While assembling the final parts to my volume pedal project, I realized that I somehow made an error in my calculations somewhere and my volume pedal doesn't rotate the pot to its entire rotation.
I don't want to change the hardware, so what options do I have for the buffer volume circuit? I can have zero resistance at tie down position, or maximum resistance at heel down position.
Any ideas? Also, I'm not sure what value pot to use in the final version.
Without knowing what you're using for a buffer (hopefully it's an op amp), if you have buffers on both sides (before and after the pot), the little bit of extra resistance probably won't hurt anything -- you could even take advantage of it and put a cap across lugs 2 and 3 on a small switch for when you need a small treble boost. Similarly, as long as you've buffered both sides of the pot with a good buffer, it shouldn't matter too much what value it is.
I don't know what kind of mechanism you're using to turn the pot, but it might be that not all hope is lost in that regard. Can you show a picture or diagram of it?
Quote from: midwayfair on December 01, 2014, 01:37:19 PM
Without knowing what you're using for a buffer (hopefully it's an op amp), if you have buffers on both sides (before and after the pot), the little bit of extra resistance probably won't hurt anything -- you could even take advantage of it and put a cap across lugs 2 and 3 on a small switch for when you need a small treble boost. Similarly, as long as you've buffered both sides of the pot with a good buffer, it shouldn't matter too much what value it is.
I don't know what kind of mechanism you're using to turn the pot, but it might be that not all hope is lost in that regard. Can you show a picture or diagram of it?
What's the difference between opamp and jfet buffers in context of a volume pedal? I'm guessing I could just build two of the same opamp buffers connected between lug 2 & 3 with 1 going to ground?
This is the mechanism I'm using. It's a string and spring system like Ernie Ball pedals
(http://tapatalk.imageshack.com/v2/14/12/01/a96f7f13ef7cb9d50b881eb295ba7e9e.jpg)
(http://tapatalk.imageshack.com/v2/14/12/01/23205698754e8c7193011ec87bdb46af.jpg)
Quote from: brand0nized on December 01, 2014, 07:12:27 PMWhat's the difference between opamp and jfet buffers in context of a volume pedal? I'm guessing I could just build two of the same opamp buffers connected between lug 2 & 3 with 1 going to ground?
For one thing, a JFET buffer will have higher output impedance and it's device sensitive. A BJT will have lower output impedance at the expense of input impedance. Quibbling over a few hundred ohms might seem petty, but an op amp doesn't really take up much room at all in a dual buffer setup and might actually be less than two JFETs (I think you won't even need a cap between stages if you tie lug 1 to Vb instead of ground). Besides, you can also very easily change it to have a slight bump in gain, to correct for the minute loss of volume that comes from buffering, or just as a boost.
You're also still fighting the input voltage issue you'd have in any other JFET pedal (I think the 2n3819 is the highest Vp I've encountered, and it's still
well under 4V, so it loses out to an op amp running on 9V). While a JFET will consistently sound better when distorted, your buffer is less likely to be distorted, so it's better to just use the op amp.
Also, at unity gain, my usual complaint about op amps (a little more noise) shouldn't apply.
Would this work? I'm doubling the same opamp buffer circuit onto a dual opamp with a 100ka in between. I've never drawn a circuit before so I don't know if this is correct.
(http://tapatalk.imageshack.com/v2/14/12/01/30a22bcb044b373ac0750f8fb4a23a2f.jpg)
The buffer is from here
http://tagboardeffects.blogspot.com/2014/08/buffers.html?m=1
why do you need a pot between the buffers?
it would only serve if you want to attenuate the signal
and why use two buffers in series?
if you wanna use two buffers, one for input and one for output, you would connect wah input where the wiper is connected and wah output on the input of second buffer
but, if you want this arrangement, you don't need 1M res to ground before second 100n cap, in fact i would say you don't even need the second 100n cap because 10u cap before pot does the ac coupling
Quote from: BaklavaMetal on December 01, 2014, 07:52:13 PM
why do you need a pot between the buffers?
