Home Forums General DIY Pedal Discussion MKC – Understanding Component Changes Reply To: MKC – Understanding Component Changes

#13983
Barry
Keymaster

Tonmann responded:

home for a few days!!!!

I usually have to run through all forums to see what hasn’t been answered and sometimes I do miss one or two questions.  If a question goes unanswered, either a bump on the topic or a PM usually does the trick.  Sometimes a bit of patience is required, if the hotel doesn’t have WIFI or they charge an exorbitant rate for internet access I can’t do anything until I get home.

I will try to answer the points without getting too “techie” as we are more interested in what effect a component change has rather than by how much of a change there is (you will need some mathematics) or why a component does what it does (techie theory).

We try to keep component values within a set of standard values (google “resistors E12 series”) as this works out cheaper for people who buy in bulk – most outlets offer a quantity discount.  We achieve this either by scaling component values (double one value, half another value) or by using values that have no or very little impact on the circuit. Sometimes scaling doesn’t work and we have to revert to “odd” value components (E24 series).

  1. C16’s main function is to prevent the DC voltage from the output of IC2A from reaching the output of the circuit, it also acts, in conjunction with the volume pot, as a high pass filter (cuts low frequencies).  The point at which it noticeably cuts low frequencies, called either the roll-off point or half power point depends on the capacitor value and the pot value; making the value of the capacitor smaller cuts more of the low frequencies.

For a 50kΩ volume pot and a 470n capacitor the roll-off point is about 6.7Hz, for a 4µ7 capacitor it is about 0.67Hz.  Since your guitar doesn’t produce anything under 80Hz (bass guitar 40Hz) a 470n capacitor is just as effective as a 4µ7 capacitor.

You would be hard pushed to get a 4µ7 film capacitor and would have to revert to electrolytic capacitors – something we try to keep out of the audio path.

A resistor between C16 and the volume pot would be used either to reduce the maximum level of the output signal or to prevent the op amp from high frequency oscillation if it were driving a capacitive load.

To reduce the maximum output level the resistor would have to be somewhere in the tens of kΩ range.

The 50kΩ volume pot is the load for the op amp i.e. resistive not capacitive so the 560R resistor is not needed.

  1. IC2B is a mixing stage which mixes the signals coming from R12, R15, R16 & R17.  The overall gain of the mixer is set by the value of R18.  Increasing the value of R18 will increase the overall gain. A 392k resistor is not in the E12 series and would provide no noticeable effect over a 390k resistor.  I used 470k to increase the gain slightly and since there were no adverse comments from the beta testers during the prototyping stage, I decided to leave it as it is.
  2. The tone control in the final stage comprises R19 – R22, P2 and C15.  With P2 at 50% rotation the frequency response is flat, at 100% rotation the mid and high frequencies are boosted and at 0% rotation the mid and high frequencies are cut.  Reducing the value of C15 will affect less of the mid frequencies, increasing the value will affect more of the mids and, to some extent, the bass frequencies.

With the value of the resistors and Treble pot as they are, a small change in the value of C15 will have no noticeable effect – you would need to double the value (10n) or half the value (2n2) to notice any real change.

  1. R16 more common value of 22k increases the gain slightly.
  2. Yes

 

Now the complicated one!!!!

  1. The output level of the second stage (IC1B) is determined by the input signal level (at pin 5) multiplied by the gain of the op amp.  P1A is doing double duty here, it sets the input level and sets the gain. Turning the pot towards 0% rotation reduces the input level and gain, conversely towards 100% rotation it increases both.

To simplify matters we can ignore C6, C7 and C8 and replace R8 with a jumper.

The input level is determined by the resistances of R7 and P1A to ground via the bias voltage VB.  The greater the resistance of P1A, the more input signal.  If P1A is at 100% rotation you would get about 90% of the signal that leaves IC1A arriving at pin5 IC1B – here’s the maths which you can ignore if you want to.

P1A / (P1A + R7) = 100kΩ / (100kΩ + 10kΩ) = 100kΩ / 110kΩ = 0.91 which is 91% of the signal from IC1A.

If P1A is set to 0% rotation it’s resistance (impedance) to ground via lug 2 is going to be about 40Ω

You would have to calculate the impedance at 80Hz of C18 in parallel with R24 in parallel with R23 plus C17 so take my word for it – it works out to about 40Ω.  I didn’t include this 40Ω in the first calculation as it becomes insignificant compared to 100kΩ.

