Voltage divider to take 5v down to ~1.5... really this simple?,

[benallison]

Is a voltage divider really this simple?

http://www.raltron.com/cust/tools/voltage_divider.asp

A 50k and 25k will get me to ~1.6, which is fine.

Is that really all it takes? The application is reducing the output of the the CV jack in my DOD FX17, which I'll be using as an expression pedal with my Line 6 M9.

[T1bbles]

Yep, it really is that simple! Life's good

But a 50k and a 22k would get you closer to 1.5v if you're bothered about getting it spot on.

Talking about spot on, you could get a 50k resistor and 25k precision trim-pot and adjust it to around 21.4k, while metering your output, and with a bit of adjusting you'd be able to get bang on 1.5v.

[PRR]

> to ~1.5... really this simple?

How MUCH 1.5V?

If you want to power a little flashlight, 50K-25K will just sag to nothing under the load of a 10 ohm bulb.

Control ports are usually "high" impedance. In the old days, sometimes 100K was one of the many standards. In the Voltage Divider, this 100K is in parallel with your proposed 25K, Together they make a problem in Parallel Resistors, and 25K||100K is 20K. Now your actual voltage divider is 50K+20K, your output is 20K/(20K+50K) or 20K/70K or 2/7ths. You put in 5V, you get out 5V times 2/7 or 10/7 or 1.43V. That may be close enough.

If not, you can fudge the "25K" to 27K or so.

Since you don't really know the control port impedance, using lower-value Voltage Divider makes the actual load matter less. However too-low will load the source port, and you may not know just what it can stand and still be accurate.

[Processaurus]

Have you actually tried hooking a 10K pot up to your expression input and measure what happens?  That's the only way to know how to tailor the CV (by knowing the max voltage that the expression pedal input sees with a 10K load across the tip and the sleeve), I don't think the internet actually knows unless somebody's done it or has a schematic of that section.  But the internet is always willing to speculate!

PS a 75K voltage divider is much too weak for feeding Line 6's input, I just stuck a meter across the expression pedal input of one of the modeler pedals, and it read 4K.  Voltage dividers aren't quite as simple as getting the ratio of resistance right, they have to be able also to provide enough current so that the load (the expression pedal input in this case) takes only about 1/10th or less of the current going down the voltage dividers resistors.  So ideally you are looking for 400 Ohm's total for your voltage divider.  That can only work well if you mod your DOD pedal, consider replacing the series 1K and 4.7K to ground, that are already forming a voltage divider right at the 0-5v outlet on Paul Marossy's schematic.  That would be a total of 400 ohms, at whatever ratio makes the correct voltage division.

His page is great resource on that pedal, it is an odd one.

Actually, now that I think about it, with the 4K load of the line 6 in parallel with the aforementioned internal 4.7K load of the DOD's CV, if it is plugged in to the Line 6, it will divide the voltage range of the dod in half, as the total load on the DOD CV is then about 2.2K.  Who knows, that might even be the right voltage (~0-2.7v).  But with an expensive pedal like that I would measure it first with the 10K load, and then measure the DOD CV (you could even use a 4K resistor as a 4K dummy load to simulate the line 6).  Adding a series resistor in between the DOD and the line 6 would attenuate the CV more.

Hope that isn't too confusing, the 2nd paragraph is the textbook way to scale the CV in the DOD down, the last paragraph guesses that it is probably ok if the CV sags a bit and interacts with the 4K load of the line 6.

[benallison]

My brain just exploded! Hahaha, and yes the Internet LOVES to speculate! That's what it's best at (along with porn).

No, everything makes sense. My M9 is in the mail, and I don't have a meter yet.  But I will measure everything to make sure my values are right, and report back.