## How Does an Attenuator Work?

A Little Simple Math and Circuit Design

**What's it All About?**

This is an attempt to explain audio power attenuators and how they work in a tube guitar amp.

Lets take a look at a simplified model of the output stage of an amplifier. It isn't correct. Don't try to built am amp using this structure, but it's enough to help us understand the role of load impedance.

In the diagram, on the left is a signal source with 10 ohms in series, representing the output tubes with their effective internal resistance. On the right is an 8 Ohm speaker and in the middle is a transformer with a 3:1 turns ratio.

The amplifier looking into the transformer, sees the load presented, as transformed by the transformer. So the 3:1 turns ratio makes the 8 Ohm load look like 24 Ohms.

The transformer serves a number of roles. It isolates the speaker output from the high DC voltages within the amplifier. It acts as a DC block, so the DC bias on the amplifier side is removed. Most importantly, the load needed by the amplifier doesn't look like the load presented by a speaker, so the transformer transforms the load to look like one that is needed by the amp.

Of course amplifiers clip when the signal amplitude is above some level. Let's say this particular circuit clips at above 10V peak to peak. So at peak amplitude, relative to the minimum amplitude, we have 10V into 10 Ohms and 24 Ohms in series. So 29.4% of the energy is lost as heat in the amplifier and the remaining 70.6% goes through the transformer to the speaker.

If you plugged in a 4 Ohm speaker, the presented load on the other side of the transformer would be 12 Ohms. So now 45% of the energy would be lost as heat in the internal resistance of the amplifier. So the tubes would get hotter, potentially leading to circuit damage. Many amps provide a an additional center tap output on the output side of the transformer, effectively doubling the turns ratio, so it can drive an load of half the impedance.

Lets consider attenuating the signal. Obviously it is going to involve putting some resistors in the way, so that some of the electrical energy is lost in the resistors before it gets to the speaker.

The amplifier looking into the transformer, sees the load presented, as transformed by the transformer. So the 3:1 turns ratio makes the 8 Ohm load look like 24 Ohms.

The transformer serves a number of roles. It isolates the speaker output from the high DC voltages within the amplifier. It acts as a DC block, so the DC bias on the amplifier side is removed. Most importantly, the load needed by the amplifier doesn't look like the load presented by a speaker, so the transformer transforms the load to look like one that is needed by the amp.

Of course amplifiers clip when the signal amplitude is above some level. Let's say this particular circuit clips at above 10V peak to peak. So at peak amplitude, relative to the minimum amplitude, we have 10V into 10 Ohms and 24 Ohms in series. So 29.4% of the energy is lost as heat in the amplifier and the remaining 70.6% goes through the transformer to the speaker.

If you plugged in a 4 Ohm speaker, the presented load on the other side of the transformer would be 12 Ohms. So now 45% of the energy would be lost as heat in the internal resistance of the amplifier. So the tubes would get hotter, potentially leading to circuit damage. Many amps provide a an additional center tap output on the output side of the transformer, effectively doubling the turns ratio, so it can drive an load of half the impedance.

Lets consider attenuating the signal. Obviously it is going to involve putting some resistors in the way, so that some of the electrical energy is lost in the resistors before it gets to the speaker.

The resistance presented by the speaker, resistor combo is now 8 + R Ohms. So it cannot be 8 Ohms, so the load is incorrect for an amplifier needing an 8 Ohm load. You could put 8 Ohms in series with the 8 Ohm load, making it present 16 Ohms. However the power to the speaker would be cut by only a half, which is -3dB, which would not sound much quieter to the human ear.

An alternative would be to put a resistor in parallel with the the speaker. This would reduce the load present to the amplifier. If you put the same value resistor as the speaker impedance in parallel with the speaker, it would halve the load presented E.G. from 8 Ohms to 4 Ohms and again only reduce the volume to -3dB. So the effect would be small.

An alternative would be to put a resistor in parallel with the the speaker. This would reduce the load present to the amplifier. If you put the same value resistor as the speaker impedance in parallel with the speaker, it would halve the load presented E.G. from 8 Ohms to 4 Ohms and again only reduce the volume to -3dB. So the effect would be small.

The solution is to do both. Put resistors both in parallel and series with the speaker. Then you can choose the right resistances such that the same impedance is presented to the amplifier and the chosen amount of attenuation is achieved.

The series resistor has to be less than the intended impedance, so that its value plus the effective impedance of the parallel part add up to the intended impedance. Then for whatever the chosen series resistor, there is only one value of parallel resistor that would lead to the chosen impedance and this combination leads to one amount of attenuation. So from the series resistance, you can compute the parallel resistor and the gain.

