EHX #1 Echo mods?

[Super Locrian]

I like this budget digital delay, but feel that it needs some improvements.

First of all, I would like to know if a different potentiometer for the delay time will give me a more useable sweep? As it is now, up to 12 o'clock is very short delay times, I'm guessing 0-60 ms. The most useable/common delay times (for me, anyway), 200-500 ms, are located roughly between 1 o'clock and 2 o'clock. Ideally, I would like to have these spread more out. Any ideas/suggestions on how to accomplish this?

Here is a gutshot of the pedal. I am also thinking about adding a trails switch, but that will have to wait for later.

[Digital Larry]

You could add resistors to each end of the pot that would limit its range to what you want.  Measure the voltage on the wiper at either extreme of the range you like, note the resistance of the pot, and then ohm's law should tell you what to put at the ends.

[Ice-9]

I'm guessing the pot is a control for the digital input to the DSP, so a voltage on the pot wiper will dictate the delay time. If that pot is a linear pot then you could replace it with a log pot or vice versa if it already is a log type stick in a linear. Worth a try anyhoo.

[Super Locrian]

You could add resistors to each end of the pot that would limit its range to what you want.  Measure the voltage on the wiper at either extreme of the range you like, note the resistance of the pot, and then ohm's law should tell you what to put at the ends.

The pot is marked B5K (linear taper, right?). The voltage at the wiper goes from 0 to 3.24V. The useable range for me would be from appr. 1.5V to 2.8V. I'm not smart enough to find out what value of resistors I need, please spell it out for me.

[Ice-9]

You could add resistors to each end of the pot that would limit its range to what you want.  Measure the voltage on the wiper at either extreme of the range you like, note the resistance of the pot, and then ohm's law should tell you what to put at the ends.

The pot is marked B5K (linear taper, right?). The voltage at the wiper goes from 0 to 3.24V. The useable range for me would be from appr. 1.5V to 2.8V. I'm not smart enough to find out what value of resistors I need, please spell it out for me.

That is exactly as I had expected, the voltage across the pot is ground to the 3.3v (3.24v measured). This is the operating voltage of the DSP or Micro that it is sending a control signal too. Replace this pot with a log or revers log pot to get a better spread of control. The value is not likely important but no need to go too far away from the 5k value. Ie if you fitted a 10k linear in place, as it is a digital control it would operate exactly the same.

[Super Locrian]

That is exactly as I had expected, the voltage across the pot is ground to the 3.3v (3.24v measured). This is the operating voltage of the DSP or Micro that it is sending a control signal too. Replace this pot with a log or revers log pot to get a better spread of control. The value is not likely important but no need to go too far away from the 5k value. Ie if you fitted a 10k linear in place, as it is a digital control it would operate exactly the same.

So it is better then to replace the pot with a 5K log pot than to experiment with adding resistors to the existing pot?

[Digital Larry]

Replacing the pot with a different taper will give you the same extremes, just spread out differently over the rotation.  Putting resistors in the top and bottom legs will constrain the range to what you said you wanted, at the expense of never being able to get above or below it.

And mostly since I just spent half an hour calculating this.

The Lower leg resistor comes out to 5724 ohms. Use a 5.6k
The upper leg resistor comes out to 1640 ohms.  Use 1.6k.

If you want me to explain it, that is going to cost you.   

[Super Locrian]

And mostly since I just spent half an hour calculating this.

Wow, had no idea it was so timeconsuming to calculate! Thank you very, very much, sir!

[Digital Larry]

Here's the 2 equations:

(5000 + A)/(5000 + A + B) = 2.8/3.24
(A)/(5000 + A + B) = 1.5/3.24

I found an online solver.  This is very cool.  Check it out!  Results a bit different.

http://www.quickmath.com/webMathematica3/quickmath/equations/solve/advanced.jsp#c=solve_advancedsolveequations&v1=%285000+%2B+A%29%2F%285000+%2B+A+%2B+B%29+%3D+2.8%2F3.24%0A%28A%29%2F%285000+%2B+A+%2B+B%29+%3D+1.5%2F3.24&v2=A%0AB

General idea is:
3 resistors in a row, from 0 to 3.24 volts.
Resistors, starting at ground, are [0 volts][Resistor  A][1.5 volts][5000][2.8 volts][Resistor B][3.24 volts]
Given the voltage between the left 2 resistors is 1.5, and between the right 2 is 2.8, what are A and B?

[Digital Larry]

By adding a DPDT switch and another pair of resistors, you could have switchable ranges.   

[Super Locrian]

By adding a DPDT switch and another pair of resistors, you could have switchable ranges.   

Nah, that's too elaborate. I did change the pot to an audio taper 10K, but the sweep is still not how I want it. So I'll try adding a resistor, but I'm not sure where to solder it, between the outer terminal and the wiper or between the PCB and the outer wiper?

[Digital Larry]

[ between the PCB and the outer lug?

Since you seem most keen on eliminating the lower range, lift the pot lug currently connected to ground and insert the resistor there.

[Super Locrian]

I soldered in two 4.7K resistors in series to the outer lug of the 10K audio pot, and now the sweep is much more useable. Now the first half of the turn is slapback instead of double-tracking, and it is much easier to dial in the medium delay times. Thanks for the patient help guys, really appreciate it! Next project will be to change the bypass to trails instead of true bypass, any pointers to how I might accomplish this?