Help me get 9 volts from a 12v supply

[ozzu2000]

Hi!

I've recently built a valvecaster, with an internal 12 volt regulator.
And now I want to add a booster circuit just before the tube input (an LPB).
The only problem is that the LPB works with 9v and my regulator is 12v (which I need for the valvecaster).

So the question is, how can I get 9v to the LPB, using the existing 12v supply?
I think I need a voltage divider, but don't know how to calculate the resistor values...

(I dont know if this is needed to calculate the resistors, but the impedante on the lpb, measured from the V+ to the ground is 1060K)

Thanks

[JRB]

You could use a voltage divider or a voltage regulator.
For the voltage regulator you could use a LM7809.
For the voltage divider you use the following formula to calculate the resistors:
Uout = R2/(R1+R2)*Uin

[ozzu2000]

Ok, that gives me the value of R1 for a given R2. So, If I use an R2 of 100k, that will give me an R1 of 33.33k.
But, will it work the same if I use R1=33ohm and R2=100ohm?  What about R1=0.33M and R2=1M ?

How do I choose a base value to start with?

Thanks

[wilrecar77]

Though you could use a voltage divider to supply the LPB, I believe it will work just fine on 12 volts. Unless your capacitors in the LPB aren't rated for at least 12 volts, nothing will blow up or go wrong.

[therecordingart]

Ok, that gives me the value of R1 for a given R2. So, If I use an R2 of 100k, that will give me an R1 of 33.33k.
But, will it work the same if I use R1=33ohm and R2=100ohm?  What about R1=0.33M and R2=1M ?

How do I choose a base value to start with?

Thanks

Current is the difference. I = E/R  

If you have 12v/33R + 100R= .090 A or 90ma. Then 12v/330R + 1k = .009A or 9ma. Etc.

[ozzu2000]

Though you could use a voltage divider to supply the LPB, I believe it will work just fine on 12 volts. Unless your capacitors in the LPB aren't rated for at least 12 volts, nothing will blow up or go wrong.

All caps can handle 12v fine. I was warried that the transistor might behave diferently with 12v. Wont it?

[chptunes]

The only difference is likely to be just a little more clean headroom before overdrive occurs.

Thought you might like this explanation of the LPB over at Beavis Audio:  http://beavisaudio.com/techpages/HIW/HIW.png

-Corey

[therecordingart]

Like others have said, you should be fine running it at 12volts. If you go the voltage divider route I'd take the time to learn how to Thevenize and Nortonize a circuit and appliy it to your voltage divider. Being able to Thevenize/Nortonize a circuit insanely insanely helpful, and doing it to a simple voltage divider is good practice.

[PRR]

> if I use R1=33ohm and R2=100ohm?  What about R1=0.33M and R2=1M ?

That's the right question.

The sad answer is: for "good" predictable voltage you pick divider values MUCH less than load resistance.

What is the equivalent "load resistance" of an LPB?

Well, it's got a 830K and a 10K.

Forget the 830K for a moment because the 10K probably swamps it.

The 10K goes to a transistor. If we assume the transistor is a short, the 10K sucks full 9V. If the transistor is open, the 10K sucks nothing. But short or open is useless. We would generally expect the transistor to split the 9V more-or-less equally with the 10K. So this path is 10K of R and about 10K of Q, 20K. More-or-less.

Check the assumption. If the 830K takes much of the 9V, we can write 20K||830K = 19.5K. If it takes less than 9V, it will act like more than 830K. So 20K for all practical purpose.

How much is "MUCH less"? Say 10X. Pick R2=2K. Compute your R1. In a worst-case, this 10X factor leads to about 10% error of output voltage. You compute 9V and you get 8.1V. That's often acceptable error. (Most LPBs ran on battery which runs-down to 7V or less.)

In this case, because your goal is near what you start from, it is a little better. R1 computes as 667 ohms. The Thevenin equivalent of 2K||667 is 500 ohms. A 20K load on 500 ohms is really a 40X ratio and 3% error of approximation.

BUT: that 2K alone pulls 10X the current of the whole LPB! This is washing the dog with a fire-hose... lots of waste.

In THIS case, the 5mA total is completely insignificant next to the 300mA load of the tube heater. So what the heck, do it.

In a more thrifty application, you might note that the 3% error is "too good". Bringing everything up by a factor of 3, to 6K and 2K, gets 10% approximation error and much less current drain.

When a load is *fixed*, does not vary with signal (should be true of the LPB), you can go further. Write R2=LPB (no shunt), R2= ~~20K, then compute R1 as about 6K. The "problem" is that we don't have an exact value for the LPB. Only that it must be more than 10K, less than infinity, and probably "around" 20K. While you could analyze the LPB for a better guess, small variation in resistors and transistors will make significant variation in current. (It is not real stable against production tolerances.) In this case, since everything in a LPB can take 12V, and we just want 9V for "the sound", _I_ would build the LPB, with a 100uFd cap at the "9V" point, then 10K or 5K to the +12V. A hot iron and a handful of red-stripe resistors will get the "right" value much faster than scribbling math on paper, and it WILL be right (for that perticular LPB build).

[ozzu2000]

wow! Thanks! That's a very good explanation   

[ozzu2000]

Maybe I could try with a pot, set it until I have 9v and then measure the resistences to use on the divider?