Voltage Division and Simple RC Filters
V out will always be smaller than V in. As an ideal circuit, the input and output relationship is perfectly linear. The attenuation is proportional to the ratio of R2/(R1 + R2).
A potentiometer is the realization of the simple voltage divider circuit as a simple mechanical device that allows for an infinitely variable adjustment of the R1 and R2 ratio. The name "potentiometer" comes from the term "potential electromotive force," a term which is synonymous with voltage. Perhaps this was chosen since "voltmeter" was already in use as the device used to measure voltage.
Important design considerations:
R1 adds series resistance to Vin, raising the source impedance.
R2 loads Vin. If the total R1 + R2 is less than ≈10x the source impedance of Vin then you must add the source impedance to R1 for the formulas to be accurate.
Along the same lines, the "load" attached to the Vout terminal should have an input impedance that is ≈10x R1 + R2.
Generally it is best to keep R1+R2 as low as the source impedance allows so that the resulting output impedance of the divider is only a modest increase in impedance over the source. Driving the input of a voltage divider with a very low impedance voltage source (such as the output of an op amp) allows for a minimum impedance design (see Douglas Self Small Signal Audio Design).
Useful formulas:
Vout = Vin * ( R2 / (R1 + R2) )
R1 = R2 (Vin - Vout) / Vout
R2 = (-Vout)(R1) / (Vout - Vin)
If R1 = R2, then Vout = Vin / 2
If a balanced attenuator is required, I recommend the "U" configuration and placing it close to the balanced input.
Design starts with calculating the minimum input impedance allowable for the source impedance. This can be found by either checking a manufacturer's data sheet, an owner's manual, a schematic, opening the unit up and drawing a schematic, or through testing. For a general purpose pro audio device (source unknown), assume a 10k minimum input impedance.
Next choose the loss desired, such as -6dB, -20dB, -40dB, etc. In voltage loss, those correspond to divisions of 1/2, 1/10, and 1/100, respectively.
Use the unbalanced 2 resistor circuit and formulas to find resistor values for your desired voltage division. Following low impedance design practices, keep R2 as low as the minimum impedance allows.
To convert from the unbalanced design to the balanced "U" circuit, simply divide the unbalanced R1 by 2. This is why there are 2 "R1" resistors in the Balanced "U" Pad drawing above.
Calculations will probably result in non-standard resistor values. I will select R2 as a standard value then calculate R1. Tolerances for both input impedance and actual voltage loss need to be calculated. Once the tolerances are set, try to round the calculated R1 value up or down to the nearest standard resistor values and check if the final design is within the set tolerances.
A common alternative to the "U" configuration is the "H" configuration. The "H" is reversible and "matched," i.e. it is appropriate for maximum power transfer. When transmitting voltage as a signal with negligible current draw, we are not trying to transfer power. For attenuating signals (such as line level audio over XLR cables), the "U" configuration is more appropriate.
Comparison to simple RC filter circuits:
Replacing the upper resistor with a capacitor creates a high pass filter. The basic idea is the same as a dc voltage divider, except the capacitor can be considered a "frequency dependent resistor." As the frequency of an ac voltage goes down, the "resistance" of C appears to go up. At dc (frequency = 0 or very close to 0) the capacitor completely blocks the voltage, similar to the effect of making R1 huge in the first voltage divider circuit. At high frequencies, the "resistance" of C is very small and the output is very close to the input. We call this "frequency dependent resistance" by a special term: reactance. Capacitors have capacitive reactance, whereas inductors have inductive reactance. The symbols for reactance are XC and XL.
Why and how we solve for a ≈30% reduction
Simple RC filters have a non-linear element to them that adds complexity the simple two resistor divider model. The non-linear element is the exponential charging and discharging behavior of capacitors. The graph of y = x(1-e^(-1/x) shows the logarithmic curve of a cap charging up, where e is Euler's number, y is the charge on the cap, x is the capacitance, and the resistance and time in seconds are both given as 1 (and have already been eliminated from the given equation). Further reading on this property of capacitors: RC Time Constant Wikipedia
To approach simple RC filters as "ac voltage dividers" becomes problematic when we want to pick some arbitrary division or attenuation for a given input. With two resistors, it is very easy to select any arbitrary division you like. Once a capacitor is introduced, with its curvy rates of charge and discharge, solving for specific amounts of attenuation along the curve is possible but much more complex.
Rather than trying to deal with the complexity of the entire curve, it is conventional to simply focus on the point where R = C. If we focus on that point, we can then return to our relatively simple Vout = Vin (R2/R1+R2) formula, update it for use with capacitors, and we will find a useful "rule of thumb" that can be applied for quick filter calculations.
When we set R = C there will be a specific frequency that will be attenuated by -3dB by passing through the filter. The -3dB frequency the often referred to as the "cutoff frequency." "Cutoff frequency" does not always means the -3dB point, but in most cases you can make the assumption unless something else is specified such as -1dB.
