Has anyone run dual LM386 for more power

[Headshot]

Thanks for the great info Dano!   Looking forward to that Tube Cricket.

[suprleed]

Great thread.  I've been stewing over 386 amp projects for a while now.  The hard part is trying to settle on just one project!  Very helpful information here.  Thanks.

[brett]

+1 for the LM380
and an efficient speaker will always be required if real punch is required.

Doesn't the 386 will push out about 0.6 W into 8 ohms (and around 0.4W into 4 ohms.)?  So paralleling two 386s into an 8 ohm load will still yield only 1 W (@9V).

So bridging or paralleling 386s doesn't work.... does it?

[Jasonmatthew911]

I actually made a mini amp with 2 LM386-4's in B.T.L. , with a NJM4558 Pre-amp section going into it, adding a  Volume control, Tone control, and a Gain control that goes from Clean to Overdrive and finally to Hi-Gain distortion, running it with 6-AA cells for 9V or adapter...I have Clip-on Heat sinks on all the chips and an 18V Zener Diode in the Power section, so that I can use 18V safely without going over, to avoid damage to the IC's with an 18V max limit...I'm using a 5W- 3", $30, 8ohm speaker, but it in a small solid wood enclosure, and at 9V I get at least 1W - 1.5W of power out of it, it sounds louder than an Orange Micro Crush and Mini Fender Twin that I compared it with, but then when I ran it with 18V supply it gave me so much more headroom, it got louder and cleaner, with 18V supply I got 2W power easily, but you need to use the 386-4 that can handle 18V, it also says that this chip will give you 1W power with 18V, so with 2 Bridged I'm getting 2W...The only problem is you need bigger heat sinks with 18V's, I'm getting more out of it with the 18V, you will get the same amount of volume as a Micro Cube, but when the Volume gets too high it starts making a resonating noise while I play, the only thing I can think of is that I need bigger heat sinks for my chips....If I can perfect this design on 18V, I will have a little 6" cube amp that sounds as loud and as good as a Micro Cube, but much smaller, with good loud enough clean, overdrive, and distortion...Does anyone here now a good heat sink part # that I could get bigger than the little clip-on wings for 8-DIP chips, maybe something I can put on with some compound?

[R.G.]

And here's my reply to that from the posting by itself:

The Dual Inline Package has some sincere power dissipation problems. Most of the heat is *conducted* out the leads, not convected out the plastic.

I say this as background for what I'm about to say.

You're probably right that it's going into thermal overload and shutting down. You may get some relief from heatsinking, but what you are doing is trying to get a small device working right out at the very edge of its capability. This is not a recipe for long term reliability.

It is better to use a bigger device well inside its capabilities. If you used ONE TDA2030 or LM1875, you'd get 4-5W on 18V, and you can do the same bridged trick to get even more from the same 18V. And both of these chips are in the TO-220 package which not only dissipated about 2W in free air, but is made for attaching to an external heat sink, which are also easily available and cheap.

So my advice is to either (1) back down the power supply voltage until the thermal cutouts quit or (2) get a bigger horse; use a TO-220 power amp chip.

To this thread: don't get stuck on the LM386. It happens to be an easy solution to the specific problem of a small, DIP power amp with no need for lots of power. It is NOT all that extensible once you hit its limits. The LM1875 and TDA2030 are GREAT solutions for the next step up in power and thermal issues. They'll go to a decent 20W, more if bridged. They are designed and packaged for heatsinking.

[Jasonmatthew911]

First of all, I recently made a 2W amp (Which I posted about a few days ago) with a 4558 Pre-amp section going into (2) LM386N-4 IC's running on 18V...What is the current drawn normally and about how long would (2) 9V Alkaline batteries last as opposed to (2) 9V Lithium batteries with 1200mAh in series for 18V?....I know that I can use a bunch of AA cells, but I don't have room in the enclosure for so many AA's, so I'm trying to use (2) 9Volts in series instead, but I figured that I would have to use Lithium batteries if I want them to last very long...I know that most 9V Alkaline give you about 500mAh and Lithium gives about 1200mAh, but would (2) regular 9V batteries be enough to run the 4558 with (2) LM386's at 18V for at least an hour or 2, and how long would the Lithium's last???