it would only serve if you want to attenuate the signal
and why use two buffers in series?
if you wanna use two buffers, one for input and one for output, you would connect wah input where the wiper is connected and wah output on the input of second buffer
but, if you want this arrangement, you don't need 1M res to ground before second 100n cap, in fact i would say you don't even need the second 100n cap because 10u cap before pot does the ac coupling
I'm making a volume pedal not a wah, hence the volume attenuation
Quote from: midwayfair on December 01, 2014, 07:38:33 PM
(I think you won't even need a cap between stages if you tie lug 1 to Vb instead of ground). Besides, you can also very easily change it to have a slight bump in gain, to correct for the minute loss of volume that comes from buffering, or just as a boost.
A few questions
1. Where is Vb?
2. Which cap would be omitted?
3. Which resistor would I change to change the output volume?
Brandon, Vb would normally be created with a separate voltage divider. Right now, you're creating it with the two 1M resistors after your input cap. Open up, say, the Greenbean schematic to see another way to do it. I think they're 47K in that pedal, but 10K would be better. It creates a more stable bias voltage than the way yours is drawn. It also gives you a "faux ground" -- the op amp's middle point (4.5V) -- which is handy for reducing parts as we'll see in a moment.
If your volume pedal doesn't have a switch, you don't need the 1M before your input cap. That's a pulldown to drain off any voltage that leaks out through the cap.
After you create the Vb point, you can omit the 10uF and 100nF caps, as well as all the 1M resistors, between pins 1 and 5, and tie the pot's pin 1 to Vb instead of ground. It will still be a normal volume control that way, but since you aren't connecting it to ground, you don't need to decouple the cap to keep the op amp working properly.
Quotein fact i would say you don't even need the second 100n cap because 10u cap before pot does the ac coupling
If the pot is going to ground, he would need to isolate pin 5 from the ground connection, so the other cap would still be necessary as drawn.
Quotewhy use two buffers in series?
So that the volume pot doesn't load the guitar or the following circuit. EDIT: even if he were building a wah pedal (not sure how you got that impression), it would still be a good idea to buffer the volume pot from the output of the wah to avoid affecting the cutoff frequencies. And once you're using an op amp, you might as well make use of the other half of the op amp and buffer the output.
2. My mistake, i thought ypu were building a wah, prolly had few too many beers
Vb is between 1M resistor which run from Vcc to +input of opamp then another 1Mres from +input to ground
I believe you could ommit the 100nf cap before 2nd opamp input, but i may be wrong.
3. If it's a vol pedal, then the pot is used for volume change, no res change necessary
Quote from: midwayfair on December 01, 2014, 08:37:52 PM
Brandon, Vb would normally be created with a separate voltage divider. Right now, you're creating it with the two 1M resistors after your input cap. Open up, say, the Greenbean schematic to see another way to do it. I think they're 47K in that pedal, but 10K would be better. It creates a more stable bias voltage than the way yours is drawn. It also gives you a "faux ground" -- the op amp's middle point (4.5V) -- which is handy for reducing parts as we'll see in a moment.
I'm looking at the Greenbean schematic, and I'm not sure what is going on to create Vb.
(http://tapatalk.imageshack.com/v2/14/12/01/d80339094e43fe66e7161add4db8f5e3.jpg)
Quote from: midwayfair on December 01, 2014, 08:37:52 PM
After you create the Vb point, you can omit the 10uF and 100nF caps, as well as all the 1M resistors, between pins 1 and 5, and tie the pot's pin 1 to Vb instead of ground. It will still be a normal volume control that way, but since you aren't connecting it to ground, you don't need to decouple the cap to keep the op amp working properly.
On my original schematic, I would be omitting everything between pin 1 and 5 but the potentiometer (the circled parts)?
