3. Hooking Up the IC

[gaussmarkov]

In this installment, we put the IC on the board and check the voltage levels of its pins:  IC Hookup.  We are getting close to a working circuit.

[gaussmarkov]

just to be clear ... this part of the tutorial is at http://diystompboxes.com/projects/ic.html

[andmej]

There's a little mistake in this section.

After the "Resistors" subtitle it reads "place the LM471 so that pin 1 is in E21 and pin 8 is in F21" instead of "place the LM741 so that pin 1 is in E21 and pin 8 is in F21".

It made me confuse.

[gaussmarkov]

There's a little mistake in this section.

After the "Resistors" subtitle it reads "place the LM471 so that pin 1 is in E21 and pin 8 is in F21" instead of "place the LM741 so that pin 1 is in E21 and pin 8 is in F21".

It made me confuse.

oops!  thanks for pointing out that typo.  i also changed the subtitle to "IC and Resistors" because that is more accurate, isn't it?

[bipedal]

Great tutorial here -- thank you for the time and energy put into creating this.

I have a general theoretical question on op-amp feedback loops (in this tutorial example, the loop running from IC pin 6 back to the negative input at pin 2).

I believe the C3 cap in the loop helps to "soften" the signal being fed back into the op-amp by removing some high frequency content...

What is the purpose of the 1M resistor (R5) in that loop, and how might raising or lowering that value impact the op-amp's output level and amount of clipping from this stage?

I have two competing guesses:

1)  A lower R5 value could result in more output and likely more IC distortion, though too low could lead to oscillation.  Increasing R5 value to more than 1Meg would reduce amplification, and also potentially dull the sound as more signal in the feedback loop is "pushed" through the C3 cap.

2)  A higher R5 value would increase the impedance of the feedback loop, reducing the signal level feeding back into the op-amp while increasing the amount of signal sent on to C4 and the post-IC part of the circuit.  End result -- more output, more distortion gained through the diodes.

Are either of these on track?

Thanks,

- Jay   

[gaussmarkov]

Hi Jay!

Great tutorial here -- thank you for the time and energy put into creating this.

You are welcome.  I'm glad that you are getting something out of it.

I have a general theoretical question on op-amp feedback loops (in this tutorial example, the loop running from IC pin 6 back to the negative input at pin 2).

I believe the C3 cap in the loop helps to "soften" the signal being fed back into the op-amp by removing some high frequency content...

Let me amplify    what you are describing to see if we are on the same page.  C3 allows more high frequency signal to feedback to the inverting (negative) input than if C3 were not there.  The higher the frequency the more signal gets through.  As a result, the higher frequencies are amplified less than the lower frequencies.  If that does not make sense, read on and see if the rest of my answer helps with this.

What is the purpose of the 1M resistor (R5) in that loop, and how might raising or lowering that value impact the op-amp's output level and amount of clipping from this stage?

I have two competing guesses:

1)  A lower R5 value could result in more output and likely more IC distortion, though too low could lead to oscillation.  Increasing R5 value to more than 1Meg would reduce amplification, and also potentially dull the sound as more signal in the feedback loop is "pushed" through the C3 cap.

2)  A higher R5 value would increase the impedance of the feedback loop, reducing the signal level feeding back into the op-amp while increasing the amount of signal sent on to C4 and the post-IC part of the circuit.  End result -- more output, more distortion gained through the diodes.

Are either of these on track?

Number 2 is on track.  Have you read about "divided feedback" for operational amplifiers?  I found the article on allaboutcircuits.com helpful.  Here's a summary:  op amps set up with a feedback loop try to keep the voltage at the two inputs equal.  In this case, because the feedback comes out of the junction of a voltage divider, the op amp has to raise the the output above the noninverting input enough to get the divider to give the same voltage at the inverting input.  If you raise the resistance of R5 then the voltage at the junction is lowered and this requires more output from the op amp to get the input voltages equal again.

I hope this helps!  All the best, Paul

[bipedal]

Very helpful article on the allaboutcircuits page -- thanks for the reference.

It isn't necessarily the standalone value of R5 that impacts gain in this type of circuit -- it's the ratio between R5 and the total resistance of the gain pot + R4, since that section of resistors acts as a voltage divider.  Makes sense.