So when P1A is set to 0% the level of signal reaching pin 5 IC1B is about 0.004% of the signal leaving IC1A

P1A / (P1A + R7) = 40Ω / (40Ω + 10000Ω) = 40Ω / 10040Ω = 0.0039 which is 0.004%

 

To all intents and purposes this means that the input signal is turned completely off which is something we try to avoid unless it is intentional.

By adding R8 to the circuit we have increased the minimum resistance when P1A is set to 0% rotation. If R8 is set to 2k2, you would now get about 18% of the signal from IC1A reaching the input of IC1B.

(R8 + P1A) / (R8 + P1A + R7) = (2.2kΩ + 40Ω) / (2.2kΩ + 40Ω + 10kΩ) = 2.24kΩ / 12.24kΩ = 0.18 which is 18%

You could have ignored the 40Ω of P1A and still arrived at roughly the same result.

The gain of IC1B is determined by the resistances of R10, R9 and P1A to ground via the bias voltage VB.

In fact, it is the ratio of R10 to R9 and P1A (ignoring C8, C7 and the 40Ω resistance of VB).

Setting P1A to 0% rotation would yield a gain of about 5

R10 / (R9 + P1A) = 470kΩ / (15kΩ + 100kΩ) = 470kΩ / 115kΩ = 4.09

Because the non-inverting input (pin 5) is used as the signal input we have to add 1 to the answer giving a total gain of 5.09.

If we set P1A to 100% rotation it’s resistance to ground would be 0Ω this would yield a gain of about 32

R10 / (R9 + P1A) = 470kΩ / (15kΩ + 0Ω= = 470kΩ / 15kΩ = 31.3

Again, adding 1 to the answer gives a total gain of 32.3

Although the above is the basics it is not the whole story, for a deeper analysis the impedance of C7 at least would have to be considered – something which would lengthen this post considerably.  I hope it is enough to say that by adding a resistor between R9/C7 and lug 3 of P1A the gain, more-so at higher frequencies will be reduced when P1A is pushed towards 100% rotation.

So if you are still awake I hope this has sufficiently explained point 6.

  1. As explained under point 2 increasing R18 increases the gain.  From point 6, the output signal of an op amp, IC2B in this case, is the input signal multiplied by the gain.  For IC2B this is quite complex as there are four different signals feeding IC2B (R12, R15, R16 and R17) you would have to calculate the signal level and gain factor for each branch and then add them together to arrive at the output signal level. As with any op amp if there is too much input signal and/or too much gain the op amp will clip.  The “magic number” for an op amp connected to a 9V supply is about 3V.

If there were just one component I would play around with in this circuit it would be R18.  Although a “socket and see” approach is OK it would be better with a 500k (1M acceptable) trim pot to replace R18.

  1. The main purpose of a capacitor such as C14 is to prevent the op amp from oscillating at high frequencies, usually outside the audio spectrum (for pedal building purposes I set this at 10kHz).  Larger values cut more high frequencies so you could use it to tame high audio frequencies.

Just to inflict a little more pain here’s the maths (you will need a calculator for this one).

Xc = 1 / (2π f C)

Xc is the impedance (or resistance) in Ohms

2π can be substituted with 6.3

f is the frequency in Hertz

C is the capacitance in Farads

With a bit of maths juggling you can calculate any of the three parameters Xc, f or C by moving it to the left side of the equals sign and what was originally on the left of the equals sign into the brackets.

If I want to find the roll-off frequency when R18 = 470kΩ and C14 = 820pF

f = 1 / (2π Xc C)                The value of Xc is the same as R18

f = 1 / (6.3 x 470kΩ x 820pF) = 1 / (6.3 x 470 x 10^3 x 820 x 10^-12) = 1 / (6.3 x 470 x 820 x 10^-9)

= 1 / (2.4 x 10^-3 ) = 417Hz

A long time ago I did offer to do a quick tutorial on scientific calculators for those people who slept throughout maths class – nobody was interested!

As you can see that’s quite a large low pass filter if I halved the value of the capacitor (390pF) the roll-off point would be double the frequency (868Hz). I could have halved the value of R18 and the roll-off point would have doubled but then the overall gain would have halved as well.

I hope this very long post is of some use.