Problem with Blends

[seten]

I just built a split n blend and a mini jfet blend with 2n5458's (yes flipped them around for the split n blend) and 100kB pots - tried both of them with an mxr micro flanger and they both worked but there was a huge volume drop in the middle of the blend pot. Figured that was because the micro flanger inverts the phase and built a phase inverter with a TL072 exactly like this one but minus the 5pf cap:
http://www.muzique.com/images/buff10.gif

I put one between the return and the pot on both circuits.. Now in the middle theres a little less of a volume drop but its still most certainly there and I think now there's less bass on the dry side of the rotation as well - on both circuits.

Seems like either 1) 25458's dont work in these circuits 2) I messed up both blend circuits in the exact same way 3) I messed up the phase inverter circuit or 4) I don't understand something thats going on here.

I've been double checking the circuits and recreating my results for a very long time so #4 seems most likely - any ideas?

http://tagboardeffects.blogspot.com/2012/02/split-n-blend.html
http://tagboardeffects.blogspot.com/2012/01/mini-blend-jfet.html?m=1

[seten]

Okay, built a new phase inverter with a tl071 and same issue. Tried to test the phase inverter and not sure how - 4.5v bias voltage on the noninverting input, 1v right after the input cap (putting it before would mess it up bc caps block dc?), and tested voltage right before the output cap and got 7.7v which doesnt make sense to me - I thought 1v inverted should make the output 1v lower than the bias voltage so 3.5v?

Okay, back to the blends - just realized that if i just connect the send directly to the return on the split n blend theres still a huge volume drop in the middle of the blend pot. Same thing on the mini jfet blend (although that makes sense because one of the sides is unbuffered?). Whats going on.

[PRR]

> #4 seems most likely -

I don't see YOUR circuit?

[anotherjim]

When blending 2 outputs with a pot, the centre volume depends on the value of the pot and the load seen by the wiper of the pot since this forms a potential divider. Generally, using a pot value too large is going to lead to centre drop. Often, some drop is wanted since 2 similar sources would be addition boosted by the 50:50 blend compared to on their own.

If the sources are from a low output impedance driver (an opamp) then a blend pot can be modest value (10k) but you have to make sure the coupling capacitors out of the amplifiers are large enough (at least 1uF with a 10k pot) or you will get bass loss. If the blend pot was 10k, then the wiper load can be around 100k without any noticable centre drop. You can introduce centre drop if the blend is too loud by lowering the wiper load resistance. This would be the input resistor of the next stage or an additional resistor to 0v if the blend wiper is your output and you can assume that is a high impedance guitar input is what its feeding.

[seten]

>
I don't see YOUR circuit?

Do you mean a picture?

When blending...

Think I get it - Imma come back to this later this week but in the meantime is my phase inverter working right? If so what am I missing, why wouldnt the output be 3.5v?

[R.G.]

It is hugely difficult to come up with helpful insights without an accurate schematic of your circuits.

[anotherjim]

If you have 4.5v or close to it on the non-inverting input, you should have that voltage on the opamp output. If you apply a test voltage to the inverting input resistor for a +1v response it will have to be 5.5v since 4.5v is the reference - then the output should in invert to 3.5v.

[seten]

It is hugely difficult to come up with helpful insights without an accurate schematic of your circuits.

Sorry, I thought I made it clear in the original post how all the schematics I linked were hooked up but now I see I did not. I need to move some things around anyhow, tomorrow I'll get everything situated and try to describe it all clearly.

If you have 4.5v or close to it on the non-inverting input, you should have that voltage on the opamp output. If you apply a test voltage to the inverting input resistor for a +1v response it will have to be 5.5v since 4.5v is the reference - then the output should in invert to 3.5v.

Okay, yeah thats what I thought - let me make sure I was testing it correctly and get back to yall. Two things I can think of off the bat -
1) should I actually disconnect the input and output caps? They cant do anything to the circuit if one leg is just floating right? (EDIT unless its been charged ? but even that shouldnt matter in this case because it would just dump its charge into the circuit and then settle at the correct readings?)

2) It doesnt matter if I'm getting all these voltages (9v power, 4.5v bias, 1v test) from the same 9v power supply correct?

[seten]

Okay, first things first I need to figure out this inverter. Just built a brand new one and im getting the same results. 1.2M resistor between (-) input and output, and a 1.2M resistor on the (-) input where the other end of the resistor is an input. The one on prototyping board uses a TL071 and the one in the socket is a TL072.

http://imgur.com/gallery/kYHdxlY

9.5v to +VCC
4.4v to + input
Readings:
4.4v + input
4.4v - input
3.9v - input resistor
4.4v output

Now, add 1v test voltage to - input resistor
Readings:
4.4v + input
4.4v - input
7.8v output

[PRR]

> add 1v test voltage to - input resistor

What happens if you "add" zero Volts to - input resistor?

Leaving a lead open-circuit is not the same as "zero".

[seten]

> add 1v test voltage to - input resistor

What happens if you "add" zero Volts to - input resistor?

Leaving a lead open-circuit is not the same as "zero".

I was taking the bias voltage off another circuit so to be safe I just made a voltage divider so now the bias voltage is 4.7V.
input - output
0V - 8.9V
1V - 8.4V
2V - 7.4V
3V - 6.4V
4V - 5.4V
4.7V - 4.7V
6.4V - 3V

I think I see whats going on here - was I just supposed to bias the input to 4.7V as well? So then a 1.7V test charge would sum to 6.4V, 4.7-1.7 = 3. The more I think about it, that would make complete sense and clear up some things I was wondering about too. Do I connect it directly to the bias voltage or through a resistor?

