How to tame a high voltage output

[Mike B]

What you are asking is how do I get the voltage I need from what I have.

An LPad is a way to create any voltage you want from what you have while maintaining the same load impedance.

If you measured the voltage you have now in a loaded condition (source connected to load of known impedance) the LPad calcs we just did apply nicely.  They don't apply if you measure open circuit voltage.  Every source has an impedance too.

You could just add series resistance and that would drop the voltage too, but the load impedance would change.

The why of it is just Ohms Law.  And Kirchoff's current and voltage laws.

Voltage Law - the sum of the voltage drops in any closed loop is zero.

Current Law - The current leaving any junction is equal to the current entering.

Tough stuff eh?

If you have 6 volts and you need 1.5, then the 1/4 stuff we did is good.  If you have 6 volts and you need 3, then the 1/2 stuff is good.

If you need something else, you figure it out and I will review your findings.

[Paul Birkeland]

That will get you 1/5.

For 1/4 you would need 75K and 33K.

You got it!

You are correct, the 33K and 100K of the pot will make 25K.

[Paul Birkeland]

For those who need more clarification, I would refer you to the following resource:

http://www.goldpt.com/mods.html

[Loquah]

Can someone please confirm the equation to calculate the value of the series resistor (i.e. what's its mathematical relationship to the input impedance, parallel resistor and / or target impedance)?

[Paul Birkeland]

The place to start is how many dB you'd like to pad at the input.  This determines how much voltage is to be attenuated (20Log(Vout/Vin))=dB.

If we take 6dB of attenuation as an example, then we would need to attenuate half the voltage.  If we didn't care about impedance, we could use a pair of 100K resistors in series with the pot to attenuate the signal.  If we wanted to keep the 100K input impedance, then we would want a pair of 50K resistors in series with the pot and a pair of resistors across the pot which would make that parallel combination 50K (100K resistors will do that).

[Grainger49]

Take a look here:

http://www.siteswithstyle.com/VoltSecond/12_posistion_shunt/12_Position_Pure_Shunt.html

There is a lot of information on the page but what you want is Figure 2.2.1 Trick 1.

[Loquah]

Thanks for the great link Grainger.

Can someone answer the question of how the series resistor value is calculated? Is the series resistor just there to keep the load on the source the same?

[Paul Birkeland]

The series resistor both reduces the voltage and helps set the impedance.

[chard]

I just googled L-pad calculator and found a nice fill in the blank calculator that came to this solution. Series resister 75k, shunt resister 33k in parrallel with 100k pot. This circuit will attenuate 75% of your input signal.
   Because the 33k resister is in parrallel with 100k pot their combined resistance is about 25k. This circuit is a voltage divider network with a series resister of 75k, which is labeled R1 and a shunt resistance of 25k, which is labeled R2 and an overall imput impedance of 100k or R1+R2 . The voltage divider rule is: voltage out= voltage in x R2/R1+R2.  In your case 6.8volts x 25k/100k=1.7v
  To use the calculator you will have to know the level of db reduction you want. In your case you want a 12 db reduction which is a  signal reduction of 75%.

[Loquah]

I just googled L-pad calculator and found a nice fill in the blank calculator that came to this solution. Series resister 75k, shunt resister 33k in parrallel with 100k pot. This circuit will attenuate 75% of your input signal.
   Because the 33k resister is in parrallel with 100k pot their combined resistance is about 25k. This circuit is a voltage divider network with a series resister of 75k, which is labeled R1 and a shunt resistance of 25k, which is labeled R2 and an overall imput impedance of 100k or R1+R2 . The voltage divider rule is: voltage out= voltage in x R2/R1+R2.  In your case 6.8volts x 25k/100k=1.7v
  To use the calculator you will have to know the level of db reduction you want. In your case you want a 12 db reduction which is a  signal reduction of 75%.

I think that was the piece of the puzzle I was trying to grasp - thank you. I understood the general concept and also found various calculators, but I was trying to understand the exact maths that went into the value for the series resistor. I was under the impression that it was the difference between the starting input impedance (e.g. 100K) and the resultant impedance of the input impedance + parallel resistor. It seems it's slightly more complex so I'll just rest with the formulas provided.

Thanks everyone for your help!

[Grainger49]

One thing that makes this odd is that we don't hear voltage reduction in a linear way.  Cutting it in half then cutting to 1/4 doesn't sound the same.  Our volume pots are logarithmic not linear.