Active splitter in bass guitar

AnalogKid

Joined Aug 1, 2013
10,986
The thing about a source follower is that it has an asymmetrical output impedance. The FET pulls the output up with a relatively low impedance, but the only thing pulling the output down is the 10K source resistor. Depending on the values for the output level controls and the loads on their wipers, driving 2 or 3 outputs with one FET could lead to an output waveform that changes shape when the pots are adjusted. For this reason, if you are going to use FETs, stay with one per output.

personally, I would put more effort into finding an appropriate opamp, one with acceptable bandwidth and distortion, and a static current under 500 uA.

ak
 

Thread Starter

Amalgam

Joined Sep 4, 2019
27
The thing about a source follower is that it has an asymmetrical output impedance. The FET pulls the output up with a relatively low impedance, but the only thing pulling the output down is the 10K source resistor. Depending on the values for the output level controls and the loads on their wipers, driving 2 or 3 outputs with one FET could lead to an output waveform that changes shape when the pots are adjusted. For this reason, if you are going to use FETs, stay with one per output.

personally, I would put more effort into finding an appropriate opamp, one with acceptable bandwidth and distortion, and a static current under 500 uA.

ak
I'm a total newb when it comes to electronics. I haven't put anything together before, or even soldered, but I need these for my bass build to work. I also see this as a great learning experience. I've started to understand how circuits work, but creating a unity gain per output (1, and 2) channel circuit is beyond me. I can find an opamp, but I don't know what the circuit would look like. Mainly my goal is to put these together and then test a few variations to understand it more fully, and look up parts of circuits to understand what each do. Of course they have to work great too, or the rest of the bass build will suffer.
 

Alec_t

Joined Sep 17, 2013
14,280
What do you think of Bordodynov's post?
He makes a good point. Biasing the gate at about half the supply voltage would reduce distortion and allows a wider input signal range without clipping. It also accomodates JFETs with a range of cut-off voltages better.
 

Thread Starter

Amalgam

Joined Sep 4, 2019
27
I feel kind of stuck and lost. I'm not sure which design I should use, or whether I should use JFET's or op amps. I don't want the output waveform to change shape when I adjust the potentiometer's. That's for sure. So, it sounds like if I go the JFET route I'd need to use one per channel (1 pickup -> 2 channels). The OPA145 looks like it could work with Bordodynov's design with two JFETs, and only use up to 950uA per splitter (2x475uA). I think I could live with that, but if I could lower the amount of current by quite a margin (not sure how much these JFETs need?) and still have them work, it might make sense to go with the OPA1641 at 1.8 mA for each channel. It looks like the LSK189 is out of the running, but I either can't interpret the datasheet, or much the information isn't listed in a way I can understand it.

So, JFETs or op amps? Probably two JFET's per splitter if used, and can the OPA1641 work by using a voltage divider to supply it with less current? That'd be 3.6mA per splitter vs 950uA with the OPA145.

If I was going for op amps, it looks like the OPA1692 uses 650uA.

One more thing. If something has a noninverted input and an inverted input, does that mean that someone would just choose which would work best for them, or am I missing something?
 

Alec_t

Joined Sep 17, 2013
14,280
can the OPA1641 work by using a voltage divider to supply it with less current?
No. That's the standby current of the opamp, which you can't alter. You're stuck with a minimum of 1.8mA per channel.
As for the FET current, that will be (Vg-Vcut)/Rs, where Vg is the gate bias voltage, Vcut is the cut-off voltage of the FET (check the datasheet) and Rs is the source resistor value. E.g for Vcut= -4V and Vg =+4.5V the current would be 850uA with a 10k source resistor.
If something has a noninverted input and an inverted input, does that mean that someone would just choose which would work best for them
Most opamps have both. If you wired an opamp as a unity gain buffer the output would connect to the inverting input and your signal would drive the non-inverting input.
Regarding distortion, don't bass guitarists usually deliberately add that with a fuzz box? :).
 

Thread Starter

Amalgam

Joined Sep 4, 2019
27
Thank you both! I'm looking into this some more, but this has cleared things up a lot, and it's nice to figure out the values myself in the future too. One thing I should point out though is that I'll be using an 18v system. Not sure how that changes things. I'll look into voltage dividing tomorrow.
 

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Amalgam

Joined Sep 4, 2019
27
*QUESTION BARRAGE IMMINENT*

My resistors and capacitors are rated at 200v or more, so that shouldn't be an issue. My switches are rated much higher than that, I believe, and the EQ's are rated to accept 18v.

So I looked at voltage dividing and it looks like I could set R1 to 3Meg, and keep R2 at 1Meg to get the voltage down to 4.5v on an 18v system.

What does R3 do? It keeps current from travelling backwards? If that's true is that all it does? What's the relationship between it and R1, and R2, or do they happen to have the same value (1Meg)?

Looking at Bordodynov's diagram, which I'll be referencing from here on out, is there a way to only use two of the four channels that the AD8244 offers without having current go to the extra channels? If not, I'd probably rather use a couple AD145's instead as they would use similar power levels of the four channel AD8244.

How would I calculate how much heat these resistors would generate? If I'm using a JFET or other component in this project, or another, does it make more sense to leave the voltage as it is if it's rated for it, or is the whole point of lowering the voltage so that the component more accurate?

And just to double check and satisfy my curiosity, if I were to look into op amps instead of using JFET's, would the diagram look the same/similar?
 
