Low Power Class AB Amplifier

Thread Starter

maker_2023

Joined Nov 20, 2023
199
Hi:

I found a schematic for a 1 watt amplifier. I did a Spice simulation
and it looks like it only produces 490 mW into a 16 ohm speaker.
Is that correct? How could I modify the circuit to produce 1 watt
into an 8 ohm speaker?

Thanks,

M
 

Attachments

LowQCab

Joined Nov 6, 2012
5,101
If the selected Components of the Amplifier can withstand the additional Current,
then changing the Speaker to ~8-Ohms will result in almost ~1-Watt of Power
with no other Circuitry-changes.

It would be much more efficient to simply use 2- 16-Ohm Speakers instead of 1- 8-Ohm Speaker.
.
.
.
 

Thread Starter

maker_2023

Joined Nov 20, 2023
199
If the selected Components of the Amplifier can withstand the additional Current,
then changing the Speaker to ~8-Ohms will result in almost ~1-Watt of Power
with no other Circuitry-changes.

It would be much more efficient to simply use 2- 16-Ohm Speakers instead of 1- 8-Ohm Speaker.
.
.
.
Thanks LowQCab
 

MrChips

Joined Oct 2, 2009
34,628
The formula for power is

P = V x V / R

For example,
if V = 4 V rms and R = 16 Ω, P = 1 W rms
if V = 4 V rms and R = 8 Ω, P = 2 W rms

Multiply 4 V rms by 1.4 to arrive at peak voltage, i.e peak voltage = 5.6 V
Peak-to-peak voltage = 5.6V x 2 = 11.2 V
Hence you need at minimum, 12 V power source into the amplifier.
 

Thread Starter

maker_2023

Joined Nov 20, 2023
199
The formula for power is

P = V x V / R

For example,
if V = 4 V rms and R = 16 Ω, P = 1 W rms
if V = 4 V rms and R = 8 Ω, P = 2 W rms

Multiply 4 V rms by 1.4 to arrive at peak voltage, i.e peak voltage = 5.6 V
Peak-to-peak voltage = 5.6V x 2 = 11.2 V
Hence you need at minimum, 12 V power source into the amplifier.
Hi MrChips! Thanks.

M
 

MrChips

Joined Oct 2, 2009
34,628
12 V peak-to-peak into 8 Ω load is 6 x 6 / 2 / 8 = 2.25 W rms.
You are not going to get this from the class AB amplifier. It would be more like 1.5 W.
 

Thread Starter

maker_2023

Joined Nov 20, 2023
199
The formula for power is

P = V x V / R

For example,
if V = 4 V rms and R = 16 Ω, P = 1 W rms
if V = 4 V rms and R = 8 Ω, P = 2 W rms

Multiply 4 V rms by 1.4 to arrive at peak voltage, i.e peak voltage = 5.6 V
Peak-to-peak voltage = 5.6V x 2 = 11.2 V
Hence you need at minimum, 12 V power source into the amplifier.
Is this correct? The peak output from the circuit in post #1 is 2.82 which is 2.82 * 0.7071 = 1.99 vrms.
Prms = (1.99)^2 / 16 ohms = 247.5 mW rms?

M
 

Thread Starter

maker_2023

Joined Nov 20, 2023
199
I don't read .asc files. Is 2.82 peak or peak-to-peak?
If it is peak voltage, then yes, Power is 1/4 W rms.
Yes, 2.82 is the peak.

I find that a lot of posts show a circuit and claim that the output is a certain value
and when you simulate the circuit the output power is not even close. I am trying
to find a simple class ab that will produce 1 watt rms into an 8 ohm speaker.

M
 
Top