Amplifying sound output with LM386

Harry's circuit will produce only slightly more power than an LM386 amplifier IC because they are similar. The power is low because the supply voltage is too low.
Higher current? How? Ohm's Law says that the amount of current is determined by the voltage swing divided by the speaker impedance. Your voltage swing is puny.

Your supply is only 5V so the peak voltage into an 8 ohm speaker is only about 1.5V from an LM386 for a peak current of 1.5V/8 ohms= 188mA then the AC RMS power is only 0.14W.
Harry's circuit has bootstrapping so its peak voltage is about 1.9V then its output power is only 0.22W which is still not much power.
The ISD1700 drives both wires of the speaker so its voltage swing is doubled so its output power is about 0.6W.
A car amplifier produces 15W into a speaker by using the 13.8V battery and a 4 ohm speaker (then the current is double what an 8 ohm speaker will draw). It also drives both wires of the speaker so its voltage swing is doubled.

High power car sound systems use a voltage booster circuit to produce 100VDC then the amplifiers produce massive amounts of power.
Many cars today use speakers that are only 2 ohms for high current.

So there's a separate equation then for audio power? because I was thinking ohms law... 5V/say a 10 ohm resistor = 500ma. 500ma times 5 = 2.5W

So my easiest solution here is to raise the voltage of the LM386's VS pin for more audio output wattage? If so, what should I use for minimum voltage? I'll be running my circuit on batteries alongside other circuits running through a 7805 voltage regulator.

Oh another question.. I've seen circuits where an output of an amp is connected to the base of an NPN transistor for more power I guess.

Would I be able to get more power with a one-transistor setup that handles higher power such as the one included in the schematics of the ISD1700 datasheet (except those transistors aren't commonly available here in Canada)?

I question this because you say the output from ISD1700 at AUD/AUX output is 1V peak to peak and thats less than the amplifier's output? I don't get it.

I thought by using an amplifier IC like the LM386, I'd get more power than a single transistor, but maybe I'm wrong

mik3ca said:

So there's a separate equation then for audio power?

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There is a graph in the datasheet for the LM386 that shows output p-p voltage swing vs supply voltage. Into an 8 ohm speaker the swing with a 5V supply is 3V and with a 9V supply the swing is 6V.
An equation for output power is RMS voltage swing squared divided by the speaker impedance. Then with a 5V supply the output is 3V divided by (2 x the root of 2)= 1.06V RMS and for a 9V supply the output is 6V (2 x the root of 2)= 2.12V RMS. You can calculate the current then calculate the power or you can use the equation V RMS squared divided by the speaker impedance. Then with a 5V supply the power is 0.14W and for a 9V supply the power is 0.56W.

So my easiest solution here is to raise the voltage of the LM386's VS pin for more audio output wattage? If so, what should I use for minimum voltage? I'll be running my circuit on batteries alongside other circuits running through a 7805 voltage regulator.

Click to expand...

The graph of output p-p voltage swing vs supply voltage on the datasheet for the LM386 shows that increasing the supply above about 10V does not increase the output power. Another graph shows that the dissipation (heating) increases instead.
The minimum input for a 7805 regulator is 7V but a 9V battery drops to 6V or less during its life so use a "low dropout" 5V regulator that works fine when its input is as low as 5.5V.

I think you should look at Texas Instruments list of modern audio power amplifier ICs and select one that uses class-D (switching instead of linear for low heating) and has a bridged output (drives both wires of the speaker) for much more output power.

A single transistor is a heater not an amplifier. It operates in "class-A" (look in Google) and 3/4 of the battery power produces heat when it has maximum undistorted output. Its heating continues wasting battery power even when there is no input signal.
A linear amplifier like the LM386 and Harry's amplifier operates in class-AB and its heating is about 4/10ths to half of the battery power but almost no heating when there is no input signal.
A class-D amplifier is always efficient and cool.

I managed to pull off a class-e amplfier with just a NPN transistor (collector to VCC through inductor and emitter grounded and base to input via capacitor). I used a 100K pullup resistor on the base. I hooked it to an FM transmitter and I was able to jam my old TV up.

I took the idea from this page which shows a schematic: http://www.classeradio.com/theory.htm

would such a class-e amplifier setup work with an ISD1700 AUD/AUX output?

I have never seen a class-E amplifier. Your link show it having severe distortion and the inductor smooths the waveform, something like class-C. It is a high frequency RF amplifier, not an audio amplifier.

I took a stab at trying to amplify the signal class-d style. The red waveform is the sawtooth input, the green waveform is the audio input (which here is a sinewave signal at V2) and the blue is the output. I don't know why I get this.

Now if I instead connect the inverting comparator input to the out pin of NE555 then the output is amplified, however running at the same frequency as the signal frequency from the NE555, but the output from NE555 is a square wave.

Is there any way I can amplify the original signal and still keep its frequency?

audioguru has already told you what you need for more power, a higher supply rail - different classes of amplifier don't provide any more more, they are still limited by the exact same laws.

In your class-D amplifier, the opamp (+) input is biased at 0V so its output is also 0V (the output is actually trying to go negative). Bias it at the same voltage as the average voltage of the (-) input.

Ok, so it seems a bias here is helpful as I receive a better (yet not perfect) waveform. I'm not sure if I have to apply a bias to both inputs or just the sawtooth input when hooking it up to the ISD AUD/AUX output.

I noticed as I was measuring the output at each inductor that there were more waves than usual (Do they call this ringing?)

The red preamp seems to produce the output at the right frequency and better amplification but somehow I need to make that waveform rounder. I wonder if my transistors are incorrect or if the inductor values are incorrect or am I missing something?

Since you are using a slow old opamp instead of a fast modern comparator then the red signal is a rounded awful looking "rectangular wave".
Your LC lowpass filter is resonating at its tuned frequency instead of being a filter. Look at the amplitude of the output from the filter, it is much higher than the power supply voltage and it even swings to a negative voltage when there is no negative supply voltage for the circuit.

audioguru said:

Since you are using a slow old opamp instead of a fast modern comparator then the red signal is a rounded awful looking "rectangular wave".

Click to expand...

Which comparator that LTspice has would you recommend I use to test with?

Your LC lowpass filter is resonating at its tuned frequency instead of being a filter. Look at the amplitude of the output from the filter, it is much higher than the power supply voltage and it even swings to a negative voltage when there is no negative supply voltage for the circuit.

Click to expand...

So I must be completely on the wrong track? Is there a better way to implement the filters?

Your supply voltage is still only 5V so the output power of your class-D amplifier will be very low.
I do not know if LTspice has a comparator. I have used LM393 dual comparators and LM339 quad comparators. I have used excellent switched capacitor lowpass filter ICs.
A Butterworth lowpass filter alignment has a flat frequency response before its cuts high frequencies. Use an active filter that has opamps, resistors and capacitors.