Lowering voltage

[SpareRibs]

Hello,
       Looking at Ohm's Law I cannot understand how to calculate a given voltage and stepping it down to
a desired voltage. Is there a formula to calculate the amount of resistance needed to get from ?Volts to ?volts

[nashvillebill]

If you are using a resistor to do the voltage drop, the current flow through the resistor determines the voltage drop.  You've got to know the rest of the circuit to determine the current that will be drawn through your dropping resistor.

That's why the formula is V=IR.  The voltage drop V equals the current flow I times the resistance R.

[nashvillebill]

So let me give you an example.  Suppose you have a 12 volt battery.  Now you hook up two resistors in series.  One resistor, which I will call the load, will be 200 ohms.  The other resistor, which I will call the dropping resistor, will be 100 ohms.  Okay, these are in series.  So the total circuit resistance will equal 300 ohms (it's in series, so 100+200).  Since this is connected to the battery, the circuit drops 12 volts across the two series resistors.  The current can now be calculated: V=IR, or 12=I(300).  Divide the 12 by 300 and that gives a current flow of 0.040 amps.  Each resistor, since this is series, has 0.040 amps going through it.  The 200 ohm load resistor will have a voltage drop across it of: V=IR or V= .040 (200) or 8 volts.  The 100 ohm dropping resistor will have a voltage drop across it of V=IR or V= .040 (100) or 4 volts.  (Sanity check, 4 volts for the 100 ohm resistor plus 8 volts for the 200 ohm resistor should equal our 12 volt power supply.)

Now let's keep the same 12 volt power supply, and the same 200 ohm load resistor.  But let's change the dropping resistor to 400 ohms.  Same series circuit, but now the total circuit resistance will be 600 ohms.  The current flow through the circuit will be 0.020 amps.  The voltage drop across the 400 ohm dropping resistor will be 0.020 times 400, or 8 volts.  The voltage drop across the 200 ohm resistor will now be 4 volts.

Go the other way around, and keep a given dropping resistor--say the original 100 ohm resistor--but change the load resistor to 1100 ohms.  The circuit now has 1200 ohms total resistance, so only 0.010 amps will flow through it.  1 volt will drop across the dropping resistor, and 11 volts across the load.

AS you can se, dropping resistors are not always an accurate way to drop voltage. 

[SpareRibs]

Hello,
       I should have known it would turn into a rubics cube. I just wanted to know if a voltage of say 230 volts
going into of 1/2 of a 12AX7 thru a 270K resistor to a 6V6 tube and the resistor steps it down to 80 volts, how do I bring it up to 100 volts ? Is there a simple formula that would give me a value of resistance that could increase voltage 20 volts ?

[Enzo]

Wello now there you gave us more details, so we can start to see a context.   A schematic would be even better though.

You are asking a basic Ohm's Law question now.

Ohm's Law just states the relationship between voltage, current, and resistance in a circuit.  If you know two of those, you can calculate the third with simple arithmetic.

You need to be a bit more precise in your terms.

Look at the Fender Champ 5E1 I have linked here:
http://bmamps.com/Schematics/fender/champ_5e1_schem.pdf

The high voltage comes through the choke and gets to the power tube at 305 volts.   Now moving left through a 22k resistor we arrive at 260v for the preamp tubes.   But we cannot say that a 22k resistor will always make that same difference.  But you have two of the three pieces of information.  You know it is 22k (22,000 ohms)  and you know we went from 305v to 260v.  That is 45v.  22k made a 45v drop.

Ohm's Law says V=IxR, which also means I=V/R   So plug in the numbers.  We have 45v and 22,000 ohms.  45/22000= 0.002A    We usually will write that small current in milliamps rather than whole amps.   So 0.002A is just 2ma.


Ohm's Law then told us if 2ma flows through 22k, we ought to see about 45v across the resistor.

From experience, I generally expect ROUGHLY 1ma per triode in a preamp tube.  The only place our 2ma could be going is to the two triodes of the single 12AX7.