The resistance of the parallel section Rb = Rp || Rload = (Rp*Rload)/(Rp + Rload)

The voltage across the load relative to the input voltage V is Vload = Rb/(Rb+Rs)

The gain is 10 log base 10 (Vload/V)

So now we get to do some simple mathematics.

We'll call the series resistance Rs and the parallel resistance Rp. Rload or just is the speaker impedance and is the same as the overall impedance R since the goal is the match the speaker impedance. So R = Rload.

We'll also invent Rb, the effective resistance of the Rp in parallel with the speaker. Which is:

Rb = Rp in parallel with Rload = (Rp*Rload)/(Rp + Rload)

The voltage across the load relative to the input voltage V is Vload = Rb/(Rb+Rs)

The gain is log base 10 (Vload/V)

In order to match the impedance Rs+Rb = Rload. So we much choose the Rp that makes this the case.

Rs + ((Rp*Rload)/(Rp + Rload)) = Rload

So we can rearrange to compute Rp from Rload and Rs thusly:

((Rp*Rload)/(Rp + Rload)) = Rload - Rs

Rp * Rload = (Rp + Rload)(Rload-Rs)

Rp * Rload = Rp*Rload - Rp*Rs + Rload^2 -Rload*Rs

Rp*Rs = Rload^2 -Rload*Rs

Rp = ((Rload^2)/Rs)-Rload

So for whatever Rs we choose we can compute Rp as ((Rload^2)/Rs) - Rload and the impedance will be matched. We can graph Rp against Rs. This graph is computed for Rload = 8:

The resistance of the parallel section Rb = Rp || Rload = (Rp*Rload)/(Rp + Rload)

The voltage across the load relative to the input voltage V is Vload = Rb/(Rb+Rs)

The gain is 10 log base 10 (Vload/V)

So now we get to do some simple mathematics.

We'll call the series resistance Rs and the parallel resistance Rp. Rload or just is the speaker impedance and is the same as the overall impedance R since the goal is the match the speaker impedance. So R = Rload.

We'll also invent Rb, the effective resistance of the Rp in parallel with the speaker. Which is:

Rb = Rp in parallel with Rload = (Rp*Rload)/(Rp + Rload)

The voltage across the load relative to the input voltage V is Vload = Rb/(Rb+Rs)

The gain is log base 10 (Vload/V)

In order to match the impedance Rs+Rb = Rload. So we much choose the Rp that makes this the case.

Rs + ((Rp*Rload)/(Rp + Rload)) = Rload

So we can rearrange to compute Rp from Rload and Rs thusly:

((Rp*Rload)/(Rp + Rload)) = Rload - Rs

Rp * Rload = (Rp + Rload)(Rload-Rs)

Rp * Rload = Rp*Rload - Rp*Rs + Rload^2 -Rload*Rs

Rp*Rs = Rload^2 -Rload*Rs

Rp = ((Rload^2)/Rs)-Rload

So for whatever Rs we choose we can compute Rp as ((Rload^2)/Rs) - Rload and the impedance will be matched. We can graph Rp against Rs. This graph is computed for Rload = 8:

But we also care about the gain.

The gain can be computed at -10Log10(Load_Power/Amp_power)

So ploughing on, lets find those values:

The amp power P is delivered into the unit. The total current though it is equal to the series current through the series resistance.

So

From basic electrical theory: P = VI, V = IR so P = I^2R, so I^2 = P/R

So

I = SQRT(P/R) , We are given P - It's the power of the amplifier. R is the overall impedance we want (4, 8 or 16 ohms usually)

Ps is the power in the series resistor, Rs is the series resistance. This tells us the the power capacity of the series resistance required:

Ps = I^2Rs

The volts across the series resistor are:

Vs = IRs

The volts across the whole thing are:

V = IR

The volts across the parallel part and load are:

Vp = V-Vs

Vp = V-IRs

Vp = V-(SQRT(P/R)xRs)

The power in the parallel resistance Pp is:

P = I^2R

I = V/R

P = V^2R/R^2

So Pp = Vp^2 / Rp

The impedance of the load equals the impedance of the whole system. So the power in the load is:

Pload = Vp^2 / R

The gain of the attenuator in decibels is 10 Log10 (Pload / P)

Graphed against Rs on the X axis from 0 to 8 Ohms, the gain on the Y axis looks like this for an 8 Ohm load:

The gain can be computed at -10Log10(Load_Power/Amp_power)

So ploughing on, lets find those values:

The amp power P is delivered into the unit. The total current though it is equal to the series current through the series resistance.