(http://tapatalk.imageshack.com/v2/14/12/01/3680874611c54b387960b404504d6749.jpg)
Vb is created with a voltage divider at the junction of R18&19 (top, near #4)
I'm no EE, but isn't it a bit of overkill to buffer the output of a volume pedal? Why not just use a single opamp and minimum of parts?
As for the lack of full rotation, that's not a bad thing, saves busting the pot. Maybe try using a larger value pot with a series resistor to take up the difference.
One other thought, Proel volume pedals have a pot on the side that sets the heel down level, so you can treddle btween full volume and a pre-set minimum (fully off or just a lower volume)- quite a handy feature.
Quote from: Guitarmageddon on December 02, 2014, 01:38:56 AM
Vb is created with a voltage divider at the junction of R18&19 (top, near #4)
I'm no EE, but isn't it a bit of overkill to buffer the output of a volume pedal? Why not just use a single opamp and minimum of parts?
As for the lack of full rotation, that's not a bad thing, saves busting the pot. Maybe try using a larger value pot with a series resistor to take up the difference.
One other thought, Proel volume pedals have a pot on the side that sets the heel down level, so you can treddle btween full volume and a pre-set minimum (fully off or just a lower volume)- quite a handy feature.
I have no idea why I would use two, actually, but it seems like it won't hurt to have it.
I plan to use this between drives and delays so I can do swells, so I need heel down position to be zero volume.
Good point about messing up the pot, didn't think about that. But in my design, I have stoppers to stop the foot rocker right where the pot reaches maximum rotation.
Quote from: Guitarmageddon on December 02, 2014, 01:38:56 AM
I'm no EE, but isn't it a bit of overkill to buffer the output of a volume pedal? Why not just use a single opamp and minimum of parts?
I'm no EE either, but turning down the volume pedal slightly does add series resistance, which could load down a following circuit. A single op amp is just as big as a dual op amp, and all you're doing it connecting two more pins and using Vb for the volume control's ground. The parts count is identical.
Brandon, you're correct on the circled parts. Needless to say you should breadboard it first to make sure you're happy with it.
Quote from: midwayfair on December 02, 2014, 04:43:36 AM
Brandon, you're correct on the circled parts. Needless to say you should breadboard it first to make sure you're happy with it.
Yup, plan to breadboard, but I need to order parts because I'm not deep enough into guitar electronics to have opamps laying around.
Is Vb created with just the two 10k resistors or with D5 and R17 also?
with just R18 and R19
Quote from: brand0nized on December 02, 2014, 06:59:28 AM
Is Vb created with just the two 10k resistors or with D5 and R17 also?
The junctions in the power section are labeled, as are voltages throughout the circuit. Look at the circuit again. Do the parts you mentioned connect to the label Vb?
Quote from: midwayfair on December 02, 2014, 01:32:55 PM
The junctions in the power section are labeled, as are voltages throughout the circuit. Look at the circuit again. Do the parts you mentioned connect to the label Vb?
Ah, I see. Just wondering because those two parts stand between the first 10k and 9v
Quote from: brand0nized on December 02, 2014, 04:05:42 PM
Ah, I see. Just wondering because those two parts stand between the first 10k and 9v
See if you can work out what they're doing. The 5817 is the easiest one to figure out. The 10R is a little less obvious.
Quote from: midwayfair on December 02, 2014, 04:21:52 PM
See if you can work out what they're doing. The 5817 is the easiest one to figure out. The 10R is a little less obvious.
From what I just read on tagboardeffects, the diode is for polarity protection and also drops the voltage a little.
And the 10r is for dropping the voltage a little to get accurate voltage to the transistor?
Quote from: brand0nized on December 02, 2014, 04:30:43 PM
And the 10r is for dropping the voltage a little to get accurate voltage to the transistor?
This link might help. Do you see something analogous? http://sim.okawa-denshi.jp/en/CRlowkeisan.htm
That will show you the "what," but you might be able to guess the "why."