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Alec_t

Joined Sep 17, 2013
14,280
So I looked at voltage dividing and it looks like I could set R1 to 3Meg, and keep R2 at 1Meg to get the voltage down to 4.5v on an 18v system.
Assuming we're talking about the post #25 circuit, it will work quite happily over a range of supply voltages. If R1 = R2 then the bias voltage for the input is half the supply voltage and is held at that by C1 and C6. You don't need to get the bias voltage down to 4.5V.
As far as AC input signals are concerned, the R1/R2 junction is at a fixed voltage. R3 prevents the input signal via C2 being clamped to that fixed bias voltage.
In general if you have, say, four opamps in one IC then the IC supply current will be much the same whether all four opamps are used or not.
The heat dissipated in a resistor is given by V^2/R, or I^2*R, where V is the voltage across R and I is the current through R.
 
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Bordodynov

Joined May 20, 2015
3,177
I am sorry.
I deduced the wrong current of the lower circuit.
The first scheme allows you to skip the run restrictions of a larger signal compared to the second. The second has the best dynamic characteristics.2019-10-11_13-07-36.png
 

Thread Starter

Amalgam

Joined Sep 4, 2019
27
Looks like I'll be going with the OP282. I looked up a bunch more around this spec, but everything is way worse. I'm going to be using 2.7 Meg resistors for R1, and R3, or R4, and R8 according to the post above this one because I can't find 3 Meg MELF Vishay resistors in stock at Digikey or Mouser.

This should work well with 250k/500k potentiometers? I'm not sure which one I'll use yet, but both channels of each of these will go to a stacked pot. If this is good, I'll go ahead and order all the parts.

Oh, and earlier when I said AD145, I meant OPA145, but that still uses more current than I'd like. So I'll go with the OP282. It'd be nice if the OPAx145 dual channel JFET could be sused, but I can't find it stocked, or listed anywhere, though a datasheet exists.
Assuming we're talking about the post #25 circuit, it will work quite happily over a range of supply voltages. If R1 = R2 then the bias voltage for the input is half the supply voltage and is held at that by C1 and C6. You don't need to get the bias voltage down to 4.5V.
As far as AC input signals are concerned, the R1/R2 junction is at a fixed voltage. R3 prevents the input signal via C2 being clamped to that fixed bias voltage.
In general if you have, say, four opamps in one IC then the IC supply current will be much the same whether all four opamps are used or not.
The heat dissipated in a resistor is given by V^2/R, or I^2*R, where V is the voltage across R and I is the current through R.
Thanks. This gives me a much better of picture of how it works.
 
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Bordodynov

Joined May 20, 2015
3,177
It makes sense to use high-resistance potentiometers if the input impedance of the following circuit is large enough. At least this input resistance must be at least as high as the resistance of the potentiometer.
 

Thread Starter

Amalgam

Joined Sep 4, 2019
27
It makes sense to use high-resistance potentiometers if the input impedance of the following circuit is large enough. At least this input resistance must be at least as high as the resistance of the potentiometer.
I guess what I mean is that I'm wondering how the 100k resistor fits in. That would be on the ground of the potentiometer, or would the potentiometer take it's place as it's basically a variable resistor right, and it has ground?

Edit: Or is that part of the circuit making sure that the current goes through the potentiometer?
 

Thread Starter

Amalgam

Joined Sep 4, 2019
27
I'm not exactly sure how to understand that. U1, U2, and U3 are the potentiometers? It looks like OUT0 would have the best characteristics? I think I can always add a 500k resistor to the first and third terminals to make it a 250k pot. I'm not sure how to find out loads. I've heard of it, but don't know what it is.

I'm going to be using an Aguilar OBP-3:
OBP-3 FEATURES

Bass control: plus/minus 18 dB @ 40 Hz

Mid control: plus/minus 16 dB @ 400Hz or 800 Hz. User selectable frequency center

Treble control: plus/minus 16 dB @ 6.5 KHz

Input impedance: 1 meg ohm

Output impedance: 100 ohm 9 or 18 volts operation

Distortion: .019 into 10k ohm load

Noise: -95 dBm unweighted

Battery life: 324 hours

Wire lead length: 6.50 inches

Low noise discrete FET input stage, center detent 50 K linear pots,with a dual concentric pot available
And a East BTB-02 which uses a class A FET:
The BTB-01 consists of a single stack knob with Bass (boost only), Treble (boost & cut) and Pull for Bright. Its low noise high quality circuit is based on the middle knob of the J-RETRO 01. Its input stage uses a Class A FET for low noise and high quality sound. A gentle pre-shape contour, with slight low end and high end boost, gives an immediate lift to the sound.
 
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Bordodynov

Joined May 20, 2015
3,177
As long as we communicate in different languages. As deaf and blind. I have tried to tell you how the adjustment characteristic of the potentiometer resistance and the input resistance of the subsequent unit (amplifier, tone control, filter, etc.) affects. Putting a resistor parallel to the potentiometer is a bad idea. You will not improve the adjustment characteristic.
 

Thread Starter

Amalgam

Joined Sep 4, 2019
27
I appreciate you taking the time to help me with the project I'm working on. I know it's hard as I do not have any electrical background and very little knowledge. What you can show in pictures or diagrams is somewhat lost on me because of this. However, I understand what you mean now, and I think I have it.

I wasn't planning on adding resistors by the way. I was just confused about R9, and R10 from the previous schematic. I take them to be a representation of the volume potentiometers now. Please correct me if I am wrong.

I am still a bit confused as to what the 100k resistor at the end does. Is it to guide the current into the potentiometer? Again, I appreciate what you're doing for me. I just don't know much about circuitry.
 
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