I also wanted to know the same if I used the 4558 Pre-amp going into an LM380 (2.5W IC) instead...What would the current drawn normally be with 18V in this situation, and would (2) 9V Alkaline batteries even be enough to power this for a little while, or how long would (2) 9V Lithium batteries last running on 18V with the LM380 power section?.....I'm not very good with math, and I'm not sure how many mA these chips draw at 18V???

If anyone can give me these answers or at least a pretty close estimate as to how long the batteries will last or if they even have enough Amp hours to power these chips at 18V, I would really appreciate it...Thanks again for any help.

[PRR]

> how many mA these chips draw at 18V???

Very little when idle and silent.

When roaring, depends on the load.

For the classic (non-Bridge) connection, figure the load on the power supply is similar to a resistor six times the total impedance, useful load plus chip losses.

Take LM386 loss as 3 ohms, LM380 maybe 2 ohms.

IF the load is 8 ohms, add 2 or 3 and round to 10 (makes math easier).

10 ohms times 6 is 60 ohms.

18 volts applied to 60 ohms is 18V/60= 0.3 Amps or 300mA.

That's at full roar. Power consumption goes down as you play softer, but efficiency drops fast also. If you don't play mostly at FULL ROAR, your battery cost will be much less with lower voltage.

> 9V Alkaline give you about 500mAh

It may be able to carry 50mA for 10 hours. It will sag and go flat a lot quicker at 300mA. Maybe less than an hour to 5V/batt, 10V total. At that point you might as well have used a single 9V batt, total 150mA drain, play for 2 hours, or four hours if you brought two batts. If you are busking, 4 hours is a LOT more loose change than under-one-hour.

The real goal is not voltage or power, but LOUDness per battery dollar, right?

9V radio batteries are about the worst buy.

Watts-to-LOUD is largely about CONE AREA. Paddling a yacht with a teaspoon is inefficient. A light Ten inch cone will blow-away a three-inch cone on the same power, or do the same job with LESS power, less battery cost.

Voltage, power, and impedance are related. Since AA is cheaper per watt-hour than 9V, C is cheaper yet, and D is the economy size, you would like to run on four C or D cells. That's 6V DC, which is at best 2.1V RMS. With a good (large) speaker One Watt is pretty loud. 2V in 4 ohms is one watt.

[Gurner]

One of the biggest "aaaaahaaa" moments I've had in my hobby electronics pastime, is when the penny dropped that an audio amplifier IC, didn't in itself muster up the power (by that I mean, I'd previously been scouring the datasheets to see which ones could pump out 1w, or 2W etc), but the maximum power achievable actually comes from the supply voltage and the load. That is, the power specification you see for an audio amplifier IC relate to it's  maximum supply voltage handling ability ....& more significantly, how well the IC package can dissapate the heat generated - if you could find someway of sumping the heat, there'd be nothing to stop you pumping out 10W from your LM386! In other words, the IC doesn't generate any power.....this gets a little clouded in that just about all audio amplifier IC apply a voltage gain to the signal (& @200x the lm386 can potentially apply the most voltage gain of any audio amplifier IC out there!) , but nevertheless, the important thing to remember is that the IC itself is just a conduit to convert signal voltage to 'current delivery' into the load.

Wrt current draw, here's my take (this is for single ended amplifiers like an LM386 etc).....

when running an LM386 at 9V supply - and assuming a supply that doesn't droop - the best signal level you can expect to achieve across the load is something like 8V ...any more & you'll get clipping. But that's a peak to peak figure, we need RMS - & being of a lazy ilk, I use this calculator ...

http://www.daycounter.com/Calculators/RMS-Calculator.phtml

that calculator wants 'Vpk' as an input, which is half the peak to peak level, so that'll be 4V.....which works out at 2.828V RMS.

so with a 9V supply, the maximum unclipped signal you can put across a load is 2.828V RMS...the rest is easy. If your load is 8 Ohms, the currents draw will be 2.828V/8  ....or 353mA.

applying the same numbers to an 18V supply, well the maximum signal achievable will be somthing in the order of 17V pk-pk, or 8.5Vpk, or 6.01V RMS. therefore into the same 8 Ohm load the current draw will be 751mA.

Bearing in mind your average PP3 is only good for 550maH, then you really need to look at sourcing power from mains once your requirements get above a few hundred milliwatts.