So

From basic electrical theory: P = VI, V = IR so P = I^2R, so I^2 = P/R

So

I = SQRT(P/R) , We are given P - It's the power of the amplifier. R is the overall impedance we want (4, 8 or 16 ohms usually)

Ps is the power in the series resistor, Rs is the series resistance. This tells us the the power capacity of the series resistance required:

Ps = I^2Rs

The volts across the series resistor are:

Vs = IRs

The volts across the whole thing are:

V = IR

The volts across the parallel part and load are:

Vp = V-Vs

Vp = V-IRs

Vp = V-(SQRT(P/R)xRs)

The power in the parallel resistance Pp is:

P = I^2R

I = V/R

P = V^2R/R^2

So Pp = Vp^2 / Rp

The impedance of the load equals the impedance of the whole system. So the power in the load is:

Pload = Vp^2 / R

The gain of the attenuator in decibels is 10 Log10 (Pload / P)

Graphed against Rs on the X axis from 0 to 8 Ohms, the gain on the Y axis looks like this for an 8 Ohm load:

So putting it all together

R = Rload

Rs is the chosen series resistor

Rp = ((Rload^2)/Rs)-Rload

gain = 10 Log10 (Vp^2/RxP)

The power dissipated in Rp is Pp = (V-(SQRT(P/R)xRs))^2 / Rp

The power dissipated in Rs is P(Rs/R)

Now we have these equations, it's easy to compute the right resistances Rs for the desired gain and compute the necessary Rp to then match the impedance. Then you can compute the power dissipated in the resistors so you know what resistors to buy.

After a few experiments, it was found that -20dB was perfect for turning a too-loud 25W tube amp with a 12" speaker down to volume levels that are useable in a home or apartment, restoring the ability to use the range of the master knob to set the exact volume you require. You might want less. So -10dB might be useful also.

Computed values for some common useful values are:

4 Ohm, -20dB: Rs=3.6 Ohms. Rp=0.444... Ohms. [Shameless Commercial Plug: You can buy a 4 Ohm attenuator here.]

8 Ohm, -20dB: Rs=7.2 Ohms. Rp=0.888... Ohms. [Shameless Commercial Plug: You can buy an 8 Ohm attenuator here.]

16 Ohm, -20dB: Rs=14.4 Ohms. Rp=0.1.777... Ohms. [Shameless Commercial Plug: You can byte a 16 Ohm attenuator here.]

4 Ohm, -10dB: Rs=2.736 Ohms. Rp=1.848 Ohms.

8 Ohm, -10dB: Rs=5.471 Ohms. Rp=3.698 Ohms.

16 Ohm, -10dB: Rs=11.985 Ohms. Rp=5.36 Ohms.

Next up we'll try to find out how to turn these into a real attenuator, with the associated problems of mapping to actually available resistors, meeting heat dissipation specs, avoiding unwanted frequency selective parasitics and building a high quality circuit.

See also:

What Ohms? Looking at the impedance requirements of amplifiers

R = Rload

Rs is the chosen series resistor

Rp = ((Rload^2)/Rs)-Rload

gain = 10 Log10 (Vp^2/RxP)

The power dissipated in Rp is Pp = (V-(SQRT(P/R)xRs))^2 / Rp

The power dissipated in Rs is P(Rs/R)

Now we have these equations, it's easy to compute the right resistances Rs for the desired gain and compute the necessary Rp to then match the impedance. Then you can compute the power dissipated in the resistors so you know what resistors to buy.

After a few experiments, it was found that -20dB was perfect for turning a too-loud 25W tube amp with a 12" speaker down to volume levels that are useable in a home or apartment, restoring the ability to use the range of the master knob to set the exact volume you require. You might want less. So -10dB might be useful also.

Computed values for some common useful values are:

4 Ohm, -20dB: Rs=3.6 Ohms. Rp=0.444... Ohms. [Shameless Commercial Plug: You can buy a 4 Ohm attenuator here.]

8 Ohm, -20dB: Rs=7.2 Ohms. Rp=0.888... Ohms. [Shameless Commercial Plug: You can buy an 8 Ohm attenuator here.]

16 Ohm, -20dB: Rs=14.4 Ohms. Rp=0.1.777... Ohms. [Shameless Commercial Plug: You can byte a 16 Ohm attenuator here.]

4 Ohm, -10dB: Rs=2.736 Ohms. Rp=1.848 Ohms.

8 Ohm, -10dB: Rs=5.471 Ohms. Rp=3.698 Ohms.

16 Ohm, -10dB: Rs=11.985 Ohms. Rp=5.36 Ohms.

Next up we'll try to find out how to turn these into a real attenuator, with the associated problems of mapping to actually available resistors, meeting heat dissipation specs, avoiding unwanted frequency selective parasitics and building a high quality circuit.

See also:

What Ohms? Looking at the impedance requirements of amplifiers