Quote from: midwayfair on December 02, 2014, 04:55:58 PM
This link might help. Do you see something analogous? http://sim.okawa-denshi.jp/en/CRlowkeisan.htm
That will show you the "what," but you might be able to guess the "why."
I'm not sure I'm understanding this properly, but I'm going to guess it's a noise filter before the 9v hits anything else in the circuit?
This is the updated double buffer with Vb
(http://tapatalk.imageshack.com/v2/14/12/02/285451fe354aef11da0ea2940febdbcb.jpg)
Remove the Vb connection from Pin 5.
You need a resistor between pin 3 and Vb. Remember, the
op amp thinks that Vb is "ground." If you connect something directly to Vb, it's like connecting your signal to ground -- no sound will come out. The resistor is there to bias the op amp (you don't need it on pin 5 because ... well, because it works without it! Someone else can explain "back bias" hopefully).
Quote from: brand0nized on December 02, 2014, 06:48:55 PM
Quote from: midwayfair on December 02, 2014, 04:55:58 PM
This link might help. Do you see something analogous? http://sim.okawa-denshi.jp/en/CRlowkeisan.htm
That will show you the "what," but you might be able to guess the "why."
I'm not sure I'm understanding this properly, but I'm going to guess it's a noise filter before the 9v hits anything else in the circuit?
For our purposes, yes. The more complicated answer is that it helps smooth out rippling in the power supply.
Quote from: midwayfair on December 02, 2014, 07:20:42 PM
You need a resistor between pin 3 and Vb. Remember, the op amp thinks that Vb is "ground." If you connect something directly to Vb, it's like connecting your signal to ground -- no sound will come out. The resistor is there to bias the op amp (you don't need it on pin 5 because ... well, because it works without it! Someone else can explain "back bias" hopefully).
Got it, so would that resistor value matter?
Quote from: midwayfair on December 02, 2014, 07:20:42 PM
For our purposes, yes. The more complicated answer is that it helps smooth out rippling in the power supply.
Ah, I think I learned about this when looking up building a power supply. (Settled with a onespot after my research)
Would I need that filtering and anything in this buffer circuit?
Quote from: brand0nized on December 02, 2014, 07:57:16 PM
Got it, so would that resistor value matter?
It does matter. Scroll down to the op amp buffer here: http://www.muzique.com/lab/buffers.htm
And good power supply filtering and polarity protection is always a good idea!
Some great teaching going on here. Many thanks, Jon! I love that you try to make him get the answer on his own, instead of just telling him. I can definitely say, you will learn more this way.
I'm also interested as I have plans for a volume pedal (build in an old wah shell though) as well.
Anyways, two questions came up reading this:
*) Does the pot have to have such a big resistance? Iirc, I was told that 10k would suffice for a buffered pedal (or does it matter if it is too big?)
*) The opamp is still connected to 9V and GND via pins 5 and 4, right?
(still have to learn about opamps at the end of this semester... we're dealing with basic transistor circuits atm)
1. i would say that that the value isn't of much importance, but i may be wrong
2. yes it is
Quote from: m-Kresol on December 02, 2014, 08:48:42 PM
Some great teaching going on here. Many thanks, Jon! I love that you try to make him get the answer on his own, instead of just telling him. I can definitely say, you will learn more this way.
Totally agree! I'm learning a lot, thanks to Jon. This has turned into an excellent tutoring session!
Quote from: midwayfair on December 02, 2014, 08:36:27 PM
Quote from: brand0nized on December 02, 2014, 07:57:16 PM
Got it, so would that resistor value matter?
It does matter. Scroll down to the op amp buffer here: http://www.muzique.com/lab/buffers.htm
And good power supply filtering and polarity protection is always a good idea!
The resistor going to Vb sets input impedance?
And would the power filtering in the greenbean schematic work fine?
Quote from: brand0nized on December 03, 2014, 04:27:14 AM
The resistor going to Vb sets input impedance?
