Determining Desired Resistance Empirically a Good Idea?

[The New Steve H]

In the past, trying to calculate resistances on paper was a pain. I can figure out an ordinary voltage divider, because the resistance of the known half is obvious, but when you're dealing with tubes and a bunch of other crap in the pipeline, it gets very hard to figure out what I'm dealing with. Tube datasheets are confusing, and resistances vary with temperature and so on.

Yesterday I was trying to change some preamp voltages, and I came up with this solution. I took a 250K pot and soldered two wire leads on it. I put it in place of the resistor I was changing, and I attached a meter. I adjusted the pot until I got the voltage I wanted, and then I checked the resistance and used resistors close to that value.

My question: is this a clever solution, or should I really be able to do this on paper? I am going to put leads with alligator clips on that pot so I can put it to use without soldering. I do have some concerns about frying myself on the uninsulated lugs, but that's nothing new.

It also made it easier to determine the effective resistance of the other stuff in the current path, instead of adding it up and praying I was right.

[boots]

I have been dealing with this situation myself recently, and I think sometimes it is faster and easier to just tack components together "experimenter-style" to find the answer you need. More than one way to solve a problem. I guess that's the difference between a technician and an engineer.

And yes, it can be a bit like driving without a seat belt, because you have to be very aware of what not to touch when you have clip leads all over the bench.

I have a breadboard ("proto-board") but it's intended for use with solid state circuits, and I'm scared that it can't handle tube voltages, so I just use the clip lead method.

[The New Steve H]

I have a breadboard, too, but man, that's a lot of work. And I wouldn't even know how to attach 20-gauge and 18-gauge wire. The holes are so small.

I am thinking I should put together a little box for the pot, to cover up the lugs. Slap a knob on it, attach a magnet so it will stick to metal surfaces while I use it. Done!

[M Fowler]

https://taweber.powweb.com/amptechtools/subs.htm

[The New Steve H]

Wow. Maybe if I do this for a hundred years, I'll guess everything everyone else already knows!

[EtherealWidow]

Didn't know other people thought about these things too. That's nice to know. $35 is stupid though.

[M Fowler]

Just use the photo as reference and build your own, would be some good projects for the work bench.

Mark

[martin manning]

$35 is stupid though.

Seems reasonable to me... See if you can build one for less.

[surfsup]

Steve if you know the starting voltage and the "voltage you want" why not just do the math? Knowing these voltages implies you have a scope (on AC) or if this is DC, a DMM. So if you know the math for the voltage divider, I would think that is the easiest way.

But I may have misinterpreted your post.

[John_P_WI]

An r c substitution box is my most used tool in tweeking.

[ToneMerc]

$35 is stupid though.

Seems reasonable to me... See if you can build one for less.

I agree 100% Martin; no parts chasing, no assembly time and it's a clean finished product. You gotta love the wonderful world of the Wal-Mart, I want everything cheap mentality.



TM

[The New Steve H]

Steve if you know the starting voltage and the "voltage you want" why not just do the math? Knowing these voltages implies you have a scope (on AC) or if this is DC, a DMM. So if you know the math for the voltage divider, I would think that is the easiest way.

But I may have misinterpreted your post.

It just seems like the numbers are slippery. They seem to change when the amp heats up, and when you have a big line of things between the test point and ground, it can be confusing to determine all the values and add them up.