Opamp recovery stage for BMP tonestack. How much gain?

[octafish]

So I think the subject says it all. I'm drawing up an opamp diode clipping pedal and have added a BMP stack, how much gain does the original discrete BMP recovery stage have? Cheers.

[amz-fx]

A gain of 4x should be plenty.  The transistor recovery stage in the BMP is very slightly more than this (~4.4)

If you are using a non-inverting opamp gain stage, a 10k on the inverting input and a 33k for the feedback will give you a gain of 4.3x

regards, Jack

[brett]

Hi.
I'm not diasgreeing with Jack.  He knows a lot more than me.  But here's an alternative view.

you might note that the "mids scoop" in the BMP tone section is about 15dB deep at 1000Hz.  At lower and higher frequencies it is about 10dB deep.  So if you wish to recover all of the losses, you might aim for somewhere between 10 (for 10dB) and 33 (for 15dB).  I haven't thought about this before, and have always use a factor of 10.

Also, you'll need more than 10k on the inverting input of an op-amp is that's your gain device.  Try 100k for the input resistor and 1M in the feedback loop.  The BMP tone section needs to see 100k or more of input impedance in the next section to behave normally.  (see Duncan's tone stack calculator for details).

Oh yeah, one more thing.  A volume pot will hide most of the gain differences (ie just turn it up a bit if the gain is lower).

cheers

[stm]

...you might note that the "mids scoop" in the BMP tone section is about 15dB deep at 1000Hz.  At lower and higher frequencies it is about 10dB deep.  So if you wish to recover all of the losses, you might aim for somewhere between 10 (for 10dB) and 33 (for 15dB).  I haven't thought about this before, and have always use a factor of 10.
...

May I say that 10 dB's are like 3x gain, and 15dB's are near 5x gain, thus, the 4x gain on average compensates for the losses of a BMP tonestack.

When doing dB to VOLTAGE GAIN calculations you use the following formula:  Voltage gain = 10 ^ (dB / 20).  (Using a factor of 10 instead of 20 is for POWER GAIN calculations).  As such, Jack's suggestion is correct, however you might still want to have more gain if you want your pedal to have higher boosting capabilities.

...Also, you'll need more than 10k on the inverting input of an op-amp is that's your gain device.  Try 100k for the input resistor and 1M in the feedback loop.  The BMP tone section needs to see 100k or more of input impedance in the next section to behave normally.  (see Duncan's tone stack calculator for details).
...

The OpAmp is configured with POSITIVE gain. As such, signal input is on the (+) input which is already high impedance, ant the feedback resistors go around (-) input, output and Vref.

A final note: The components on the BMP tonestack that are tied to GND should be tied to Vref (Vcc/2) instead so as to properly bias the (+) OpAmp input.

Regards.

[brett]

Hi.
Thanks stm.  New Year's resolution... remember difference between dB and log tenths of volts.
I'd like to say that the E engineers don't help by using sound pressure units (dB) for electrical power (and the reference point (eg dB=0) appear arbitrary ??). 
Why not use the SI terminology for working in log base 10, p (as in pH, -log10[H ion concentration]).  Charts, etc of pW or pV would then be self-explanatory.  Maybe???  Am I missing something???
thanks again stm