And would the power filtering in the greenbean schematic work fine?
Yes to both. :)
Quote from: midwayfair on December 03, 2014, 12:51:29 PM
Quote from: brand0nized on December 03, 2014, 04:27:14 AM
The resistor going to Vb sets input impedance?
And would the power filtering in the greenbean schematic work fine?
Yes to both. :)
DING DING DING! Woohoo!
Alright, here is the buffer schem with power filtering and the input impedance cap.
(http://tapatalk.imageshack.com/v2/14/12/03/bb2cede10b72d24b061e5b08a4737590.jpg)
Why isn't the pot value so important?
Quote from: brand0nized on December 03, 2014, 07:18:13 PM
Why isn't the pot value so important?
Output impedance is very small from an op amp -- and the input impedance of the next op amp stage is extremely high.
To think about this, remember that everything is a voltage divider. Just like your potentiometer. But the potentiometer isn't the only resistance. There's also the resistance at each audio pin of the op amp.
Let's take two situations for the output impedance of stage 1, and we're going to use a 10K pot, which is about as small as you usually see for volume controls.
In situation 1, our output impedance is, say, 50 Ohms. That's about right for some op amps. 50R is a mere 0.5% of our 10K potentiometer. This means that at full on the pot, we're losing less than 1% of our signal. Likely inaudible loss.
In situation 2, our output impedance is a 5K Ohms. This isn't an uncommon amount for a transistor being used as an amplifier (it's the transistor's collector resistance + a few hundred for the transistor itself). In that case, our 10K volume pot would only be twice as big as the output impedance! We could easily lose quite a bit of signal.
You can see that even a few hundred ohms would make the choice much more difficult. But in an op amp, the resistance is so small that even if you made that 10K pot 10 or 100 times larger, you're not changing too much.
That makes perfect sense! I've ordered 100k and 25k, so I'll just try both.
Also, pin 4 still needs to go to +9v and pin 8 still needs to go to ground right?
EDIT: pin 4 should go after the filtering?
Quote from: brand0nized on December 03, 2014, 07:48:06 PM
Also, pin 4 still needs to go to +9v and pin 8 still needs to go to ground right?
EDIT: pin 4 should go after the filtering?
Look at the pinout in the datasheet again.
Quote from: midwayfair on December 03, 2014, 08:18:04 PM
Quote from: brand0nized on December 03, 2014, 07:48:06 PM
Also, pin 4 still needs to go to +9v and pin 8 still needs to go to ground right?
EDIT: pin 4 should go after the filtering?
Look at the pinout in the datasheet again.
Yup, just confirming.
Quote from: brand0nized on December 03, 2014, 07:18:13 PM
Alright, here is the buffer schem with power filtering and the input impedance cap.
(http://tapatalk.imageshack.com/v2/14/12/03/bb2cede10b72d24b061e5b08a4737590.jpg)
Why isn't the pot value so important?
just one thing, your power supply portion of the schematic doesn't have any filtering. It has a small value resistor which can help reduce noise and interference, but for power supply filtering you would use capacitors. Every power supply has some ripple, that means that DC voltage isn't excatly a constant value but it fluctuates.That fluctuations can introduce noise.
Here's a picture to help clarify things
Have in mind that the numbers on the pictures are purely for demonstration, i haven't done any math, so don't take them as is.
(http://i61.tinypic.com/esoyma.jpg)
Top left picture shows your power supply fluctuations, according to the picture, it fluctuates between 9 and 8.5V every milisecond. We need a cap here.
Why capacitors? Capacitors are devices used to store charge/energy. As such they need some time to to fully charge/discharge, as you can see in the top right picture.
The trick is to use capacitor which needs much longer time to charge/discharge than the actual time of your power supply variation. That way, when your PS starts dropping from 9 to 8.5V rather fast, capacitor discharges much more slowly thus making the voltage much more stable as you can see in the bottom left picture. We can say that when PS voltage starts dropping, the capacitors gives it's own accumulated charge to compensate.