[anotherjim]

I think you've picked up opamp theory from classic dual supply teaching where Vref is 0v. In single supply, we raise Vref to be in the middle (usually) of that single supply voltage.
So you would connect a test voltage between Vref and the input resistor. You do not add another resistor.
Note that your supply providing a test voltage cannot have it's 0v common to the circuit under test or it will short your Vref to 0v.

Isn't it easier though, to continue to work with the 0v of a single supply from test purposes? All you need do is adjust your expectations to acount for the +4.5V refererence.
So working from 0v, a +1v input is -3.5v below reference. The amplifier multiplies this by -1 which outputs +3.5v above reference which is +8v above 0v. Note that a real opamp may not swing its output fully to either the +supply or 0v unless it is a "rail to rail" output type.

[seten]

Gotcha, that makes sense - so I am supposed to bias the input to vref as well? I was using this schematic
http://www.muzique.com/images/buff10.gif
and I dont see any indication of that but i figure maybe thats one of those things that seems obvious to people with experience. Otherwise, the input would be a small guitar signal AC voltage which is centered around 0v, so the outPut would be centered around about 9 volts (in my case 8.9V) - sure its the same signal (just inverted) and would probably work since my power supply is 9.5V, but we want it to be centered around the middle of the opamps operating range, especially since if our power supply is exactly 9v and the output signal is centered around 9v then in the upswing of the signal itd be out of the operating range entirely. am i wrong about any of that?

[antonis]

@seten: If I promise nobody will steal your circuit, would you post a schematic..?? 

[anotherjim]

No, you don't have to supply bias to the inverting input. What happens is the amp gets Vref from the non-inverting input and moves the output voltage so that Vref is applied to the inverting input via the feedback resistor.

[seten]

@seten: If I promise nobody will steal your circuit, would you post a schematic..?? 

Hahaha sorry okay so I'm going to be using these two blends on the left and right channels of a flying pan http://byocelectronics.com/skilletschematic.pdf
but when it didnt work originally I switched over to trying to use the blends individually on other pedals (including my mxr micro flanger) to make sure it was a problem with the blends and not the flying pan circuit, but I now understand I should be troubleshooting them together with the flying pan. I didnt post it earlier because I thought the issue was with the blends and phase inverter themselves so I didnt think it was necessary to post any schematics other than those (which are all in the original post). Now that I know my phase inverter is working correctly and now that I know how the pot value would affect my problem I'm gonna mess around with that and get back to yall if (when) I need more help.

No, you don't have to supply bias to the inverting input. What happens is the amp gets Vref from the non-inverting input and moves the output voltage so that Vref is applied to the inverting input via the feedback resistor.

Ahhh gotcha - just watched some youtube videos and had some aha moments and I definitely get it on some level but I think I just need to keep watching youtube videos until it sinks in - all the Escher hands drawing themselves stuff with opamp feedback is hard to wrap my brain around.

The thing I described in the last post is still bothering me though and I think I'm close to getting it - say I apply a 0.01V DC test voltage to the input, that would make the output a little under 9V. What makes that different from when the guitar signal swings positive to 0.01V? Is it because the test voltage is from a power supply thats doing everything it can to keep it there at 0.01V and therefore will "overpower" the circuit in a way the guitar signal would not?

[anotherjim]

The thing I described in the last post is still bothering me though and I think I'm close to getting it - say I apply a 0.01V DC test voltage to the input, that would make the output a little under 9V. What makes that different from when the guitar signal swings positive to 0.01V? Is it because the test voltage is from a power supply thats doing everything it can to keep it there at 0.01V and therefore will "overpower" the circuit in a way the guitar signal would not?

This is why the input from the guitar is coupled via a capacitor. The input resistor will have Vref on it at the inverting input end and because there is no DC path through the input resistor, it will have Vref at both ends (no current through it so no voltage drop).

When guitar signal is applied via the cap, it will cause current flow in the input resistor which would modulate the voltage at the inverting input pin but the amplifier reacts to this by adjusting its output voltage to cancel out the input current via the feedback resistor. This is why it inverts the output.

If the input and feedback resistors are the same resistance, then the output voltage moves the same amount as the signal at the input cap - so you have an inverting amplifier with x1 gain. If you make the feedback resistor larger, the output will have to swing further to cancel the signal at the inverting input so you have more than x1 gain.

If an inverting opamp is all working properly in a perfect world, you will never detect a signal at the inverting input pin. The negative feedback from the output will always be cancelling it out.  It will try its best to keep the inverting pin equal to the DC voltage applied to the non-inverting pin.

[seten]

The thing I described in the last post is still bothering me though and I think I'm close to getting it - say I apply a 0.01V DC test voltage to the input, that would make the output a little under 9V. What makes that different from when the guitar signal swings positive to 0.01V? Is it because the test voltage is from a power supply thats doing everything it can to keep it there at 0.01V and therefore will "overpower" the circuit in a way the guitar signal would not?

This is why the input from the guitar is coupled via a capacitor. The input resistor will have Vref on it at the inverting input end and because there is no DC path through the input resistor, it will have Vref at both ends (no current through it so no voltage drop).

When guitar signal is applied via the cap, it will cause current flow in the input resistor which would modulate the voltage at the inverting input pin but the amplifier reacts to this by adjusting its output voltage to cancel out the input current via the feedback resistor. This is why it inverts the output.

If the input and feedback resistors are the same resistance, then the output voltage moves the same amount as the signal at the input cap - so you have an inverting amplifier with x1 gain. If you make the feedback resistor larger, the output will have to swing further to cancel the signal at the inverting input so you have more than x1 gain.

If an inverting opamp is all working properly in a perfect world, you will never detect a signal at the inverting input pin. The negative feedback from the output will always be cancelling it out.  It will try its best to keep the inverting pin equal to the DC voltage applied to the non-inverting pin.

THANK YOU I finally get it now. Thanks so much everyone!