The continuing fascination with LM386 is curious....if you wish to double your output power & keep your circuit supply at 9V & your speaker at 8 ohms, then simply use a BTL IC....a TDA7052 will do nicely. BTL ICs have differential outputs which essentially doubles your achievable signal levels across the same load....to cut a long story short....that translate to 4x the power delivery for the same given situation vs a single ended IC like the LM386. Better still, there's no mandatory requirement to grow a droopy 'tache & wear flares/platform shoes (as you do when using an LM386)

[defaced]

if you could find someway of sumping the heat, there'd be nothing to stop you pumping out 10W from your LM386!

Peltier?

[PRR]

> just a conduit to convert signal voltage to 'current delivery' into the load.

Yes.

And the LM386 is a skinny pipe.

> if you could find someway of sumping the heat, there'd be nothing to stop you pumping out 10W from your LM386!

No, there's often another limit. CURRENT. If too high, either the transistor melts (for any amount of cooling) or its gain drops to uselessly-low. In many chips, the current is monitored, and cut-back if excessive. The LM386 apparently is too lame to pass much current: push harder and nothing happens.

Fer example, TDA2003 is safe for about 4 Amps. In the ST 1998 datasheet schematic, Q14 watches R10 and shuts-down the amp if R10 current is too high.

The LM386's power in low impedance does NOT increase as you raise supply voltage past some point.


In 4 ohms, 6V to 12V is all just 0.36 Watts. In 8 ohms, yes it rises from 6V to 12V, but does not rise from 12V to 18V. If you work the numbers, it won't go past 0.42A to 0.44A of peak load current. AND the higher supply voltage translates DIRECTLY into more heat. And this heat is totally a waste on the battery.

The "optimum" for a 18V rating (say 8V peak) and 0.4A peak comes with a 20 ohm load and is 1.6 Watts.

If you want to fool with higher voltages, TDA2003 and kin are MUCH better "conduits", fatter pipes.

They also cool better.

> lm386 can potentially apply the most voltage gain of any audio amplifier IC out there!

It's a lot of gain for 50 cents. More than we can usually use. However TDA2003 datasheet shows gains of 500, and I think there's more in there. At Gv=500 the response may roll-off by 3KHz, but at that level of gain (and hiss and distortion) maybe you want a low-pass.

> use a BTL IC

FWIW: transistor losses are bad, a simple stage has two transistor losses, a bridge has four transistor losses. BTL makes sense when you "can't" change supply voltage. Car battery IS 12V-14V. Good for 2W in 8 ohms or 4W in 4 ohms. 4W in a car is not much. You could always convert it, but that complication is usually reserved for 60W-200W trunk-thumper amps. Instead, a BTL is "almost like" a 24V supply, 16 honest Watts (22 watts boasting). That's enough to overcome road-noise.

[Gurner]

if you could find someway of sumping the heat, there'd be nothing to stop you pumping out 10W from your LM386!

Peltier?

Thank heavens this was picked up upon .... someone might have took me literally (vs me overzealously emphasising a point) ......because, even after implementing some scifi supercooling arrangement, they could have ended up very disappointed with the final performance of their LM386.

So please everyone, don't try to squeeze 10W out of your LM386  ....simply put your ear closer to the speaker.

[Jasonmatthew911]

> how many mA these chips draw at 18V???

Very little when idle and silent.

When roaring, depends on the load.

For the classic (non-Bridge) connection, figure the load on the power supply is similar to a resistor six times the total impedance, useful load plus chip losses.

Take LM386 loss as 3 ohms, LM380 maybe 2 ohms.

IF the load is 8 ohms, add 2 or 3 and round to 10 (makes math easier).
<quote>

If you have 2 - 9V batteries in series for 18V will that bring up the available 500mAh to 1Ah?...Or is that only the case if you parallel the batteries?

10 ohms times 6 is 60 ohms.

18 volts applied to 60 ohms is 18V/60= 0.3 Amps or 300mA.

That's at full roar. Power consumption goes down as you play softer, but efficiency drops fast also. If you don't play mostly at FULL ROAR, your battery cost will be much less with lower voltage.