Why -3dB? That point is analogous to the equal resistors in the purely resistive voltage divider, and thus is it is relatively easy to calculate once you understand the formulas.
To consider the other options, you can return to the curvy functions produces by a charging and discharging capacitor. It is possible to calculate the -1dB point, the -2dB point, etc, but since these points all lie upon a curved line, it is arguably more complex to solve for these other points. The -3dB point is arguably the simplest point to calculate, and it is useful in real life applications.
-3dB is approximately equal to a 30% reduction. The next section shows how that number is calculated.
Voltage Division with Impedance
When a reactance and a resistance are combined, the result is an impedance.
Our voltage divider formula using capacitors has the same structure as the divider with just resistors, but it must be expressed in terms of impedance.
Z stands for impedance. Ztotal means the combination of C1's reactance and R1's resistance (analogous to the expression R1 + R2 we used previously).
Ztotal = Z1+Z2
Z1=Xc
Z2= R1
Vout = Vin (Z2 / Ztotal).
To find impedance of any component, we must add the resistance of the component to the reactance of the component. This is calculated using a Cartesian plane (aka "xy" graph) where the resistance is plotted along the horizontal, or "x" axis, and the reactance is plotted on the vertical, or "y" axis. We cannot sum them directly (x+y), but we can sum them as a vector, which means we will measure the length of the line segment from the origin (0,0) to our (x,y) point. (Consult a physics textbook for a deeper explanation.)
Our line segment forms the hypotenuse of a triangle with length x and height y. The Pythagorean Theorem says the length of the hypotenuse will equal the square root of the sum of x2 + y2:
Ztotal = √(R12 + XC2), where XC is the capacitive reactance of C1.
To show the analogy to the purely resistive divider, we will solve Ztotal for when R1 = 1 and XC = 1.
Ztotal = √(12 + 12)
Ztotal = √(1 + 1)
Ztotal = √2
Z2, since it is a resistor, will be = √(12 + 02), = 1.
Let's plug Ztotal = √2 and Z2 = 1 into the voltage divider equation for impedance:
Vout = Vin (1 / √2)
If we estimate √2 to be 1.414, then the equation can be become:
Vout = Vin * 0.707, which can be interpreted as "the output voltage will be about 70% of the input voltage."
As an expression of decibels, multiplying by 1/√2 is expressed as -3dB. A much neater and easier to digest expression.
How to solve for the -3dB frequency
You can solve for the "-3dB cutoff frequency," shorted to f-3dB, by using this formula:
f-3dB = 1 / (2πRC)
This formula is derived from where C's reactance = R's resistance at a desired frequency (hopefully we now understand why that's meaningful).
If the frequency is known, and you just need to solve for either R or C, they conveniently just swap after switching things around with algebra:
R = 1 / (2πCf-3dB)
C = 1 / (2πRf-3dB)
Once you have the -3dB point figured out, the attenuation decreases as the input frequency rises, and increases as the input frequency lowers.
If you're confused as to why the "2π" constant just jumped in the equation, it expresses 360º of an ac signal in radians. The "C" value has an inherent relationship to frequency that is expressed in the capacitive reactance formula which only differs from the above "R = ..." formula by putting XC in place of R, as in:
XC = 1 / (2πCf-3dB)
So the filter mechanism could also be expressed as:
When XC = R for a specific input frequency, the output will be -3dB less than the input (as shown above).
[Next topic: The graph of y = x /(sqrt(x^2 + 1)), x domain is 0-1, this shows what happens if you replace R1 with a linear potentiometer. Explains the limitations and challenges of simple adjustable RC filters.]
Replacing the lower resistor of a voltage divider with a capacitor creates a low pass filter. The capacitor is still presenting a low reactance to high frequencies and a high reactance to high frequencies, but now its role in the voltage divider has flipped around. Now low frequencies, down to dc, are passed along fine since C "looks" like a big resistor, whereas high frequencies "see" C as a very small resistor and are attenuated accordingly.
All of the above formulas apply, but now attenuation decreases as the input frequency lowers, and increases as the input frequency rises.
Experiment/Lab
Items needed:
9V battery
Alligator clips
Two 1k 1/4W resistors (or any two resistors of the same value, greater than 1k)
Voltmeter
Voltage division is a very important concept. If you can understand voltage division, you can understand basic amplifiers, and basic amplifiers are the heart of most audio circuits.
So far, it is expected that you understand what a power supply/voltage source is, at least at a very basic level. For the purposes of our first voltage division, we just need something like a 9V battery. It can deliver 9V to a 2k load without "loading" too much (see the batteries and resistors page).
