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Chip Amps Amplifiers based on integrated circuits 
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28th December 2012, 04:01 AM  #1 
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Join Date: Aug 2012

OPA549 max psu rating for 4 ohm load?
I want to use OPA549 into 4 ohm load so what is the best possible psu rating since the max is at +/ 30V can i go till +/25v
Here the load is 4 ohm thats what Im very much concern about... 
28th December 2012, 04:15 AM  #2 
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Join Date: Nov 2006
Location: Indiana

Can you provide a link to the manufacturer's datasheet?

28th December 2012, 04:17 AM  #3 
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30th December 2012, 06:40 AM  #4 
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Join Date: Nov 2006
Location: Indiana

It looks like the chip is able to handle 8 Amps of output current, continuously, if properly heatsinked, and 10 Amps (max) peak output current.
Even with a continous DC output voltage of 30V across 4 Ohms, the output current would be less than 8 Amps (30 V / 4 Ohms = 7.5 Amps). I've never used that chip but unless I'm missing something it looks like you should be able to use any PSU voltage up to and including the maximumspecified +/ 30V, with a 4 Ohm load, without any danger of exceeding the maximum continuous output current rating. But you will also want to ensure that the peak output current cannot try to go above 10 Amps. You could try to find out what your speaker's lowest impedance might be, which will be frequencydependent. If they drop much below 4 Ohms at some frequency, then you could figure the max psu voltage with V / (min load impedance) <= 10. (Or you could use <= 8, to be safer.) It's often better to run chipamps at something less than their absolute maximum supply voltage. So your +/25V might be better than +/30. 
30th December 2012, 06:46 AM  #5 
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depending on the size of your power traffo, that +/ 30V might drop at full volume to +/25v .....
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30th December 2012, 08:45 PM  #6  
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Quote:
Unfortunately, becuase of the +/30V absolute maximum rating for the power supply inputs, that really complicates using the chip close to the max rating. First of all, if you use an UNregulated power supply, then your transformer should be rated for an output voltage of no more than 22 VRMS. Even with that, if your line voltage goes too high by a little more than 1 %, your rails will go over 30 V, when there is almost no load (volume turned way down, for example, or, with no source material playing). If you did use a 22 VRMS transformer, rated at 150 VA, per channel, and somehow managed to not go over the +/30V voltagerail limit, you could expect to be able to produce no more than about 50 Watts per channel into 4 Ohms, without clipping, even with a fairlyhefty 30000 uF reservoir capacitance on each rail. And you would mostlikely not want to push it past about 40 Watts, to have a small (10%) safety margin for clipping. See the attached screen image. [Before anyone says that the output power could be higher, don't forget that even if you assume no ripple, you have to subtract the voltage across the amplifier itself, which has to sit between the top of the output voltage and the bottom of the ripple voltage. Typically you might have a minimum of VCE0, there, across collector and emitter pins, plus the voltage across a small emitter resistor; figure at least three and maybe four volts that are unavailable for the signal voltage, no matter how small the ripple is. If the ripple does dip into the amp's voltage range, it takes gouges right out of the output signal waveform.] That screen image was produced with the spreadsheet that can be dowloaded at: Power Supply Resevoir Size That's the Excel 2007 version. If you have an older version of Excel, try the Excel 972003 version at: Power Supply Resevoir Size So, without a regulator, to be FAIRLY safe (but still not really close to being "guaranteed" safe), you could plan for a maximum of about 13% overvoltage on the power line, and use a transformer with a 20 VRMS output voltage (assuming that's the noload rating!), rated at 140 VA or more, which would give a peak rail voltage of around 26.5 Volts (with almost no load and no overvoltage on the AC line) and enable you to produce a maximum of only 40 Watts, which you probably wouldn't usuallly want to push past about 35 Watts. SO, apparently, if you want an amplifier that can do more than 3540 Watts and not violate the +/30V railvoltage limit of the OPA549, then you should use a regulated power supply. i.e. Make a supply that can produce a higher voltage than you need and then add a threeterminal voltage regulator to each rail, after the smoothing caps, to limit the voltage to something like 29 or 29.5 Volts. With 29 to 29.5 Volts, you could get around 75 Watts before clipping became imminent. But your regulator would need to be able to provide over 6.1 Amps peak (4.33 Amps RMS). Something like an LM338 should work for that. They are rated to do 5 Amps continuous and 710 Amp transients. (There might be other newer regulators that I am not aware of, that might be better. See linear.com and national.com [now autorouted to ti.com]) The regulator's input voltage (if LM338) should be 3 V higher than you want the output voltage to be, plus a safety margin for power line sags, etc, and a safety margin for transformer voltage sags under load (use more reservoir capacitance to help with that). So it looks like a transformer rated for 30 VRMS output voltage might work well for a regulated 29to29.5Volt supply, using regulators like the LM338. You can use two "positive" voltage regulators to make one positive rail and one negative rail, if you configure the power supply circuit like the one shown at: Spice Component and Circuit Modeling and Simulation OR Spice Component and Circuit Modeling and Simulation In those schematics, ignore the second set of regulators in the first one, and ignore the resistors in series with the main capacitors. And you can probably also ignore the inrushcurrent limiter and the snubbers, etc. The main thing is that you could use one secondary (of a dualsecondary transformer) for each rail, and a bridge rectifier for each rail, and an LM338 per rail, and make a +/29 to 29.5 Volt power supply, using a similar circuit. You will need to use enough reservoir capacitance per rail to try to guarantee that the minimum of the ripple voltage waveform can never dip down and violate the regulator's dropout spec. Otherwise the regulator's output voltage could get very ugly. if we assume an initial rail voltage of 30 VRMS x sqrt(2) minus 3.5 Volts for large diode drops = about 39 Volts, then subtract 10% for a possible low AC line = 35V, and we assume a 6Amp load current (we could probably reasonably use 5 Amps, instead), and we never want to go below 29.5 + 3V dropout spec = 32.5V, then our ripple voltage amplitude could be allowed to be up to 2.5 Volts, at the very most. Knowing the max load current and the desired ripple amplitude, we can estimate the needed reservoir capacitance if we also know the time between capacitor charging pulses, which is 1/2 the period of the AC line voltage (1/2 of 1/60 or 1/50). So, in this case (assuming you're in the USA), C = imax/(2xfmains*ripple amplitude) = 6/(120*2.5) = 0.02 Farads = 20000 uF, at a minimum. You might want to add 50% to that. Cheers, Tom Last edited by gootee; 30th December 2012 at 09:12 PM. 

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