Trafo choice for 15W, LM1875 based ss-amp

[Rutger]

Hi,

I ordered an LM1875 kit to use in a guitaramp (datasheet), which can poor out up to 25 watts. I have a speaker which I want to use that rates 15 watt max. To prevent the speaker from blowing out, I don't want the amp to put out more than 15W.

According to the datasheet the LM1875 poors out 20W with a +/- 18V trafo. Is that in theory or is that the real output power? In that case, what trafo do I need to use to keep the power at max 15Watt? 15V?

[Wilder Amplification]

The device will put out whatever amount of power that can be supplied from the power supply up to the output device's rating. Output devices don't "make" power. The power comes from the power supply and the output device simply controls that power.

Going off of this, we can see that when desiging any amplifier, the first things to consider is the power supply and the load impedance. After all, the power has to come from somewhere right?

That being said, since this is a solid state amplifier I think it would be a good idea to make the amp so that it is "capable of" supplying more average power than what the speaker can use. This way the amp won't have to be maxed out at the point at which it is delivering the required speaker power. Also, with SS devices output power is load dependent, which means the output power will drop the higher the load impedance. According to the datasheet, it can safely run into a 4 ohm minimum load. On a DC coupled SS amplifier, DC is present on the speaker during the periods of clipping so we want to avoid clipping at all costs to prevent overheating the speaker coil with DC current. To do this we rate the amplifier above what the speaker wants to see.

That being said, basically this device is capable of sourcing 2.25 amps into a 4 ohm load, which translates into a 9Vrms output given a +/-25V supply. However, in order for it to pass this current, the current has to be available from the power supply. Output devices don't just "make current"...they control load current from the power supply.

When figuring out output power most think in terms of average power. However, for calculating required power supply voltage/current, you have to think in terms of peak power (i.e. the instantaneous power at the peaks of the output signal sine wave), which is double the average.

Say for instance we need 25 watts average output power. We need to design the power supply for 50 watts peak power. For 50 watts peak power on a 4 ohm load we need -

Ppk x Zl = Vopk^2 where -

Ppk = Peak Output Power
Zl = Minimum Load Impedance
Vopk = Peak Output Voltage

50 Watts x 4 Ohms = 200

Find the square root of 200 to determine peak output voltage -

Square root of 200 = 14.14Vpk

Now we must find peak current -

Ipk = Vopk / Zl where -

Ipk = Peak Current
Vopk = Peak Output Voltage
Zl = Load impedance

14.14Vpk / 4 Ohms = 3.535 Amps peak

So for this design, a power supply capable of delivering 3.535 Amps peak would be required. By "peak" we mean that at the peaks of the output signal sine wave the power supply can deliver 3.535 amps.

For a transformer, I would go with a 48VAC center tapped @ 2 amps full winding current rating. This will give you 24-0-24VAC @ 4 amps rating to make a dual rail/dual polarity supply. Each supply only has to be able to source 4 amps peak current at a time since they never are supplying current to the load at the same time. The positive rail sources current on the positive cycle of the output signal while the negative supply sources current on the negative cycle.

[Rutger]

Wow, thanks alot for your extensive reply, helps me alot!
I can calculate the powersupply myself now

What I understand from the first part of your post, is that it is even better when the powerstage is able to give more power than the speaker can handel, right? Is there also a way to calculate that?

[Wilder Amplification]

Wow, thanks alot for your extensive reply, helps me alot!
I can calculate the powersupply myself now

What I understand from the first part of your post, is that it is even better when the powerstage is able to give more power than the speaker can handel, right? Is there also a way to calculate that?

Amplifiers are commonly thought of as devices that are either -

A) Putting out all of their power regardless of volume

B) "Pushing" power through the load (i.e. the speaker)

Truth be known though that amplifiers are nothing more than modulated DC power supplies...AC power inverters if you will. Just because they are CAPABLE OF delivering a certain amount of power doesn't mean they HAVE TO. The speaker draws current from the amplifier, which results in a voltage drop across the speaker that is dependent on the amount of current it draws. Combine the two and you have "power.

Some people think "Well why have an amplifier with more power than you can use?". The answer to this question is simple...HEADROOM!

The textbook definition of headroom is "power that you are not using". Well if you use an amp that maxes out at just enough to run the device you're powering you have zero headroom. However, if the amp is CAPABLE OF supplying more than what the speaker calls for, it can supply what the speaker needs to run and still have room left to go, which results in very clean output.

Think of it like this...when you drive your car down the road at 40mph, are you having to floor the accelerator just to go that speed? Not at all! You're using very little of what the engine is CAPABLE OF supplying power-wise to do 40 and you still have lots in reserve for passing power. If it were the other way around your engine wouldn't live very long.

[R.G.]

...
For a transformer, I would go with a 48VAC center tapped @ 2 amps full winding current rating. This will give you 24-0-24VAC @ 4 amps rating to make a dual rail/dual polarity supply...

That's a bit of a problem, too. Transformers feeding rectifiers get rectified to the peaks of their voltages. 24Vac gets rectified to 24*1.414 = 33.9Vdc, ignoring rectifier losses of about 0.7V per side and ripple under load. It ignores the AC line's variance too, and it could go up maybe 10%; DC follows, giving +/- 37Vdc, or 74V from the + to the - on the LM1875. Since absolute max on the 1875 is 60V, the chip might just give a *pop* and die.

You need to work it backwards. The chip can take +/-30Vdc, and we'll take that as 10% high, so we aim for a max of 27V nominal, and add back the 0.7 for the diodes. this gets to 27.7, and the RMS value of that is 19.6Vac.