> 9V Alkaline give you about 500mAh

It may be able to carry 50mA for 10 hours. It will sag and go flat a lot quicker at 300mA. Maybe less than an hour to 5V/batt, 10V total. At that point you might as well have used a single 9V batt, total 150mA drain, play for 2 hours, or four hours if you brought two batts. If you are busking, 4 hours is a LOT more loose change than under-one-hour.

The real goal is not voltage or power, but LOUDness per battery dollar, right?

9V radio batteries are about the worst buy.

Watts-to-LOUD is largely about CONE AREA. Paddling a yacht with a teaspoon is inefficient. A light Ten inch cone will blow-away a three-inch cone on the same power, or do the same job with LESS power, less battery cost.

Voltage, power, and impedance are related. Since AA is cheaper per watt-hour than 9V, C is cheaper yet, and D is the economy size, you would like to run on four C or D cells. That's 6V DC, which is at best 2.1V RMS. With a good (large) speaker One Watt is pretty loud. 2V in 4 ohms is one watt.

[Jasonmatthew911]

> just a conduit to convert signal voltage to 'current delivery' into the load.

Yes.

And the LM386 is a skinny pipe.

> if you could find someway of sumping the heat, there'd be nothing to stop you pumping out 10W from your LM386!

No, there's often another limit. CURRENT. If too high, either the transistor melts (for any amount of cooling) or its gain drops to uselessly-low. In many chips, the current is monitored, and cut-back if excessive. The LM386 apparently is too lame to pass much current: push harder and nothing happens.

Fer example, TDA2003 is safe for about 4 Amps. In the ST 1998 datasheet schematic, Q14 watches R10 and shuts-down the amp if R10 current is too high.

The LM386's power in low impedance does NOT increase as you raise supply voltage past some point.


In 4 ohms, 6V to 12V is all just 0.36 Watts. In 8 ohms, yes it rises from 6V to 12V, but does not rise from 12V to 18V. If you work the numbers, it won't go past 0.42A to 0.44A of peak load current. AND the higher supply voltage translates DIRECTLY into more heat. And this heat is totally a waste on the battery.

The "optimum" for a 18V rating (say 8V peak) and 0.4A peak comes with a 20 ohm load and is 1.6 Watts.

If you want to fool with higher voltages, TDA2003 and kin are MUCH better "conduits", fatter pipes.

They also cool better.

> lm386 can potentially apply the most voltage gain of any audio amplifier IC out there!

It's a lot of gain for 50 cents. More than we can usually use. However TDA2003 datasheet shows gains of 500, and I think there's more in there. At Gv=500 the response may roll-off by 3KHz, but at that level of gain (and hiss and distortion) maybe you want a low-pass.

> use a BTL IC

FWIW: transistor losses are bad, a simple stage has two transistor losses, a bridge has four transistor losses. BTL makes sense when you "can't" change supply voltage. Car battery IS 12V-14V. Good for 2W in 8 ohms or 4W in 4 ohms. 4W in a car is not much. You could always convert it, but that complication is usually reserved for 60W-200W trunk-thumper amps. Instead, a BTL is "almost like" a 24V supply, 16 honest Watts (22 watts boasting). That's enough to overcome road-noise.

If you have 2 - 9V Batteries in series for 18V, will it bring up the available 500mAh to 1Ah?...Or does this only apply for parallel batteries?

[davidallancole]

Only applies to parallel.  So if you parallel two 9V 500mAh batteries you get a 9V 1000mAh supply.  If you put two 9V 500mAh batteries in series you get a 18V 500mAh supply.

[Brymus]

I was reading through an old Guitar mag the other day and came across an ad for Pignose amps.
They were touting their new 40 watt combo that ran on 2 sealed lead acid batteries.
Now those must have been some serious batteries to get 40 watts output to the 10" spkr for any amount of time.

[Gurner]

I reckon those car batteries will be in series, and the amp will be using a bridged output.

Even then the only way you can get 40W is by taking the impedance of the speaker down to 6 ohms ....that translates to 2.6A ...the average car battery supplies 120AH, so you're gonna get about 46 hours playing AC/DC riffs at full whack (less in reality) before you need to reach for your jump leads &/or call roadside services.

Personally, I wouldn't buy one unless it had a cigarette lighter socket on the fascia.