Bottom right picture is the usual way of power supply filtering in pedals.
Hope this helps
Quote from: BaklavaMetal on December 03, 2014, 10:04:17 PM
(http://i61.tinypic.com/esoyma.jpg)
I see, thanks for explaining filtering, makes a lot of sense.
Can I just drop in the 10u and 47u for filtering in my schematic? And keep the diode for polarity protection?
yes to both.
Filtering added. Pin 4 after the 100u cap.
Is this correct?
(http://tapatalk.imageshack.com/v2/14/12/04/5e1de24bbb26319e4fe82e6e02c1051f.jpg)
Your power pins are wrong. Look carefully at the diagram in the datasheet.
Quote from: midwayfair on December 04, 2014, 07:52:31 PM
Your power pins are wrong. Look carefully at the diagram in the datasheet.
Darn, I remembered incorrectly. Now to redo my layout.
Where can I add a trimmer to give a slight boost?
Quote from: brand0nized on December 06, 2014, 07:44:59 AM
Where can I add a trimmer to give a slight boost?
You need more than a trimmer. You need at least a second resistor (if using a inverting amplifier) or a capacitor + 2 resistors (if using an inverting amplifier). Look here: http://en.wikipedia.org/wiki/Operational_amplifier#Negative_feedback_applications
(note that the wikipedia examples assume the op amp is running on AC, not DC. You need a large cap in series with the resistor R1 in example 1.)
Can I use this non-inverting example?
I'll read it a couple more times before I get how the voltage divider network increases gain...
Still need the huge cap to go to ground here, paralleling the resistor that goes to ground?
http://www.electronics-tutorials.ws/opamp/opamp_3.html (http://www.electronics-tutorials.ws/opamp/opamp_3.html)
Quote from: brand0nized on December 06, 2014, 07:31:00 PM
Still need the huge cap to go to ground here, paralleling the resistor that goes to ground?
Not in parallel with the resistor -- in series with it. Take a look at the first stage of the greenbean for an example of a variable gain amplifier. The resistor and capacitor coming off the inverting pin to ground controls the negative feedback. You won't necessarily want to use the values in that circuit here because they cut a lot of bass, but it will give you an example.
AMZ has a multi-purpose buffer/booster here: http://www.muzique.com/tech/opamp_multi.htm
That one uses an inverting amplifier for the second stage, which is fine as long as you have the first buffer so your pickups don't change the gain.
Here's a good handout on explaining a lot of stuff about an operational amplifier. Not much math (thank goodness), and it goes into good detail: http://web.stanford.edu/class/ee122/Handouts/2-Op-Amp_Concepts.pdf
Okay, so a voltage divider network can create gain where gain(A) = 1 + (R2/R1) where R2 is between the pin 2 (inverting input) and pin 1 (output) and R1 goes from pin 2 to ground (or to a cap which goes to ground)?
What does the cap do?
Quote from: brand0nized on December 07, 2014, 04:28:02 AM
What does the cap do?
Decouples it, and it will also set the -3dB high-pass ("bass cut") cutoff frequency of the gain together with the resistor that's in series with it, and then the gain will boost with that as the bass cutoff frequency. The formula for the -3dB cutoff of a filter of any sort is
always 1/Tau*R*C. But just use a calculator: http://sim.okawa-denshi.jp/en/CRhikeisan.htm
Put another way, the non-inverting amplifier will be an active high-pass filter. If you want to tailor the bass a little bit when you inch up the gain, you can do it that way, or you can just use a big capacitor, like 47uF or 100uF, and you'll boost everything.
Okay. I haven't gotten into another slightly simpler way: You can leave out the capacitor and just use a resistor to Vb instead of ground. You won't be able to get as much gain as the method going to ground, but if you're only trying to get a smidge more gain, that will save you the extra part and the headache of figuring out the frequency.