So the chip needs a transformer with 19.6Vac secondaries or 39.2Vct. That's not nearly standard, and he wants lower power anyway. The chip data sheet says it puts out 15W with +/-18V (about) power supplies. That says a 25.4Vct transformer is needed. This is good, as 24Vct is easy to find.

I'd use 24Vac centertapped, rated for about 1A.

You can usually get 24ct, 30ct, and 36ct in transformers fairly easily.

[Rutger]

Hm, things start tot get a little confusing now...

So what your saying is that Wilder is a bit too positive and that you need to calculate with sertain losses,right?
So +/-18V will give 15W out, thats a good thing.
Only thing is that anyone uses at least 50VA trafo's and not 24VA, has that to do with the peaks where Wilder is talking about?

[Steve Conner]

Wilder did the calculation for a hypothetical 50 watt amp, not your actual amp. Maybe just to confuse you.

RG's math is correct, and if you want 15w output power, a 30va transformer should be about right.

[R.G.]

So what your saying is that Wilder is a bit too positive and that you need to calculate with sertain losses,right?

+1 for Steve - Wilder did an example, not your particular solution.

And yes, there are always losses.

So +/-18V will give 15W out, thats a good thing.

Any time you're doing your own design, it's always smart to look at the datasheets for the parts involved. I googled "lm1875 datasheet" to find the information. There is a graph on the datasheet of power output versus power supply voltage, and I picked 15W, then looked for the power supply that gave that.

Only thing is that anyone uses at least 50VA trafo's and not 24VA, has that to do with the peaks where Wilder is talking about?

This is a slippery issue with a long history.

Music is not continuously the same level. It has peaks and valleys. Power devices on heat sinks and transformers have a lot of thermal lag - it takes them a while to heat up. So they respond to the average power out much more than the peak power out. And ultimately, for transformers, it's the heating that determines their power abilities.

A class AB amplifier (this is one of those) that can put out 15W has a well defined heat output at maximum power. At full output, it's sending out a bit over 70% of the power it takes in from the power supply. So that 15W costs you at least 21.4W from the power supply in DC. Some amps may be less efficient than that. And the conversion of AC to DC in the rectifiers heats the transformer more than the actual DC current out would indicate. It's about 1.6 times bigger for a full wave bridge power supply, which you'll almost certainly use. That means for continuous output of 15W, you need 21.4W of DC, but the transformer heats like it's providing 34W of AC.

But then the music is not continuously 15W, because of that peaks-and-valleys stuff. So it will vary, and so will the power from the transformer. So a 24VA rated power supply is probably going to be fine. If you were building an industrial strength amplifier intended to run at continuous max power, you'd want a 24VA, 1.5A transformer, giving the capability of 24*1.5 = 36W output continuously, and that could then do 15W output continuously. It's a bit of overkill, but you might want to do it. I would use either one, depending on ease of finding a transformer and cost.

Over the years, if I intend to build a power amplifier, I don't look at circuits first. I look for the power transformer and heat sinks first. If you can't get a big/good enough transformer, you can't get ANY amplifier to work; if you can't get a big/good enough heat sink, you can't get it to keep working. That's a little funny sounding to an era of people who think of power supplies last, but it's accurate. In actuality, an audio power amplifier is a wart on the power supply.

[Wilder Amplification]

RG...I do realize that rectified/filtered voltage is equal to the peak AC voltage. However I was also assuming an unregulated supply so I figured that the supply voltage would load down a few volts, which would put the main supply voltage right at +/-30VDC under load.

However...I also stated to the OP that you don't want an amp that max's out at 15 watts. It would be much better IMHO to make the amp capable of delivering more than the power needed, then only set the volume dial on it to put out the required power. It would be like spec'ing a power supply to max out at exactly the needed current to power a device.

[Steve Conner]

Yeah, but the OP specifically stated that he wanted an amp that maxed out at 15 watts.

To the OP: I'm sure a few more watts wouldn't do any harm. Besides, remember that this method gives you the maximum unclipped output power. If you overdrive the power amp and mash the output into a square wave, you'll get more than 15W.

[Rutger]

I know Wilder made an example, I just meant that many people who build a 20W version of the LM1875, already use a 50VA trafo or higher. I think that that has to do with the headroom?
I surely don't want the amp to get be overdriven so I wouldn't mind some extra headroom. But if say 30VA is plenty I can go for that, saves some money doesn't it

[Steve Conner]

Oversizing the transformer will give you a little more "headroom" because the power supply voltage won't sag so much under load.

Oversized transformers are very popular with hobbyists, but they're never used in commercial work because they cost more. A commercial design would take sag into account by starting with a higher unloaded voltage, so that the rails ended up at the required voltage under full load.

(By oversizing, I mean making the transformer bigger than it needs to be to run without overheating and burning out, which in this case is around 30VA.)

[Rutger]

Okay great, thanks alot!

An extra question: I need to feed the preamp as well. It will be a copy of the Marshall Lead 12 preamp with a Flatline compressor in front, which is basically a simple but effective ''guitareffect'' design. The preamp needs +/-18VA as well for the 1458 chip, the optocomp can be fed with 18V single rail for the TL072 chip. My guess is that it will take minor power to feed the preamp+compressor and I can take that from the same 30VA power supply. Am I right?

[J M Fahey]

Yes you are.
Preamp current is only a few mA, nothing compared to what the power amp needs.
The Lead 12 preamp is great.

[Rutger]

Thanks!
Yes I know, and it's also very easy to build.