Ross compressor question

[jmasciswannabe]

Just finished mine. I dog it, but it loses low end. Is this the norm? If not, I know to look at the output cap, but is  there anywhere else in the circuit that I might have put a wrong value that would effect the bass amount. I used tants in place of some of the electros and a nonpolar 1uf right before the trim. I went with the fixed trim 1k and 1k and added the 150k clip control. Any suggestions would be appreciated. Thanks!

[amptramp]

If this is the schematic shown at tonepad, you only have three coupling capacitors, the 0.01 µF at the input, the 1 µF going to the operational transconductance amp and 0.05 µF at the output.  The input capacitor goes into a circuit which is two series 470K resistors in parallel with (10K x hfe of the input transistor).  The transconductance amp has an input resistance of 26K typical / 10K minimum.  The output goes to 60K in parallel with the load.

Let's assume the hfe of the transistor is 100.  This gives you 1 megohm in parallel with 940K or 484536 ohms.  Using the calculator at http://www.muzique.com/schem/filter.htm

You get a 32.9 Hz turnover.

The worst case for the middle capacitor is 1 µF and 10,000 ohms or 15.9 Hz.

The output is 53.1 Hz, which goes up due to paralleled resistance being lower than 60K if there is much of a load.  So increase the 0.05 µF first to 0.1 or 0.15 or 0.22, all standard sizes, using the biggest film capacitor that fits.  Then increase the 0.01 µF input to 0.47 µF, also a standard value.  Why the much bigger increase at the input?  Because the input resistance determines noise and if you can reduce the turnover at the input, the input resistance (which is 940K here) can be paralleled by the output resistance of the previous stage, giving lower noise above the turnover point.

[JDoyle]

The input capacitor goes into a circuit which is two series 470K resistors in parallel with (10K x hfe of the input transistor).  The transconductance amp has an input resistance of 26K typical / 10K minimum.  The output goes to 60K in parallel with the load.

Not to get too nitpicky here but none of the above is exactly correct.

The input resistance should be calculated as (10k*hFE)||470k because the 1uF cap to ground at the junction of the two 470k resistors shunts all AC signals above 0.3Hz to ground, so the 940k quoted in the formula is only valid for freq. less than 0.3Hz.

The input resistance of the 3080 is 26k/10k min as stated, however the filter string on the non-inverting input needs to be taken into consideration. (Also, the 10k value is extreme, it would mean that the transistors inside the 3080 had a gain of less than 50, which isn't likely these days - use 26k, which denotes a gain of 100 for the internal Qs and would most likely be themodern day 'worst case').

The output turnover freqency is determined by [10k + (portion of level pot in series with the output (i.e. if it isn't at max output))] + [(portion of level pot not in series with output (i.e. shunting to ground))||Zload]

Regards,

Jay Doyle

[amptramp]

Correct, Jay, I read the Dyna parts list instead of the Ross that leaves out the 1 µF capacitor between the 470 K resistors in the front end.  This gives an input resistance of 319.7K to which the 10K series resistance should be added.  This gives you 48.3 Hz in place of 32.9 Hz in my first answer.  But even this number varies with hfe of the transistor and the value of 100 is just nominal.

OK, next stage: the operational transconductance amplifier acts like a differential pair with the current set by the signal fed back via pin 5.  The 1 Meg resistors and the 220K resistors can be ignored here as they carry less current than the variations of input impedance in the 3080.  At low frequencies, the 15K and the 1 µF cap turn over at 10.6 Hz.  Below this frequency, the signal input changes to common mode, cancelling the output below this frequency.  Note that all the turnover points are points where the output is 3db down and the effect still occurs to a lesser extent at higher frequencies.  The input impedance should be 26K for the inverting input in parallel with the series 2K and another 26 K (26K||28K) at frequencies below this.  Above 10.6 Hz, the 15K resistance is added in parallel with the 26K of the non-inverting input.  At higher frequencies, the 0.01 µF cap across the 15 K reduces the impedance even further, but this is not a turnover effect because it happens at too high a frequency.  The input resistance would be 13.48K at an 11.8 Hz turnover as seen at the inverting input, but the non-inverting input follows the inverting input so that the gain is essentially the difference in input voltage between inverting and non-inverting inputs, so the 1 µF cap to ground from the 15K is effectively in series with the input cap.  The two 1 µF caps are in series in this case, so the low frequency would be 13.48K and 0.5 µF or 23.6 Hz.  I may have to take another look at this to see if my approximations work, but it appears that the middle stage is not the most critical unless you really do get down to 10K inputs.  Note that the input impedance will vary with the feedback signal on pin 5.

You have given a more precise description of the output frequency response, but my calculation stands as the value for infinite output impedance.  It also shows that the output is the most critical turnover because it is the highest one even with no load.

[jmasciswannabe]

Thank you gentlemen for the information. I dropped a 200k pot in before the opamp a la keeley clip control and in the process, the bass returned. Not sure how that happened, but it did!