Power Supply to use with LM4702C power amplifier

Hi Daniel,

Typically the total (maximum) amplifier output power will approximately be 1/2 of the VA rating of the transformer.

You can surely use four 10,000uF caps; these are reservoir caps and stiffen the supply.

The muting circuitry should apply to any version of the LM4702.

It appears that you are getting into your first power amplifier design. Three really good books with lots of information (available on amazon.com) are stated below:

1. High-Power Audio Amplifier Construction Manual by G. Randy Slone

2. Designing Audio Power Amplifiers by Bob Cordell

3. Audio Power Amplifier Design Handbook, Fifth Edition by Douglas Self

A few transformer sources:

1. http://www.plitron.com/standard-toroidal-transformers/power-transformers/international/

2. http://www.toroid.com/standard_transformers/rectifier_transformers/solid_state_amplifiers.htm

I hope this helps you get started.

Best Regards,

JD

Thank you very much for your help,

I don't understand if you're talking about the total power is the RMS power based on a purely resistive load or the peak power, ideally I think that:

Prms = (Vrms)^2/RL [W] with Vrms = Vsat*(sqrt(2)/2)

Ppeak = (Vsat)^2/RL [W]

I based the power of the transformer in the RMS current:

Irms = Vrms/RL

I don't know if that's a good aproximation, but using this equations I reached that value, 800VA for 6ohm based on Irms.

Based on Ppeak I got  266 VA.

That's a huge difference!!!

Can you help me?

That's not my first amp but the other ones that I've mounted I only have tested on breadboard they are from TDA family and don't have this THD characteristics, neither this power. I'm interested in a hifi amplifier project.

I don't have any of that books but I'm very interested in reading them.

Hi Daniel,

I'm not sure I understand your definition of Vsat. The first equation Prms = (VOrms)^2/RL [W] is correct and how power is generally calculated. Vsat or the "saturation voltage" is a value added on top of the peak voltage for the power level in the equation just stated. VOpk = sqrt2*VOrms. 

200W into 6 ohms --> 35Vrms --> 49Vpk + Vsat = 5V = 54Vpk

Peak output power (ie instantaneous signal bursts) is a bit different then the maximum output power for the amplifier characterized with sinewaves with THD sweeps. The difference will depend upon the stiffness of the supply and the desired signal burst.

The stiffness in the supply is made up of the current capability of the transformer and the reservoir caps. This will determine the VDC(no-load) to VDC(full-load). Larger valued reservoir caps reduce the power supply ripple.

In terms of designing a power supply, you can start by determining the maximum continuous power that you would like, which will be based upon the rms output voltage.

If you want 200W per channel for a stereo amplifier, then your 800VA rated transformer is about right.

For a 6 ohm load, the maximum rms output voltage would need to be 35Vrms at the edge of clipping. The peak voltage associated with this output is 49.5Vpk.

The output stage saturation voltage then needs to be added in; say 5V. The supply DC voltage full-load would then need to be about 55V.

The power supply sag then needs to be added in: say 5V, making DC voltage no-load about 60V.

Converting the 60V to an rms value creates 42Vrms. Selecting transformer secondaries around this value should suffice. Typically dual secondaries are available so dual secondaries at 21Vrms is about right. You can see the transformer connections as an example in SNAA057A.

One thing to remember is that the AC line voltage coming from your house could have significant variation, so unless you use an auto-transformer the output power you will get may vary a bit.

Additionally, the transformer suppliers have a lot of application note information and sometimes have full-load ratings to help with proper secondary selection.

Best Regards,

JD

Hello and thank you very much for for your help,

For a Prms of 2*133W using a 6 ohm load, can I use a +/-45 V toroidal transformer, or it's better to use a +/-50V toroidal transformer?

|Vosat| = 40 V aprox.

Vorms = (|Vosat|/2)*sqrt(2)

Porms =(Vorms)^2/RL= 133 W aprox.

Pomax = (|Vosat|)^2/RL = 266 W aprox.

Iorms = (Vrms)/RL = 5 A aprox.

This means that I need a toroidal transformer that have a (45V/0/45V) and a 10A in secondaries, because of having two channels, needing 900VA?

Or that power is calculated taking the Porms of both channels in account and then I should buy a (45V/0/45V) with 500VA or a (50V/0/50V) with 500VA, if I choose this last option I can have 133 W of RMS power in both channels, it's better use 2*50V or 2*45V?

Thank you for your atention,

Daniel Almeida

Hi Daniel,

Based on your desire to have 133W per channel into 6 ohms, this equates to a total delivered power to the load of 266W. Class AB amplifiers are a little over 50% efficient, so a transformer with a rating of twice the delivered power is required or approximately 532VA. As previously mentioned, there are a number of variables that don't require the VA to be exact, plus the fact that sinewaves are far different then audio signals.

So, a transformer with a VA rating of 400VA - 550VA would suffice. Many people derate the VA to save cost especially in light of the fact that audio signals are not as stringent as sinewaves for bench testing.

The next step based on the 133W into 6 ohms is to determine the secondary voltage of the transformer.

VOrms = 28Vrms for 133W into 6 ohms, just before clipping. This converts into a peak voltage of VOpk = sqrt2(VOrms) = 40Vpk

The output stage of the amplifier will have approximately a 3V to 5V saturation voltage (deadzone). This is added on top of the 40Vpk to give 45Vpk. Therefore the DC voltage to obtain 133W into 6 ohms needs to be 45V full-load. If using a bench power supply, this is the value that would be used.

We then add in 1V for a bridge rectifier loss and 5V for power supply droop from the no-load value to the full-load value at maximum output power. This is 6V.

Therefore, 45V + 6V = 51V no-load. We then convert to rms to obtain the transformer secondary voltage. 51V/sqrt2 = 36Vrms.

The above calculations are for one of the DC supply voltage rails in a split-supply configuration utilizing a full wave center-tapped bridge configuration, which is a standard audio supply configuration.

Therefore, we need two 36Vrms secondaries in series and most transformers are wound with dual secondaries for you to connect them in series or parallel.

087018201 or 087035201 from Plitron at 500VA would work fine. http://www.plitron.com/standard-toroidal-transformers/power-transformers/international/#SchematicDrawings

Your +/-45V secondaries will obviously work, but based on my calculations that will give you about 230W per channel into 6 ohms.

+/-45Vrms secondaries equates to about +/-57V supply rails at full-load.

+/-50Vrms secondaries equates to about +/-65V supply rails at full-load.

I hope this helps out.

Best Regards,

JD

Thank you very much for your support,

I live in Europe and I think that I can't buy transformers from Plitron unfortunately, the Nuvotem 0500P1-2-035 and Nuvotem 0500P1-2-040 are equivalents to Plitron 087018201 or 087035201?

http://docs-europe.electrocomponents.com/webdocs/0027/0900766b80027b4a.pdf

Thank you very much for your help

The best regards,

Daniel Almeida

Thank you very much for your atention and by spending your time helping me solving this problem, I will follow your advice and I will acquire  the High-Power Audio Amplifier Construction Manual by G. Randy Slone.

The best regards, 

Daniel Almeida