Circuit_Stuff‎ > ‎

Amplifier Gain

Gain is: Output/Input

The units can be anything: (Sound dB)/voltage, (revolution/second)/amp, Waistline/cookies. 

An "Amplifier" is something that takes an input signal and uses another power source to make the input signal larger

In High School I developed a love for mobile audio and louder was better. I went off to be an electrical engineer with the intent of "learning how to make amplifiers"... 7 years, 3 student in-field jobs, 1full-time job and a bachelors in Computer Engineering, Electrical Engineering, and a Masters in Electrical Engineering later I was hired to "make amplifiers".(It's not easy getting here, but it's really awesome and totally worth the effort) The first time I hung out with a work friend I made the statement that I wanted to make amplifiers and his natural response was: "what kind? Operational, RF, Low Noise, Wideband, Audio?". I always automatically associated "Amplifiers" with Audio Amplifiers which are power amplifiers but these others are what enable the world of electronics.

An Audio Amplifier is a power amplifier meaning it takes a small signal and "buffers" it to be able to drive big speakers, aka heavy load.

The steps required to make a home amplifier:

Build a power supply capable of converting a power source, 120v in the USA, to the appropriate output rails to support the rated power. Power transformer and diode rectifiers or a switch mode power supply, then supply filter capacitors, Inrush limiter circuit,

Build a gain stage capable of selectively and linearly increasing the input signal to the desired output level. This involves an input bandpass filter that linearly passes audio frequencies and an output bandpass filter that linearly passes audio frequencies.

Build an output stage capable of delivering the rated power to the load.

All of these things require math, the math is actually fairly simple and systematic, but there’s a lot of it. Understanding the system and persevering through all of the steps will provide a decent result. 


The gain on an audio amplifier sets what input voltage(around 1Volt in) corresponds to a full scale output.  
In high school I didn't care about the amplifier maxing out... I just wanted the subs to be as loud as possible. I would set the gain to max the amplifier out at about half volume on the radio. Lets talk about what this was doing...

First, lets say the maximum input level is 1 and the maximum potential output level is 2
A gain of 2 will set the maximum output level to: Input*Gain = Output = 1*2 = 2
Gain of 2
If we want to put out the maximum output with only half the maximum input level, in other words max sub level for half volume... Then we would set the gain to 4. 
Gain of 4 
(The labels are flipped on the second graph)

But Audio amplifiers don't usually have a Gain in numbers or they are some numeric scale that doesn't map like that... The volume control does directly control the gain, but the label is simply the way an input is mapped to the power level. An AVR(Audio Video Receiver) assumes a signal always has a maximum peak to peak level of 1 volt and based on this the volume control is how much of the maximum power can be used. 

A car(mobile) audio amplifier is designed to take a varying level signal from another source, like a head unit, and translate this signal as described above. Typical mobile audio amplifiers have a gain setting labeled in volts. A head unit will describe what voltage level is the line level full scale outputs. This varies between manufacturers and is from 0.5V to 8V. The larger the voltage from the headunit the less proportional contribution an imposing noise source(Like alternator hum) will have.

Example: Head Unit puts out a maximum scale of 0.5V and the alternator induces a 0.05 Volt signal onto the signal(RCA) cables. This is the Signal to Noise ratio of 0.5(signal)/0.05(Noise) which means the signal is 10x the noise... 10x sounds significant but fundamentally there must be a difference of 7x(17dB) to be able to distinguish a sound from noise at all. 

For the same noise and an 8V line level signal: 8/0.05 = 160x SNR, a 16x improvement by increasing the input level while the noise stays the same. The gain to take 8V to full scale power will also be 1/16th that required for the 0.5x signal. The maximum output level doesn't change and is determined by V^2/R as seen in https://sites.google.com/site/0123icdesign/circuit_stuff/electrical-power with R being the total  impedance of the speakers and V being calculated by the total or RMS output power of the amplifier.

Example: What will the voltage rails be for a 500Wx1 @ 4ohms RMS amp and what gain setting should be used with a 1V head unit.

Solution: First RMS to Peak Power, RMS power for a sine wave is Peak/sqrt(2). 

500Wrms = Peak/sqrt(2)
(move sqrt(2) across equals sign)
500Wrms X sqrt(2) = Peak Power = 707Watts peak

So then what voltage will provide 707 Watts into a 4 ohm load.

V^2/R = 707Watts
V^2/(4ohms) = 707Watts
V^2 = 707Watts*4ohms
Take square root (sqrt) of both sides
V = sqrt(707 X 4) = 53.2V

So the amplifiers maximum voltage output is 53.2V. 
For an input with a 1Volt maximum the gain needs to be:
(53.2V out)/(1Volt in) = a gain of 53.2.

If the head unit was putting out 4 volts instead of 1V:
(53.2V out)/4V in) = 13.3

Ok, so what happens when I decrease the load by putting speakers in parallel?
The Voltage Rail doesn't change, but the current does! This brings in the other power equation: I^2*R.

The power supply and the output devices of the amplifier themselves must be designed for a specific current. If the power supply can't deliver enough current, the voltage rail will "droop" to whatever level it can support. If the output devices aren't sized to handle the current then they will overheat and burn up.

 The third power equation: Power = Voltage*Current (P=V*I) guides the design and sizing of the power supply. The amplifiers power supply must be able to support the continuous level output. 

Example: What is the power supply rating for a 90% efficient, 500wrms x 2 class D amplifier mobile amp?

As we just talked about the 500wrms actually requires 707watts peak to peak. 2x707 = 1414watts. Then a typical 90% class D efficiency means that 90% of the input power is delivered to the speakers. In equation form: Input Power * 0.9 = 1414. move the 0.9 across the equal sign(multiply both sides by 1/0.9) and the result is 1414/0.9 = ~1571. 

A conservatively rated amplifier is based on 12v purely from the battery, while a liberally rated amplifier is based on 14.4 which is typical voltage when the alternator is charging the battery... this 2.4v makes a 20% difference in the rated power for unregulated supplies. Finally, what is the max current pulled from the battery for this amplifier: Power = Voltage*Current so Power/Voltage = Current: 1571/14.4 = ~110Amps whereas 1571W/12v = ~131Amps. This current can be viewed as Peak. RMS currents would be 92.57Ams@12v and 77.1Amps@14.4




Comments