Quote from: midwayfair on December 07, 2014, 05:07:06 AM
Quote from: brand0nized on December 07, 2014, 04:28:02 AM
What does the cap do?
Decouples it, and it will also set the -3dB high-pass ("bass cut") cutoff frequency of the gain together with the resistor that's in series with it, and then the gain will boost with that as the bass cutoff frequency. The formula for the -3dB cutoff of a filter of any sort is always 1/Tau*R*C. But just use a calculator: http://sim.okawa-denshi.jp/en/CRhikeisan.htm
Put another way, the non-inverting amplifier will be an active high-pass filter. If you want to tailor the bass a little bit when you inch up the gain, you can do it that way, or you can just use a big capacitor, like 47uF or 100uF, and you'll boost everything.
Okay. I haven't gotten into another slightly simpler way: You can leave out the capacitor and just use a resistor to Vb instead of ground. You won't be able to get as much gain as the method going to ground, but if you're only trying to get a smidge more gain, that will save you the extra part and the headache of figuring out the frequency.
Yup, just enough gain to make up for the pot not being turned all the way.
With A = 1 + (R1/R2), I want to do R1: 10k and R2: 5kb trimmer, so that max gain will be limited and I can accurately dial in the gain to be unity at toe down position. Does that work? Also, where would the third pin of the trimmer go? Ground?
Quote from: brand0nized on December 07, 2014, 07:31:56 AM
Quote from: midwayfair on December 07, 2014, 05:07:06 AM
Quote from: brand0nized on December 07, 2014, 04:28:02 AM
What does the cap do?
Decouples it, and it will also set the -3dB high-pass ("bass cut") cutoff frequency of the gain together with the resistor that's in series with it, and then the gain will boost with that as the bass cutoff frequency. The formula for the -3dB cutoff of a filter of any sort is always 1/Tau*R*C. But just use a calculator: http://sim.okawa-denshi.jp/en/CRhikeisan.htm
Put another way, the non-inverting amplifier will be an active high-pass filter. If you want to tailor the bass a little bit when you inch up the gain, you can do it that way, or you can just use a big capacitor, like 47uF or 100uF, and you'll boost everything.
Okay. I haven't gotten into another slightly simpler way: You can leave out the capacitor and just use a resistor to Vb instead of ground. You won't be able to get as much gain as the method going to ground, but if you're only trying to get a smidge more gain, that will save you the extra part and the headache of figuring out the frequency.
Yup, just enough gain to make up for the pot not being turned all the way.
With A = 1 + (R1/R2), I want to do R1: 10k and R2: 5kb trimmer, so that max gain will be limited and I can accurately dial in the gain to be unity at toe down position. Does that work? Also, where would the third pin of the trimmer go? Ground?
Looks right in the description at least.
The third leg of the trimmer isn't needed; it's just a variable resistor.
Quote from: midwayfair on December 07, 2014, 08:56:25 PM
Looks right in the description at least.
The third leg of the trimmer isn't needed; it's just a variable resistor.
Excellent! I'll redraw the schematic when I get home.
Alright, here it is. Variable gain should be the last thing I need in this circuit.
Does all seem right?
(http://tapatalk.imageshack.com/v2/14/12/07/31bc89f1f9d5c86082568d282babf45c.jpg)
You need to hook up pin 2 of your trimmer to another leg of the trimmer (preferably 3, so the resistance increases as you turn the pot and therefore have more boost). The way you have it, the trimmer wont do anything when you turn it
Sent from my iPhone using Tapatalk
Quote from: m-Kresol on December 08, 2014, 08:48:52 AM
You need to hook up pin 2 of your trimmer to another leg of the trimmer (preferably 3, so the resistance decreases as you turn the pot and therefore have more boost). The way you have it, the trimmer wont do anything when you turn it
Sent from my iPhone using Tapatalk
Oops, you're totally right.
But other than that, all is well?