output devices correct? that seems like a lot. but using the same math from above and pushing the devices to 1.5 amps each. we would then only need 4 devices 2 per polarity. this still seems more likley. So Question #2 is is safe to run output devices that close to the listed SOA?? Question #3 am i totally off base and should forget all this and just go and get a beer? Zc
 21st November 2007, 04:17 PM #2 Jan Dupont   diyAudio Member     Join Date: Apr 2003 Location: Copenhagen, Denmark I will try to keep this simple Use Ohms law: Take your max output Voltage and devide it by the load resistance. Example: 50V into 2 Ohm load = 50/2 = 25A at 50Volts Check the SOA curve at 50V and calculate the no. of output devices needed to safely handle 25A ..... __________________ Free Schematic and Service Manual downloads www.audio-circuit.dk, Spare time company (just for fun): www.dupont-audio.com
 21st November 2007, 04:43 PM #3 PigletsDad   diyAudio Member     Join Date: Jul 2007 Location: South Worcestershire The key point to remember is that relevant voltage is the voltage on the transistor. So if you have say a 50V rail, and 2Ohm load, there are some simple cases. 1) Transistors hard on - 25A in load, saturation voltage on trannies (maybe 1V?) no problem. 2) Half voltage, resistive load - say 25V on load, 25V on transistor, 12.5A. Still probably fairly easy. 3) Highly reactive load at zero crossing - 0V on load, 50V on transistor, 25A current, nasty SOA problem. So handling low impedance resistive loads is much easier from an SOA point of view than highly reactive ones. For many loudspeakers, impedance minima have low phase angles, and the more reactive loads have a bigger magnitude, often somewhere up the sides of the dip. From an engineering point of view something that can drive 2Ohms resistive safely, and pretty much any 4Ohm phase angle may well be good enough, especially if a protection circuit of some sort prevents destruction under fault conditions.
 21st November 2007, 04:49 PM #4 Jan Dupont   diyAudio Member     Join Date: Apr 2003 Location: Copenhagen, Denmark Please remember that Voltage and Current phase can be 90 deg. off, meaning lots of Current but no Volts __________________ Free Schematic and Service Manual downloads www.audio-circuit.dk, Spare time company (just for fun): www.dupont-audio.com
 21st November 2007, 04:55 PM #5 megajocke   diyAudio Member     Join Date: Jan 2003 That would be one evil speaker manufacturer though!
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Join Date: Jul 2007
Location: South Worcestershire
Quote:
 Originally posted by ACD Please remember that Voltage and Current phase can be 90 deg. off, meaning lots of Current but no Volts
Yes, but not at impedance minima - extremal values are (usually) resistive.

If you have 90degree (or anywhere near it) 2 Ohm impedance, there will be some frequency where the impedance is much lower, and predominantly resistive, which is probably going to be even harder to handle!

 21st November 2007, 05:04 PM #7 KSTR   diyAudio Member     Join Date: Jul 2007 Location: Central Berlin, Germany For brief instants, with some odd speakers, one might even need to consider reverse current polarity beyond +-90deg, that is, momentary "negative output resistance". I often read to assume the DCR/2 at +-60deg current angles as a conservative worst case, continuous load. And have some headroom for large I*V spikes. - Klaus
 21st November 2007, 10:58 PM #8 Zero Cool   diyAudio Member     Join Date: Sep 2004 Location: MN Ok so i should be looking at max RMS ac voltage out and not DC power supply voltage then? correct? I was thinking i needed to look at the voltage across the device. 93V in this case. Ok so if this amplifier is rated 650 watts @ 4 ohms, that equates to 51 volts AC RMS across a 4 ohm resistive load and a current of 12.75 amps Figuring the 2 ohm load is about 25 amps as you said. i would need a minimum of 8 devices. double that number to handle PA duty nasty loads...makes sense Looking at the spec sheet the 60V SOA is almost 4 amps per device. 4x16 devices = 64 amps well within the specs under most operating conditions. Do i have this right? Now with those numbers each bank + and - can handle a 2 ohm load at the rated power. 8 devices x 4 amps = 32 amps max per bank. easily enough to handle a 2 ohm 25 amp draw. When figuring devices. it would make sense to me that i would need to calculate the current for one bank only as above. as only one bank conducts at a time and the whole load is across that bank during the time it is conducting. For example. lets rework our numbers a little bit. Our target output power is 200@8 and of course doubling down to 800w @ 2ohms 40V rms across 2 ohms = 20 amps and 800 watts Using the SOA chart from the Mj21194 spec sheet at 40V these devices can handle almost 6 amps. lets figure 5 to keep things simple. If we need to handle 20 amps for a 2 ohm load. then i would need 4 of these devices per bank. 8 total correct? 4 devices x 5 amps = 20 amps total. Ahhh if i am understanding this correctly. then im getting it... Thank you to everyone for working with me to help me understand. i am right on that edge where its coming together in my head. Zc
AndrewT
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Join Date: Jul 2004
Location: Scottish Borders
Quote:
 Originally posted by PigletsDad The key point to remember is that relevant voltage is the voltage on the transistor. So handling low impedance resistive loads is much easier from an SOA point of view than highly reactive ones. For many loudspeakers, impedance minima have low phase angles, and the more reactive loads have a bigger magnitude, often somewhere up the sides of the dip. From an engineering point of view something that can drive 2Ohms resistive safely, and pretty much any 4Ohm phase angle may well be good enough, especially if a protection circuit of some sort prevents destruction under fault conditions.
transistor Vce and Ic are what matters. There are three situations I design for:-
1,) consider continuous DC conditions for a short circuit.
2.) Look at driving a resistance equal to half your intended speaker impedance.
3.) Study the reactive load condition that you think applies to your worst case speaker. This is usually a transient condition over a narrow frequency range and can use the transient SOA. This may not apply to a sub-bass amplifier/speaker that could hold the worst case reactiveload/current for very extended periods.

You must look at all three conditions and pull out the relevant Vce vs Ic for each of them.

Now move on to BJT second breakdown.
At elevated voltage the BJT cannot carry it's full power rating. your 21193 is down to 200W for a 1S/100V current pulse. Maybe 1.5 to 1.7A continuous, but hopefully your fuses will be effective before this quessimate becomes significant.

Apply temperature de-rating to the SOA. MJL have a max. temp of 150degC whereas the MJ are 200degC devices. This makes a big difference to reliability when the amp is operated at elevated temperatures.

(16) David Eather, "A Practical Approach to Amplifier Output Design", Silicon Chip, February 1991, pp.14~18 and April 1991, pp.64~67.

read, read and re-read until you can follow exactly what David is doing in the hand calculated temperature de-rated SOAR.
Once you understand this you can then download Bensen's spreadsheet and let it do all the calculations for FET output stages.
If you want the BJT version send me an Email (it even has 21193 set up for you).

BTW,
8pair of MJ21193/4 @ Tc=70degC can do 1000W into 2ohm @ 31degree phase angle and 650W into 4ohm @ 60degree phase angle with +-88Vdc supply rails using +-15mF smoothing.
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regards Andrew T.

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