LM380 help

[Triodes]

hi all i dont know if there is a post answering this somewhere, but i have bin looking all over for it and couldn't find one.
im building a battery powered amp with the lm380 chip (the 14 pin model)

it will run on 2 9v batts wired for 18 volt operation.(with heat sink wings and 8ohm load)
i have learned that the bypass cap on pin 1 (4.7uf) is not needed when powering from a battery, id like confirmation on that by anyone in the know about it.

but my main question is this, if im running on battery and not a power supply, do i still need the .1uf & 470uf caps on pin 14 VSS?
are they for ripple filtering? which wouldn't be necessary with a battery supply. or are they for oscillation prevention, making them necessary no matter the supply type?. 

also i rigged up the volume control to the inverting input, so as you turn the control down it lets more of the signal go to the inverting input to cancel out the the non inverting input signal.(i thought this would be better for sound quality at lower volume levels) it works, but i noticed a bit more hum being picked up by the inputs of the chip then when there was no volume control and the signal just went to the non inverting input with the inverting input connected to ground through a .1uf cap. is this volume control method not a good idea with this chip? should i just use the conventional method with one end of the pot to ground?

thanks for any help! 

[ElectricDruid]

There's a datasheet at:

https://www.ti.com/lit/ds/symlink/lm380.pdf

The datasheet shows the pin 1 cap as "dotted", suggesting that it is optional. The datasheet circuits also show no caps on pin 14/Vss at all, so where did you see those? You didn't post a link to your schematic.

[antonis]

What Tom said..

100nF & 470μF caps are a rule of thumb for mains supplied amps for "some" reasons.
Batteries don't suffer from any kind of those illnesses..

[Triodes]

here is the schematic im using,

in the one im following its using a 220uf and a .1uf on the vcc. almost every schematic i came across using this chip, had some form of cap on the vcc even guys talking about running it on batteries, but no one clarified whether they are still needed with a battery supply or not. i just want to make absolutely sure its not critical for oscillation protection before i go making a custom board for the amp

anyway thanks for the replies!

[duck_arse]

and like they say, welcome to the forum.

fit the caps. won't kill you.

[amptramp]

I would add the caps because even though a battery would have no hum, it does have an internal impedance that may be too high once it goes beyond a certain level of discharge.  It may make the difference in the amplifier working badly when the battery voltage drops 10% and still working when the battery has discharged 20%.

High-gain amplifiers also have a tendency to oscillate unless you take steps to keep the instantaneous voltage under control even with wide current fluctuations.

You can try building the amp without the caps and add them to see what difference it makes.  If you are designing a PCB, leave space for them - you don't have to populate the caps if you don't need them but I suspect you will need them.

[anotherjim]

Wait! 2.9v? Are they coin cells?

And +1 for fitting the caps.

[Clint Eastwood]

What batteries will u be using? standard 9v pp3 blocks are not able to deliver the current for the rated output power I think. Maybe 9v nimh rechargeables of good quality.

[Vivek]

I would say that you should always have capacitors from Vcc to ground, irrespective of using a power supply or batteries, due to


A) Reserve supply when sudden bursts of current are needed to handle peaks in the music

B) Making sure that Vcc is equivalent to ground for AC signals

C) Ripple reduction


and I would feel reasons (A) and (B) are more important when one uses batteries.

[antonis]

I would say that you should always have capacitors from Vcc to ground, irrespective of using a power supply or batteries,

It depends on the presence of power supply LPF series resistance..

For battery supply, LPF is, more or less, useless so capacitors from Vcc to GND (especially electros) might cause inrush current to be reckoned with..

[Triodes]

thanks all for the help! i will add the caps to vcc

Wait! 2.9v? Are they coin cells?

And +1 for fitting the caps.

you miss read, i meant two 9 volt batteries wired to get 18 volts

What batteries will u be using? standard 9v pp3 blocks are not able to deliver the current for the rated output power I think. Maybe 9v nimh rechargeables of good quality.

i was reading that the current draw of the chip should be under 20ma, i dont see how two standard 9 volts couldn't handle that 

If you are designing a PCB, leave space for them - you don't have to populate the caps if you don't need them but I suspect you will need them.

im actually making a hand wired board with hard cardboard that i wax impregnate. i need to know all the parts that will be used so i can trace out the layout & underside connections, then drill the holes and hand wire it. its more hassle then pcb or proto board but i like doing it this way as an artform 

[Elektrojänis]

i was reading that the current draw of the chip should be under 20ma, i dont see how two standard 9 volts couldn't handle that 

That 20mA is probably Qiuescent current. That is the current it draws when there is no signal to be amplified.

