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How do I figure out the Compression Ratio from Voltage Divider
How do I figure out the Compression Ratio from Voltage Divider
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Old 3rd July 2020, 08:27 PM   #1
garybdmd is offline garybdmd  United States
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Default How do I figure out the Compression Ratio from Voltage Divider

I am building an audio compressor and I am using a voltage divider. If I lower the voltage of the voltage controlled resistor as shown in the schematic, the voltage is cut in half and this is a 6db loss of voltage. How do you figure out what the compression ratio is? is this 2:1, 6:1 etc as shown?
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Old 4th July 2020, 04:46 AM   #2
PRR is offline PRR  United States
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How do I figure out the Compression Ratio from Voltage Divider
Does your "voltage controlled resistor" really cut Voltage linearly? Or Resistance?

Set up a spreadsheet.

Vin -- Rvar ---- Rvar/(R2+Rvar) --- Vout
0V --- 20k ---------- 0.66 ----------- 0V
1V --- 10k ---------- 0.50 ----------- 0.5V
2V ---- 5k ---------- 0.33 ----------- 0.66V
3V -- 3.3k ---------- 0.25 ----------- 0.75V
3:1 input rise makes 1.5:1 output rise. We normally work in dB, so 9.5dB input makes 3.5dB output rise, 2.7:1 CR, *over this small range of inputs*.

I do not know of any physical single-part "voltage controlled resistor" with a simple predictable control law.

Also an audio compressor normally has a "threshold" below which it does not act.

And the gain law is a product of several factors including sidechain gain.

Here is a worksheet for a functional real-parts limiter (tube and opamp). The tube gain law is very curved, and varies tube-to-tube. But the limiter works feedback which makes it much less sensitive to the tube used. But not a lot of feedback so it does not squash the dynamics flat.

I see I did not compute the compression ratio. Take the range from 100mV to 1V, a 20dB rise of input. The output rises from 215mV to 550mV, 8dB. So CR is 20:8 or 2.4.
Code:
__ in _ dBin __ out _ dBout __ Gv ___ GR __ dBgr ___ CV ___ THD __ mA
__ 10 _ -30 ____ 27 _ -23 ____ 2.7 __ 1.0 ___ 0 ____ 0.3 __ 0.02 _ 19
_ 100 _ -10 ___ 215 __ -5 ____ 2.2 __ 0.8 ___ 2 ____ 2.2 __ 0.08 _ 10
_ 300 ___ 0 ___ 392 ___ 0 ____ 1.3 __ 0.5 ___ 6 ____ 4.0 __ 0.08 __ 4
__ 1V __ 10 ___ 550 ___ 3 ____ 0.5 __ 0.2 __ 14 ____ 5.5 __ 0.4 ___ 1.3
__ 2V __ 16 ___ 620 ___ 4 ____ 0.3 __ 0.11 _ 19 ____ 6.4 __ 2 ____ 0.6
__ 3V __ 20 ___ 650 ___ 4.4 __ 0.2 __ 0.08 _ 22 ____ 7 ____ 5 ____ 0.55
__ 6V __ 26 ___ 672 ___ 4.7 __ 0.11 _ 0.04 _ 27 ____ 8 ___ 20 ____ 0.4
_ 10V __ 30 ___ 690 ___ 4.9 __ 0.07 _ 0.025_ 32 ___ 10 ___ 37 ___ 0.014

Last edited by PRR; 4th July 2020 at 04:51 AM.
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Old 4th July 2020, 03:04 PM   #3
garybdmd is offline garybdmd  United States
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Thank you for the very helpful answer. I think I see how this works now. I did create a spreadsheet and that helped.

Basically I have made several different compressors and I've built them with a similar format. The R2 is variable resistor to control the ratio, then I have another resistor forming a voltage divider, and there is another "voltage controlled resistance type device" which I am showing in the attached figure. The idea is to use a device who's resistance is large compared to the resistor when the control voltage is low and small compared to the resistor when the control voltage is high.

I wanted to label the ratio potentiometer in some way, so it looks like I'll be labelling it as a maximum compression when a signal starts well below threshold to well above.

The last compressor I built uses 4 diodes and the one before that used an led/photoresistor combination. (I also made one using some transistors, but its more of an all or nothing reduction which believe it or not when used properly didn't sound too bad). The 10k resistors shown in the schematic are arbitrary for the sake of discussion. For the diode compressor I'm actually using 100k voltage divider as shown, I think I used 20k for the optocompressor.

This really helped me, thanks again.
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Last edited by garybdmd; 4th July 2020 at 03:12 PM. Reason: clarification
 

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