EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: iampoor on December 10, 2013, 07:11:50 pm
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Hi all
Been playing around with eq circuits recently....just started building some baxandal type, and I have a quick question. Since Im running off of a single supply, I know I need to bias the opamp through R7. However, what about the second stage? Do I bias the noninverting input since it is at "ground" potential? I understand that my tone controls are in the feedback loop, Im just looking for the theory behind this! I am presuming that if I injected the bias into the inverting input of the second stage, then it would create an awfully skewed tone control response, and potentially create positive feedback. Is it even necessary since the first stage is biased and I dont have any DC blocking caps between both stages?
Anyone have any insight on this?
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Virtual ground is what you are looking for. Basically a series connected pair of resistors ( with or without another opamp arranged as a unity gain buffer) between the supply rails with all points that are ground ( and Vb as well) on your circuit connected to the junction of the 2 resistors. That means that to the opamps the supply is now a split rail, the total of which is the battery, but each rail is half the battery voltage.
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AARGGHHH!!!!
http://en.wikipedia.org/wiki/Peter_Baxandall (http://en.wikipedia.org/wiki/Peter_Baxandall)
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Virtual ground is what you are looking for. Basically a series connected pair of resistors ( with or without another opamp arranged as a unity gain buffer) between the supply rails with all points that are ground ( and Vb as well) on your circuit connected to the junction of the 2 resistors. That means that to the opamps the supply is now a split rail, the total of which is the battery, but each rail is half the battery voltage.
I didnt include the power supply on the schematic, but it is a voltage divider that provides +9v 0v and 4.5v (for the VB Voltage Bias pin)
Isnt that the same setup that you are suggesting? The way I understood it, biasing the opamp at 4.5v is effectively the same as a dual supply, since it has 4.5v swing each way. Or for this circuit, will this simple arrangement not work?
AARGGHHH!!!!
http://en.wikipedia.org/wiki/Peter_Baxandall (http://en.wikipedia.org/wiki/Peter_Baxandall)
Fixed, and duly noted. :-)
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You might want to take a look at this little microphone preamp + active base/treble control circuit that I built for equalizing an old crystal mic to a modern ham radio. This uses the same basic circuit as you've shown. The video discusses the circuit and shows measurements made on the circuit:
D-104 Microphone amplifier / Equalizer for Ham Radio (https://www.youtube.com/watch?v=aLbB4XUS0lM#)
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@iampoor
I presume the bias you're talking about is marked VB, being fed to IC1A's inverting input via R7 & R19?
If so, remove that VB/R7 combo, instead attaching the non inverting inputs of IC1A & IC1B to your virtual ground. The inverting configurations here mean that IC1A will drive current through R18 to bring it to 4.5V to match the + input, and IC1B will do the same via R24, pot U$5, and R26.
As the active portion here is decoupled from the rest of the circuit by C8 and C14, it will "float" on the virtual ground, ensuring correct opamp behaviour.
For good performance, google virtual ground circuits, and take the advice you can find - you probably don't need a buffer if your divider resistance isn't too high, since you're only feeding the high input impedance on the two + inputs, but a capacitor over the bottom leg would likely be a good idea.
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You might want to take a look at this little microphone preamp + active base/treble control circuit that I built for equalizing an old crystal mic to a modern ham radio. This uses the same basic circuit as you've shown. The video discusses the circuit and shows measurements made on the circuit:
D-104 Microphone amplifier / Equalizer for Ham Radio (https://www.youtube.com/watch?v=aLbB4XUS0lM#)
Hi Alan
Thanks for the video link! Very cool and simple to understand. Im looking at the schematic right now. (Also stumbled across your other videos and website, Im a big fan of MGBs as well and restored a 1979 one similar to yours!!)
@iampoor
I presume the bias you're talking about is marked VB, being fed to IC1A's inverting input via R7 & R19?
If so, remove that VB/R7 combo, instead attaching the non inverting inputs of IC1A & IC1B to your virtual ground. The inverting configurations here mean that IC1A will drive current through R18 to bring it to 4.5V to match the + input, and IC1B will do the same via R24, pot U$5, and R26.
As the active portion here is decoupled from the rest of the circuit by C8 and C14, it will "float" on the virtual ground, ensuring correct opamp behaviour.
For good performance, google virtual ground circuits, and take the advice you can find - you probably don't need a buffer if your divider resistance isn't too high, since you're only feeding the high input impedance on the two + inputs, but a capacitor over the bottom leg would likely be a good idea.
Yes, that is correct.
Okay thank you very much! That was very clear and informative and has helped clear up some of my initial hunches. I will redraw the schematic (and include the rest of it, its a distortion pedal if anyone is curious) and will then report back!
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Okay here it is. C8/C14 are DC blocking capacitors that will allow me to bias the second stage independantlly through R2 correct?
Its an MXR Distortion+ circuit with a few mods. Still havent figured out what im doing for the input section, so please ignore that part of the schematic unless you want to make some pointers. :p
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Correct - just remember that the R1/R2 combination will act as a voltage divider to the AC input, so adjust values and overall gain as appropriate.