As a sanity check you can calculate: 20mA * 18V = 0.02A * 18V = 0.36 Watts

That chip is supposed to put out 2.5 Watts. It must draw more power than it puts out or it will be in the realm of perpetual motion devices and unlimited free energy.

Usually this kind of linear amplifiers have an efficiency that is nowhere near 100%, so it will draw quite a bit more from the power supply than it puts out.

[PRR]

...efficiency that is nowhere near 100%...

78% at very best, zero device losses.

You can read this on the datasheet. Fig 8. At 18V, 4W out, there is 1.7W in the chip (as you know, since you realize you need the fins), so total 5.7W from the power supply.

(Man, postimg sure garbled the color in the thumbnail.)

[Clint Eastwood]

With 2.5 watts into 8 ohms, you have sqrt(2.5x8) = 4.5 volts rms across the speaker, so 4.5/8 = 0.56 amp rms through it. That means 0.56x sqrt(2) = 0.8 amps peak, and so the power supply will have to deliver 0.8 amps peak current.
I suggest using an 18 volt recharcheable powertool battery, or a 19 volt from a laptop, to enjoy full power.  Or maybe lots of capacitance would work?

[PRR]

> power supply will have to deliver 0.8 amps peak current.

Yes. Momentary peak. (And showing why a stack of pocket-radio batteries may faint.)

Although on square-wave the peak lasts a full half-cycle.

The peak should come from the reservoir capacitor, it is better for peaks.

Even on square wave, half the time the current in the supply pin is zero. So more like 0.4 Amps battery drain.

(I figure 0.36A; I think the "4W 10%" condition is far into clipping so Ideal math fails us.)

Hypothetical sine wave, 0.707*0.36A is 0.26 Amps.

Typical guitar tones, even "fuzz", are not as severe as Sine.

Hmmmm... a modern 9V is rated 500mAH, 0.5AH. That implies 2 hours at max Sine; but taking power that fast won't give full advertised capacity. Still a couple batts might last a couple 40-min sets.

There are now Lithium Primary batts in the 9V shape, "5 times more capacity" (maybe more like 2X) for alarms and parking meters. At a much higher price, I am sure.

[Triodes]

> power supply will have to deliver 0.8 amps peak current.

Yes. Momentary peak. (And showing why a stack of pocket-radio batteries may faint.)

Although on square-wave the peak lasts a full half-cycle.

The peak should come from the reservoir capacitor, it is better for peaks.

Even on square wave, half the time the current in the supply pin is zero. So more like 0.4 Amps battery drain.

(I figure 0.36A; I think the "4W 10%" condition is far into clipping so Ideal math fails us.)

Hypothetical sine wave, 0.707*0.36A is 0.26 Amps.

Typical guitar tones, even "fuzz", are not as severe as Sine.

Hmmmm... a modern 9V is rated 500mAH, 0.5AH. That implies 2 hours at max Sine; but taking power that fast won't give full advertised capacity. Still a couple batts might last a couple 40-min sets.

There are now Lithium Primary batts in the 9V shape, "5 times more capacity" (maybe more like 2X) for alarms and parking meters. At a much higher price, I am sure.

wow, i didn't realize the draw would be that much. here's the thing though, we have a CRAP ton of 9v batteries lol thats why i wanted to use two 9v batts wired to get 18v to power it. i think i will still go with 9v batts even though they are not ideal just cuz i have so many of them that need to be used lol but i might also install a 18v input so it can run off wall power when im not away from wall power.
anyway, thanks a bunch guys for your help!

[antonis]

we have a CRAP ton of 9v batteries

Put as much as you can in parallel..

[Clint Eastwood]

Yes, it would be a waste not to use those batteries.
When putting them in series and parallel, I would measure and select batteries with similar no-load voltage, this way you have units with similar charge status, and current will be distributed more evenly.

[anotherjim]

Batteries in parallel mean that weak ones become a load on the strongest ones. To solve this you can put a series diode in the + of every parallel string with the anodes common to the circuit +supply. Schottky diodes like 1N5817 will lose less voltage. When a battery weakens it will drop out as its diode won't forward bias and the stronger ones take over. This continues automatically until no battery can supply enough for the load. It will also prevent a bad battery from getting back-charged by the good ones which can be damaging.

The diodes prevent the battery's own capacitance from providing a proper low impedance AC path so fitting sufficient supply capacitors will be very important.