Easy Question

[boyersdad]

Hi gang!

Wanting to build a blend circuit for my Big Muff, and have decided on using Sean M's minimal B. Blender. The schematic can be found at: http://seanm.ca/stomp/minblend.html

My question is, Vr in this case, is half the Vcc, and not ground right? I can make a simple voltage divider from the Vcc?
In this instance, would two 4.7k's be okay for the divider?

Thanks!

[earthtonesaudio]

Looks like Vr is 1/2 supply. 

I would use a larger value for the divider network, 10k or more.  Lots of stompboxes use 100k for each one.  Too small will drain the battery faster, and you don't need that much current for this application (depends on the opamp used).

Don't forget the cap from Vr to ground.

[demonstar]

Just to add to the above post...

I'm not familiar with the big muff but if there is already a Vr established you may be able to use that one (Can't always do that).

As stated above different values of R are often used in a potential dividers although they give the same Vout. What the differences in the resistor values does affect is the current through the divider. One thing that must be ensured is that the current flowing out of Vr must be 1/10th or smaller of the current flowing through the divider (10% is just a nominated value by engineers I believe).

Lets assume you are setting up a divider at 4.5V  for the minimal blender using an LM741 opamp.

An ideal opamp has NO current flowing into the input. Oh, great we don't have have any current flowing out of Vr so we're sorted. (If only it was that simple!)

Our opamps are NOT ideal but instead are real. A real opamp has SOME current flowing into it's input (very small).

Right, so we need to know how much current is flowing into our opamp input so we go to the datasheet and look up bias current. For the LM741 I found a typical value of 80nA (8*10^-8A).

We now know we must have at least ten times that flowing through our resistors so using ohm's law we get...
I=V/R
I=V/(R1+R2)
8*10^-7=9/(R1+R2)

We also know we want a outpu voltage of 4.5V so we need to use the potential divider equation to get... (assuming R2 is the bottom resistor)

Vout=Vin*(R2/(R1+R2))
4.5=9*(R2/(R1+R2))

We solve the two equations below simultaneously...
8*10^-7=9/(R1+R2)
4.5=9*(R2/(R1+R2))

We get...
R1=5.625M
R2=5.625M

So we go easter egging for nearby values and in my case I happen to have some 5M6 resistors so I'd use them. If you don't have any don't worry using lower resistors are find because if you remember I said that, "the current through the divider must be AT LEAST ten times the current flowing out of Vr." By making the resistors smaller you are making the current through the divider bigger which is good for stabillity of Vr but not so good for battery life.

Hope that makes sense and helps!

[earthtonesaudio]

"the current through the divider must be AT LEAST ten times the current flowing out of Vr." By making the resistors smaller you are making the current through the divider bigger which is good for stabillity of Vr but not so good for battery life.

Hope that makes sense and helps!

Thanks for the nice explanation!
But I believe you can get the same effect of divider current=bias current*10 by using a large resistor from the bias point to whatever is being biased.  That's what the 1M resistor from the opamp's input to Vbias does in Tube Screamer type circuits... I think.

Anyway that might be a better way of getting the ratio right without worrying about draining the battery too soon.

[demonstar]

I believe you can get the same effect of divider current=bias current*10 by using a large resistor from the bias point to whatever is being biased.  That's what the 1M resistor from the opamp's input to Vbias does in Tube Screamer type circuits... I think.

I could be wrong but I think the 1M resistor from Vr to opamp input is pretty much just there to set the input impedance.

I think it may well form a filter with the decoupling cap from the centre of the divider to ground. Not definite but I think so.