A 9V battery supplies about 9V... maybe a little more, and less and less as it runs down... but what if we want or need a different voltage? For a higher voltage, we either need to pick a different battery (like a 12V battery), use a special circuit (a bit advanced for us at this point!), or put 2 9Vs in series to obtain 18V. Higher voltages from series connected batteries tend to only be available in 1.5V increments. What if we want an arbitrary voltage? What about lower voltages? Can you connect a battery in parallel and get a lower voltage? No. Two 9Vs in parallel still have 9V, but they have twice the current handling as a single 9V.
Hm. How does one make a smaller voltage? The (simple) answer is voltage division. Here's the experiment:
Join the two resistors together in series.
Take one of your two resistors of equal value (1k or higher) and twist one lead around one lead of the other resistor.
If they are 1k resistors, this is the equivalent of having a single 2k resistor, except we have a place we can measure voltage in the middle of our new "2k" resistor.
Connect, with alligator clips, the + terminal of the 9V battery to one of the unconnected resistor leads, and the - terminal of the battery to the other unconnected resistor lead.
The circuit should go from 9V + terminal to a 1k resistor, to another 1k resistor, to the 9V - terminal.
Measure the voltage across the battery. Red probe to + terminal, black probe to - terminal. This should read 9V or more. Keep the black probe to the - terminal, and move the red probe to the junction of the two resistors. Note this voltage reading.
What did you get? Was is near 4.5V, or 1/2 of the reading at the battery + terminal?
Congratulations, you have "created" a 4.5V point that is very "usable" for many, many circuits. This is a powerful skill to be able to "create" new voltages from a fixed voltage.
Now that you've done it, the next question is why did that work?
Let's use Ohm's Law to think about it.
As a "2k" resistor, we know that there's +9V on one side of the resistor, and 0V on the other.
Ohm's law says there will be current flowing in the following amount:
Current = Voltage / Resistance
Current = 9V / 2,000Ω
Current = 0.0045A
The current is the same for all of our series connected elements... in this case, that's the battery and 2 resistors.
Using Ohm's Law again, we solve for voltage across each 1k resistor, using 0.0045A as the current.
Voltage = Current x Resistance
Voltage = 0.0045A x 1,000Ω
Voltage = 4.5V
That means that from the +9 battery terminal to the junction of the two resistors there is a 4.5V drop. "Looking" from the other side, there is a 4.5V drop from the 0V, or - terminal, to the two resistor junction. Either way, we have 9V-4.5V = 4.5V at the junction of the two resistors.
These simple calculations are very drawn out, but they will help you understand the more complicated calculations.
In the Battery and Resistor as Source and Load experiment, we observed a "loading" effect that a 1k resistor can have on a 9V battery. Unloaded, the 9V may have 9.5V or so, and a load of 1k or so will cause a slight drop in voltage across the battery terminals.
For 9V batteries, I use 3Ω as the internal impedance of the battery. This is like having a 3Ω resistor permanently connected at the + terminal of the battery. It could be at either terminal, or imagined as being "in between" the terminals. The point is that it is in series with the load, and thus, with our 1k example again, we need to calculate our "output voltage" with respect to the voltage division that will occur between our 3Ω source impedance and the 1,000Ω load impedance.
Let's say that the unloaded 9V measured 9.5V on our volt meter. We are connecting 3Ω + 1,000Ω across the terminals. Let's see what the current will be:
Current = (9V) / (1,003Ω) = 0.00897A
Let's see the voltage drop across the 3Ω resistor.
Voltage = 0.00897A x 3Ω = 0.0269V.
For my fresh 9V with 3Ω source impedance, I am only experiencing a 0.0269V drop in power supply voltage when connecting a 1k load. We would read about 9.47V on the voltmeter.
Let's connect a heavier load. Let's try 500Ω. Again solving for current:
Current = 9V / 503Ω = 0.01789A
And now solving for the drop across the 3Ω:
Voltage = 0.01789 x 3Ω = 0.0537V
Still OK. The voltmeter reads about 9.44V.
A heavier load still. Now 50Ω. Current again:
9V / 53Ω = 0.17A
Voltage drop:
0.17A x 3Ω = 0.5V.
Now the "loading" is significant. Our 9.5V unloaded battery is now barely able to hold true to the "9V" title.
Heavier still. We go to 10Ω.
9V / 13Ω = 0.7A
0.7A x 3Ω = 2.1V
Now our 9.5V unloaded battery succumbs to the heavy load and becomes a 6.9V battery. The 9V is a poor voltage source for such a low load impedance. We would need a source with much less than 3Ω to drive a 10Ω load.
This brings about a rule of thumb when "driving" loads from voltage sources. The rule of thumb is: the source impedance should be about ten times less than the load impedance. For the best transfer of voltage from source to load, you may want to make this ratio even higher. A source impedance 100x less than the load will maintain nearly all of the source voltage. This concept comes up constantly in dealing with weak instrument signals, like guitar pickups and microphones.