If I'm doing my calculations correctly, per

Gain = 1M / (4.7k + Gain Pot Value) + 1

the voltage gain in this noninverting amp circuit ought to run between approx. 17 and 212?

- Jay

[gaussmarkov]

If I'm doing my calculations correctly, per

Gain = 1M / (4.7k + Gain Pot Value) + 1

the voltage gain in this noninverting amp circuit ought to run between approx. 17 and 212?

that's almost right. one has to take the capacitors into account to get exact answers.  and also the 9V supply places an upper bound on how much amplitude you can get, which is no more than a 4.5V and with this particular op amp less than that.  when you hit this limit, you get clipping in the op amp output.

with the caps there, the gain also varies with frequency, as discussed above.  at 800Hz, the minimum gain about 20.  at 8KHz, it's around 11. 

at 800Hz, op amp clipping starts around a pot value of 30K.

[Tuhkam]

I have a newbie question 
Why is it important to use an electrolytic cap as C4, is it protecting the IC in some manner? I breadboarded this and it seemed to work with a 100nF metal-film as well.

[aron]

Probably size and cost.

[newfish]

Hi,

Thanks for putting this project together.

I have a question about R3.  I've put together a simple OA overdrive on my breadboard, and it works without R3 being there (I have a jumper in place from the mid-point of the voltage divider to supply my 4.5v).

Gut feeling tells me that R3 has two purposes - firstly to limit the amount of current being supplied to the input signal (for some purpose?), and secondly to make the input signal ever-so-slightly mis-biased - and therefore more prone to clipping (at least in one direction).

Are these gut feelings correct?

Also - why do other designs (MXR Distortion+, I think), have a 1M resistor here instead of 470K?

Many thanks for your time.

Ian.

[gaussmarkov]

hi, well a quick answer is that without R3 the signal from the guitar would be overcome by the reference voltage level. all the best, gm

[newfish]

Ah.

Hadn't thought of that!

Thanks, gm.

I've also read through your website's Op-Amp section, and learned plenty to keep me going.  Many thanks!

[gaussmarkov]

good.  you know, it's really a great question because it highlights a way of thinking about circuits that is useful to learn.  we often look at a circuit and think in terms of voltage or in terms of current, but not always in terms of both.  for the signal of the guitar, it is often natural to think in terms of voltage alone.  that is helpful for thinking about R3 because you see that on one side of R3 is a stable voltage, called VB in the schem, and on the other side of R3 is the input signal coming from the TIP of the input.  so R3 must be insulating the signal somehow from the voltage level VB.  R3 does this by influencing the current.

without R3, the flow of current would be essentially infinite and the voltage supply would "flood" pin 3 of the LM741 with electrons that would keep the voltage level at VB.

with R3, VB pegs the "average" voltage level at VB and R3 slows down the flow of electrons enough to preserve the voltage wiggles from the input TIP.  sometimes the flow is faster than others and that is what makes it possible for the voltage to remain constant on the VB side of R3 and vary on the other.

because this is a tutorial, it might be worth noting that similar things are going on with R1 and the VOLUME pot at the end of the circuit.  instead of VB, the voltage level on the other side of the resistor is ground (GND).  R1, which has a really high value, is the so-called "pull-down" resistor for preventing pops when the circuit is switched into the signal path.  a very small current through R1 is enough to prevent a TIP voltage that could creep above ground level while the circuit is by-passed due to electrons leaking through capacitor C1.

at one extreme, the VOLUME pot allows us to "flood" the TIP of the output with electrons from ground so that the signal is wiped out and the output is quiet.  at the other extreme, the VOLUME pot acts as another pull down resistor, attenuating the signal somewhat and guarding against electrons leaking through C4.  this is why one does not see a pull-down resistor on the output TIP.

[pnnt]

Hello!

Where do I have to connect the negative lead coming from the battery?

[deadastronaut]

pin 4.....if its on breadboard battery negative should be on the ground rail...and put a jumper from ground rail to pin 4 ok...

look at the diagram here....

http://diystompboxes.com/projects/ic.html

and 741 ic here...

http://www.national.com/ds/LM/LM741.pdf

[pnnt]

Thanks a lot!

It's actually working

[deadastronaut]

great!....good feeling when it actually works eh?....

[jokersmile32]

Can i eliminate C6? i dont see a cap used here in any other dod 250 schematics.  I also dont see R7 used either.