Calculating audio transformer losses?
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Jeffin90620
diyAudio Member

Join Date: Jul 2007
Location: Los Angeles
Calculating audio transformer losses?

I need some assistance in designing a transformer-coupled audio amplifier.

In the attached image, I need to calculate the output voltage of U7 so that I get 2 Vrms at Zload (a 150 Ohm resistor). How do I calculate the losses from the input to R28/R29 through T3 to Zload?

Does the primary side DCR affect anything, or is it only R27, R28 and T3’s primary impedance? Do I have to go through all the Real and Imaginary calculations, or will using T3’s primary impedance as a real number get me reasonably close (say, within 3%)?

I need the same answers for losses on the Secondary side, too.

I’d prefer to do everything in LTSpice, but Transformers are not native devices and I’d just be guessing with the values if I tried to simulate one with two linked Inductors.

Jeff
Attached Images
 Transformer and Driver.gif (29.4 KB, 86 views)

 16th November 2012, 07:45 PM #2 jcx   diyAudio Member   Join Date: Feb 2003 Location: .. the only info you show is the winding R - may be the dominant loss at moderate levels, mid audio but frequency extremes are set by core and winding properties that determine Lm, leakage L, winding C
 16th November 2012, 08:41 PM #3 Jeffin90620   diyAudio Member   Join Date: Jul 2007 Location: Los Angeles I don't have that information in the transformer specs, but the requirements for my application are 300 Hz to 6 KHz (set by HPF and LPF filters), well within the transformer's bandwidth, so I'm thinking those characteristics will not be a consideration. What I need to do is calculate the maximum system gain at 1 KHz within 1 dB and for that, I need to know how to account for transformer losses. Regards, Jeff
 16th November 2012, 10:42 PM #4 jcx   diyAudio Member   Join Date: Feb 2003 Location: .. for the given resistive information you reflect the load impedance + sec R to the primary side, do the R divider math 4*150/(51+9.7+4*2.8+4*150) the coupling transformer impedance ratio is the square of the turns ratio, is a recommendation for circuit impedance but the actual values of the impedance don't directly enter the equations
 17th November 2012, 02:51 AM #5 counter culture   diyAudio Member   Join Date: Jun 2011 Transformers are typically very efficient, among the most efficient machines in general use. They're not straightforward to design though, so you'd normally look for a transformer to suit your purposes, given the output impedance of your amplifier, to handle the calculated power required in the load with a margin of maybe 10%. I don't really understand why you're looking to use a transformer. The LT1994 operates from up to 12V supply and will swing rail-to-rail with 85mA output, which is more than enough current into a 150 ohm load (19mA peak for 2vrms). You can use a dual rail supply (+/-) (less trouble than a transformer IMO) and ground the OCM pin and get symmetrical differential output, if that is your concern. All you need to do is select your feedback network resistors to give an appropriate gain to give 2vrms given the driving source. You can leave the resistances R28 & R29 in circuit and choose a slightly higher output voltage (1/3 higher for a 150 ohm load) if you are driving thru a cable with some capacitance. Perhaps you should tell us why you feel constrained to use the transformer.
 17th November 2012, 12:50 PM #6 counter culture   diyAudio Member   Join Date: Jun 2011 If you feel that you must use the transformer for isolation, then turn it round, so that the 100 ohm side is driving the 150 ohm load, this is a much better match. Now the 150 ohm load is reflected to the primary side as 37.5 ohms. You want ~19mA in the load, so you need ~75mA in the primary, still just within the capacity of the chip. You can add up all the DCRs in the primary side, this will tell you how much voltage you need to get that much current.
 19th November 2012, 10:32 PM #7 Jeffin90620   diyAudio Member   Join Date: Jul 2007 Location: Los Angeles I am designing to a MIL-SPEC requirement with the following items: > The output is transformer-coupled. > The resistive load is 150 Ohms. I am also duplicating the specifications from the original vendor (output impedance is less than 50 Ohms). I don't know why that is the case, but that is what is in their product spec. I could have easily chosen another transformer with a 150 Ohm Secondary, but that is not what is in the original piece of equipment. Regards, Jeff
Jeffin90620
diyAudio Member

Join Date: Jul 2007
Location: Los Angeles
Quote:
 Originally Posted by jcx for the given resistive information you reflect the load impedance + sec R to the primary side, do the R divider math 4*150/(51+9.7+4*2.8+4*150)
jcx,

Thanks for the formula. It seems to be the ratio of the ideal load (impedance ratio times actual load) to the sum of the 'lossy' load (the capacitive isolation resistors and the primary DC Coil resistance added to the impedance-multiplied sum of the secondary DC coil resistance and actual load resistor).

In this case, it would be 0.89 V/V or -0.98 dB, much better than what I was getting by simulating it in LTSpice (but LTSpice doesn't have a native transformer device).

Thanks,

Jeff

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