Now we also see that 260v supply feeds each triode through a plate resistor.  And at the other end of each we find 160v.  The plate resistors are 100k, which makes it simple.   In each case, 100v across 100,000 ohms.  I=100/100,000=1ma.  There are two of them so that makes sense with our 2ma flowing through the power supply.

Each tube also has a 1.5k cathode resistor, and that same 1ma flows on through those.  The schematic says 1.5v across them.  Ohm's Law says  1ma (0.001A) through 1500 ohms means:  V=IxR = 0.001x1500, whwich is exactly 1.5v, like it says.


Now the tube will generally stay at about 1ma.  You can worry about more detail later.  So if I wanted to get 290v insead of 260v, then I want to know what resistor to use.  But I also MUST know how much current there will be.  In this example it will be 2ma.   So in my example we drop from 305v to 290v, a 15v drop with 2ma.  Ohm's Law  V=IxR   or R=V/I  so 15v/0.002 = 7500.   I would need to use a 7500 ohm resistor instead of that 22k.


Likewise we had 260v power for each triode with 160v on the plate, and 100k resistor.   If I wanted to get that 160 up to 180 say, then I would see that I need to drop 80v (260-180)  R=V/I = 80/0.001 = 80k.

None of this covers whether it is a wise move or not, but that is how to calculate changes.

There is no formula in all of electronics any simpler than Ohm's Law, but it is THE most important of them, in my mind.

[SpareRibs]

Hello,
      It's the same problem with a Newcomb Pathfinder amp, converting it to a guitar amp. After all the the work
I put into it I only have 80 volts coming out of the 12AX7 to the 6V6. I had just given up on it for a while.

[Enzo]

This is where your terminology confuses me.

On your schematic, I see the 12AX7 has a power supply voltage of 232v, do you have that?   The schematic says you should have 100v on the plate of the 12AX7.  The plate resistor is 270k, but NO voltage goes to the 6V6 from there, only the signal going through the cap C6.

So are you trying to raise the 232v?  Of does that stay and you just want higher plate voltage on the triode.  Are you telling us the 100v at the plate is only 80 on yours?   I myself wouldn;t really sorry that 100v was 80v there.

[g1]

  Yes, I think that is the issue; the schematic shows 100V but he only has 80V.  Like Enzo said, don't worry about it.  In the world of tube guitar amps, that is actually close enough.  The current through that 270K is what sets the voltage there.  It will change a little bit for every individual 12AX7 you try.  If you have some other 12AX7's try them, you may find one that gets you closer to 100V.
You have 232V on one side of the 270K, and 80V on the other.  That is a difference of 152V.  Ohms law says I=V/R, so 152 / 270,000 = .00056, or .56mA  So we know the current through your tube is about 1/2 a milliamp.  But let's solve for their tube in the schematic, 232V on one side of resistor, 100V on the other side.  Difference of 132V.  132 / 270,000 = .00048 or .48mA  Very little difference, both their tube and your tube are running about 1/2 a milliamp and well within the variation we normally see with tubes (+/- 20%).
  And, like Enzo said, that DC voltage does not go to the 6V6, C6 blocks DC but lets AC (signal) through.  If there was DC getting through C6 to the 6V6, it would mean C6 was defective and (leaky).

[SpareRibs]

Hello,
       Thanks to all who commented. I have spliced the tone stack from the LTSpice diagram from R22 to C11
into the original Newcomb diagram between the first half of the 12AX7 plate to the 6AU6 grid. As difficult as
that seemed it seems to have been effective. I was just frantic about the readings because I was not sure
about what would be "good enough'' because 20 volts seemed like a lot to me. Thanks again.

[Enzo]

20 is just a number, it only matters in a context.

Out of 100v, 80v is only 20% off.   For many years, most fender schematics had a note on them telling you any voltages were approximations and might be off as much as 20%.

Your mains voltage coming from the wall outlet moves all over the place.  So if your amp has a B+ of 360v, and your mains voltage is 120v, then every volt the mains changes, your B+ changes 3v.   A 5v change in the mains means a 15v change in B+.