Frequency Compensated Voltage Divider

[Unlikekurt]

Hello all,

I'm hoping this will be simple; but of course it may be complex.
With a fixed voltage divider in the signal path, say series R of 100K and shunt R of 10K, if you put a capacitor in parallel with the series resistor, how do you determine the corner frequency at which the capacitor begins to impact the divider?

I'm just a bit confused as to how the RC calculation for F would take place due to the two interacting R's.
If it were simply a R in parallel with a C, or a simple RC High or Low pass, I understand that.  But this is just a hair vexing to me.

Thanks for any feedback.

[amptramp]

If the 100K is fed from a low impedance, the 100K is in parallel with the 10K or 9.0909090... K and you can calculate from that.

If the capacitance across the 10K is ten times the capacitance across the 100K, you get a resistive divider of 1/11 for DC and also a capacitive divider of 1/11 at higher frequencies.  Thus, you tend to get flat frequency response.  The problem is trying to figure out what the capacitance is across the 10K because that usually goes to a stage with an indeterminate capacitance or a cable which may be of variable length and capacitance.  Oscilloscope inputs usually have a switched divider at the input and each resistor in the divider has a capacitor across it that keeps the response flat at all settings, so check scope schematics like this:

http://michaelgellis.tripod.com/scope/atten.html

Check the input attenuators on this scope schematic on the left hand side:



In this case, the designer is trying to get the RC time constant of the capacitor and input resistor to match the time constant of the grounded resistor and its shunt capacitance.

[Rob Strand]

There's two cases.   

The first is when you divider is driving a capacitive load (normally the capacitive load is fixed and beyond your control)  and you want to add a capacitor across the input resistor to make the response flat again.  This is what you do in oscilloscopes (amptramp's example), CRO probes, Voltmeters and to some degree guitar volume controls.

The second is when you add a capacitance across the input resistors to get some treble boost.   In the simple case the amount of boost is always equal to the attenuation of the divider.  So if you divider  by 10 (20dB) the treble boost will give you +20dB boost.   To limit the boost you need to add more resistors.

As for the calculations.   Suppose the capacitance is C, the input resistor is R1 and the output resistor is R2.

The point where the treble boost starts to rise is (at a low frequency),

  fL  = 1 / (2 * pi * C * R1)

The point where the treble boost starts to flatten off is  (at a high frequency),

  fH  = 1 / (2 * pi * R)   where R = R1 in parallel with R2.

You will find the ratio of fH to fL is exactly the reciprocal of the amount of division on the resistive divider.

[PRR]

> how the RC calculation for F would take place due to the two interacting R's.

What they said.

In short: there are _two_ bends in the curve. One is nearly R1*C, the other nearly R2*C.

And of course there are two ultimate gains: unity, and cut-down.

Plot the two gains, plot the two frequencies, draw a slant.

[PRR]

> Plot the two gains, plot the two frequencies, draw a slant.

simple math:
10k+5nFd is 3185Hz
100k+5nFd is 318Hz



sim:
-3dB @ 3480Hz
+3dB @ 322Hz

hmmmm.... about 9% error on the high corner. 

better math:
"10K" is really 100K||10K = 9.090k
9.090k+5nFd is 3503Hz

The 3503/3480 and 318/322 discrepanies are due to the two corners being fairly close together. (And also cursor-jerk in my simulator.) About 1% "error". This is "almost never" significant in musical audio design.

[Rob Strand]

The 3503/3480 and 318/322 discrepanies are due to the two corners being fairly close together. (And also cursor-jerk in my simulator.) About 1% "error". This is "almost never" significant in musical audio design.

If you have a step EQ of only say 3dB then you can see the 3dB points can't actually be correct - the 3dB point don't exist.   However, when you have a 20dB step the 3dB points are a good approximation - the approximation we use most of the time.

Take the case of a -3dB voltage divider.  Divider R1=4.142k, R2=10k and C=3.842n.   The expected frequency points are using accurate calculations are 10001Hz and 14144Hz.   

In spice you first find where the step *levels off* at each extreme, not the sloped part.  In this case you get 3dB (of course).   The key step is to find (say using spice cursors) the (single) frequency in the *centre* of the sloped part ie. at 3dB/2 = 1.5dB,  that should be 11893Hz.   In linear space 1.5dB is 10^(1.5/20) = 1.189. 

Then you use the center frequency to calculate fL and fH,

fL = fc / 1.189 = 11893/1.189 = 10000Hz
fH = fc *1.189 = 11893*1.189 = 14140Hz

So that method does let you use spice *and* get the correct frequencies.