Elliott Sound Products | Voltage Dividers & Attenuators |

© 2002 - Rod Elliott (ESP)

Page Created 18 Dec 2002

Page Created 18 Dec 2002

Based on the number of requests for help I receive from people wanting to know how to connect a volume control, or convert speaker level to line level, I must conclude that voltage dividers (or attenuators) are not well understood.

A volume control is, in most cases, nothing more than a variable attenuator. Exactly the same formula applies to determine the output level for any given input level, and there is nothing mysterious about any of these building blocks. This small article will de-mystify the voltage divider in any of its forms, and will be useful for the beginner and accomplished hobbyist alike.

Although simple, there are a great many uses for the humble voltage divider, and indeed, without it many of the circuits we take for granted would not exist.

A voltage divider is created whenever you have two resistors (or impedances) in series, with the signal 'take-off' point between the two. Although there are literally hundreds of different possibilities, I shall only look at the standard connection for a voltage divider, and this will be all that is needed in the vast majority of cases. A traditional voltage divider is shown in Figure 1, and this is the form taken by volume and balance controls, general purpose attenuators, and similar configurations.

Basic Voltage Divider

This circuit is used for both AC and DC, and performs identically in either case. The voltage division is given by the formula ...

V_{d}= 1 + (R1 / R2)

Where V_{d} is the voltage division ratio. So, using two 1k resistors for example, voltage division is 1 + (1 / 1) = 2 ^{[1]}. 1V input will result in 0.5V output, and this holds true for DC, AC (RMS), or AC peak (as measured on an oscilloscope).

Any combination of R1 and R2 will create a voltage divider, and a pot (used for volume, for example) will still obey the same rule, except that the wiper (the moving contact in the pot) allows an infinite number of voltage division ratios.

Let's assume that you have a 25V RMS signal (the speaker output of an amp, for example), and want to reduce that to 1V RMS at maximum power from the amp. The voltage divider obviously must divide the voltage from 25V to 1V - or a 25:1 ratio. If we make R2 1k as before, R2 must be (25 - 1) × 1k = 24k ^{[2]}.

Hold on a minute - how did I arrive at that? The formula (in the form I have used here) can be transposed easily - it follows from the original that ...

V_{d}- 1 = R1 / R2

allowing you to easily determine the values needed for any voltage division ratio you need.

To convert voltage division to dB (attenuators are commonly referred to in dB), you need to apply the dB formula ...

dB = 20 × Log_{10}(V_{d})

... where V_{d} is the voltage division ratio determined as above. The two voltage dividers we used as examples above will give ...

dB = 20 × Log= 6dB_{10}(2)^{[1]}

and ...

dB = 20 × Log= 28dB_{10}(25)^{[2]}

... respectively (close enough).

The only other point to consider is the loading on the previous circuit, and the power dissipation in the voltage divider resistors. A voltage divider used to convert speaker level to line level (as shown in the second example) could just as easily use 1 Ohm or 1 Megohm for R2 (instead of 1k). The voltage divider / attenuator will still work exactly as before, so why would 1k be "better" than any other value? The answer is not especially simple, and it comes down to a compromise (all too common in electronics).

Let's look at the case of R2 = 1 ohm first. R1 will be 24 ohms, and there will be 24V across it (you *must* understand this concept - draw it out on paper to make sure that you do!). Power dissipation will be 24² / 24 = 24W. This is power that the amp must supply, and the resistor will get hot. Using 1 ohm for R2 is obviously not a good idea. Ok, how about 1Meg?

R2 will now be 24 Megohms - not an easy value to find! We will also come up against another issue - output impedance.

It is essential that any attenuator or voltage divider is driven from a low impedance source, or the load of the divider itself will reduce the available voltage (the formula shown will appear to be in error). Likewise, the load (the impedance connected to the output) must be high compared to the divider output impedance. It is generally considered that the signal source should have an impedance of at most 1/10 that of the attenuator, and the load should have an impedance (at least) 10 times the attenuator's output impedance.

The output impedance is the parallel combination of R1 and R2, so again using the examples above, we can determine input and output impedances.

Z= 2k_{in}= R1 + R2 = 1k + 1k^{[1]}

Z= 25k_{in}= R1 + R2 = 24k + 1k^{[2]}

Z= 500 ohms_{out}= (R1 × R2) / (R1 + R2) = (1k × 1k) / (1k + 1k)^{[1]}

Z= 960 ohms_{out}= (R1 × R2) / (R1 + R2) = (24k × 1k) / (24k + 1k)^{[2]}

These figures tell us that the maximum source impedance should be 2k / 10 = 200 ohms ^{[1]} and 25k / 10 = 2.5k ^{[2]}, and the minimum load impedances should be 500 × 10 = 5k ^{[1]} and 960 Ohms × 10 = 9.6k ^{[2]}. As it turns out, these are easily achieved by all common circuits used in audio.

It must be understood that even with a 'safety factor' of 10 as described, there will still be an error when the voltage divider is driven from any source impedance above zero, or is loaded by any circuit whose impedance is less than infinite - i.e. all voltage dividers will be in error to some (often significant) degree, unless the source and load impedances are included in the calculation. The load impedance is effectively in *parallel* with R2, and the source impedance is in *series* with R1.

Returning to the R1 = 24Meg, R2 = 1Meg impedances, it is obvious that the divider will be very easy to drive from any common circuit (due to the minimal loading), but the output impedance is much too high. The final voltage divider will be loaded excessively by any load impedance less than around 9.6 Megohms, and because of the high impedance, high frequency losses will be excessive if a cable is used at the output of the attenuator (the capacitance of the cable will be sufficient to shunt HF signals to earth, instead of allowing them to reach the source). Even stray capacitance in the attenuator itself will have an effect! For this example, a more sensible choice might be to use 24k and 1k (or perhaps 240k and 10k), which maintain fairly 'friendly' impedances for both input and output.

It is outside the scope of this little article to cover capacitive voltage dividers (or Resistor Capacitor dividers), but they are commonly used in high impedance circuits. Project 16 (Audio Millivoltmeter) does show a perfect example of this technique, which eliminates stray capacitance effects (at the expense of higher than normal input capacitance).

In fact, a voltage divider can be made using capacitors or inductors - but only for AC. These are much less common than resistive dividers, but still work in much the same way (they are somewhat harder to design though).

As is obvious from the above, the humble attenuator or voltage divider is not so humble after all. The maths are simple, and it is easy to convert any high voltage to a lower voltage. The divider technique is not suitable for *power* circuits however. It is used to reduce voltages, *not* current or power (although both are affected, that is a side effect, and not the real purpose).

Make sure that the resistance values you use are 'sensible', and do not impose excessive loading or introduce excessive output impedances, and it is hard to go wrong. Sensible (in this context) is something that comes with experience, but the guidelines given here should be more than enough to get you under way.

Page created and copyright (c) 18 Dec 2002