[PRR]

> the average car battery

They are often 50 Amp-hours. At 12V this is like 600 Watt-hours. Taking 50% efficiency, this is 30 Watts for 10 hours.

Any recent car radio makes 16W-22W into 4 ohms with just the 12V supply. Most have four such channels, like 70 Watts total honest output. They can play LOUD for several hours (but when they get weak, the battery is too flat to re-start the car).

You don't carry two car batteries.

There are smaller lead batts for emergency lights, security alarms, etc. All the way down to a D-cell good for 2V 2.5AH. (I have seen even smaller lead cells.) LA1270 12V/7.0AH is readily available around $20.

[Gurner]

I was assuming in this day & age, if someone is gonna design an amp of that power to run off batteries, then Class D would be the only game in town....and at full whack, you can get 90% efficiency.

[R.G.]

Some techie background is in order.

Power delivered by DC is easy: it's P = voltage * current; this is the same as power being the square of the voltage divided by the load resistance, or the square of the current times the load resistance; these are all equivalent statements.

Power delivered by AC waveforms is more complicated. The voltage changes all the time. It was found long ago that if you took the square of the instantaneous voltage at all points of the waveform, added those, divided by the waveform period, and then took the square root, you came up with a number which was the practical equivalent to a DC voltage in terms of heating a resistor (that is, delivered real power).

This is, of course, what is meant by "RMS", being Root of the Mean Square. It's different for every waveform. For a sine wave, the RMS value of a wave with a peak voltage of Vpk is Vrms = Vpk/1.414. 1.414 is approximate; the real number is the square root of two.

The reason this is important is because if you know the peak value of a sine wave across a resistor, you know how much heating it's doing (that is, how much heating is delivered).

For any non-bridged amplifier working from a power supply of Vpower, if the amplifier was theoretically perfect, it could only swing its output between 0V and Vpower. So at the output, Vpk would be Vpower/2. In real amplifiers, you never get here, because the output devices cannot make the output voltage go to fully Vpower or 0V. Notice that if the amplifier is powered from a bipolar power supply, you get the same result, but the power supply voltage Vpower is equal to the sum of the + and - power supply voltages.

So once you know the power supply voltage for your amplifier and also its load resistance, you know how much power it can even possibly put out if everything is perfect (as it never is.)

For a standard class B output amplifier working from 9V into 8 ohms, the peak AC voltage out can't be more than 4.5V; the RMS can't be more than 3.18Vrms. Then the power into 8 ohms can't be more than (3.18)*(3.18)/8 = 1.26W. For a 4 ohm load, it can't be more than 2.53W.

If you bridge such an amplifier, you get 5.06W into 8 ohms and 10W into 4 ohms. The problem is that this is not a real amplifier. It can't swing that big a power supply voltage, and the output transistors can't swing fully to the power supplies, and they also start to burn up. You can't get the heat out.

The worst case condition for a class B amplifier heating is NOT at full power out. It's more like half power. At the max dissipation point, the amplifier has to get rid of about 40% of the maximum possible power output.

When you pick a speaker load, you pick what power outputs are possible for any given power supply to the amplifier.  If you have a defined speaker load and a defined power supply, you have defined how much power you can get out. If you use the amplifier in bridge, this is exactly the same as two other situations: one is having each amplifier running a load of half the speaker load, the other is the real speaker load driven from twice the power supply. Real output transistors and drive conditions will give you less than the theoretical maximums.

[PRR]

> in this day & age

Define "this".

See my last post. I'm Firmly in the 1960s, when not in the 1880s.

> Class D .... 90% efficiency

Yeah, well, if we really wanted efficiency we'd run wires from the guitar into every listener's brain. That only takes micro-Watts, not many-Watts.

And 50% to 90%, battery life is only 1.8 times better.

And some plans here appear to do 25% efficiency at best. Raising pressure (voltage) is not always the answer; sometimes you need a fatter pipe.

R.G. has the math, of course. That's been well-known since the 1870s when power systems developed, since the 1920s when audio engineers gained some control of their power. You can even simplify by saying Sine RMS voltage is 1/3rd of total DC voltage (within half-dB error). But even that math may not be popular here.

None of this is new. Part-Watt, several-Watt, and dozens-Watt markets have existed in radios, TVs, and cars for decades. There's chips engineered FOR these uses.