# Calculate necessary attenuation (in dB) to achieve a given output voltage

I need to connect an audio unit with an output of 2V (and an output impedance of 560 ohms) to an amplifier with input sensitivity of 0.985V. Since the 2V output is too high for the amplifier, I need to attenuate the signal, but I'm having trouble calculating how many dBs would be necessary to lower the signal from 2V to, say, 0.9V. How should I calculate this?

• Welcome to EE.SE. Good question. You need a slightly over 6dB voltage drop.
– user105652
Jul 26, 2016 at 1:01
• @Eduardo What's the input impedance of your amplifier? Jul 26, 2016 at 6:01

This is a fairly routine calculation, the attenuation you need can be expressed as a voltage gain, i.e., the ratio of the input voltage to the output voltage:

$$G = 20\log_{10}\left(\frac{V_{out}}{V_{in}}\right) = 20\log_{10}\left(\frac{0.985}{2}\right) \approx-6.15~\text{dB}$$ or if you want to limit the output to a maximum of 9 V, $$G = 20\log_{10}\left(\frac{0.9}{2}\right) \approx-6.94~\text{dB}$$ You will want to match the 6-7 dB attenuator to your load impedance if you still want to obtain maximum power transfer.

The following circuit will give an attenuation of 6.5 dB and have an impedance of approximately 560$~\Omega$. simulate this circuit – Schematic created using CircuitLab

• Another method is to simply create a voltage divider that reduces the signal by a bit more than 50%, then choose resistor values so that the small signal impedance is close to 560 Ohms. Jul 26, 2016 at 0:17
• @AndrewW. Yes, that's true but he asked about how much attenuation he needed, not how to accomplish it. A resistive attenuator is just as simple to implement though and will maintain the same system impedance more accurately than a voltage divider. Jul 26, 2016 at 0:20
• Awesome @Captainj2001, thanks for the comprehensible explanation. I saw this formula on some pages but wasn't sure about it, so I thought would be better to ask and avoid blowing speakers because of a flawed attenuator. Since we're on it, in case I don't match the impedance, will the difference be significant? Because I'm considering buying a "plug-like" attenuator, so I have no control over its impedance. Jul 26, 2016 at 0:32
• @Eduardo You might consider making your own or buying one that has an impedance close to the 560$~\Omega$ system impedance you're using. If you use one that doesn't match the system impedance it will not attenuate the voltage as you expect. For instance if you buy one for a lower system impedance the voltage will still be above your threshold, if you buy one for a higher system impedance it will attenuate your voltage more than you expect. Jul 26, 2016 at 1:06
• @Captainj2001 The attenuator I've found is passive, so it's pretty likely it has a very low impedance. I'm considering buying a version with a higher attenuation value (e.g. 9 or 12 dB) to compensate for the impedance and also because 0.9V is to drive the amplifiers to full volume, which is something I don't want because of distortion. Is there a way to calculate the output voltage considering impedance? Jul 26, 2016 at 1:22

When we work in a system of defined impedance, we talk of attenuators. So RF tends to be all ports 50 ohms, telephony is all ports 600 ohms. Attenuators are designed expecting those impedance on the ports, and only giving the right attenuation when that is the case.

When an output drives an input without an attenuator, the voltage the input receives is half the open circuit output voltage of the preceding stage, because of the loading. If an output drives an attenuator that is not loaded, the attenuator output will rise to twice its nominal design output voltage.

The value of attenuation is well defined, as both input and output impedances are the same, the attenuation figure is the same whether calculated for voltage, current or power.

When we work in typical audio systems, with low output impedances, and high input impedances, we tend to talk of voltage dividers. Voltage dividers are designed load the preceding stage only lightly, and to be relatively unloaded by the stage following.

When an output drives an input without a voltage divider, the voltage the input receives is essentially equal to the open circuit output voltage of the preceding stage, because it is not loaded. If we put a finite load on a voltage divider, the output voltage will drop a little.

As the input and output impedances are different, the only sound computation for attenuation should be the power attenuation. Unfortunately this means the voltage ratio and the current ratio cannot be computed from the attenuation figure alone, but must include the port impedances for calculation. As a high impedance system might be 20k, or 1Meg, this makes life difficult. As these systems tend to respond to the input voltage, we tend to talk about voltage levels and ratios, rather than attenuation. Attenuation figures are sometimes given for just the voltage ratio, which most audio practitioners will understand, but purists will object to. A voltage ratio of 0.5 is just a tad over 6dB.

To pot-down (ie reduce in voltage by a potentiometer, an adjustable voltage divider) an output to a half (roughly what's needed here) we'd use two nominally equal value resistors. In series, they must be higher resitance than the driving stage, in parallel they must be lower resistance than the input they are to drive. In the case of 560 driving, 20k driven, that's about 3300 ohms, which gives a better than 10:1 ratio of the divider to the port impedances. simulate this circuit – Schematic created using CircuitLab

Due to the relatively small margin between voltage divider impedance and the port impedances, this equal resistor voltage divider gives 42% output voltage, a little lower than the 50% original target. R2 can be reduced and R3 increased slightly to get the exact ratio required. Alternatively, R2 and R3 could be replaced by a 5k or 10k potentiometer, to make the ratio adjustable.