Dropping Resistor Values for Two-Channel Single Ended Amp Using 6V6 or EL84

[mwelch55]

I am planning to build a two-channel single ended amp using either 6V6 or EL84 for power tubes.  I have the schematic drawn, but the power supply dropping resistors are only wild guesses.  The voltages in the schematic are what I want them to be.  How do I determine the dropping resistor values needed to get the voltages on the schematic, other than trial and error?

[shooter]

with math and tube datasheets, PT datasheet, some voodoo, couple guesses
might be some insight from this thread;


 https://el34world.com/Forum/index.php?topic=32182.0

[kagliostro]

As thumb rule you can consider consumption around  3mA 1mA (see later post by Merlin) for each triode in gain configuration and around 10mA for each triode configured as CF

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Give a look here

https://www.ampbooks.com/mobile/vacuum-tubes/

One thing you must consider, an EL84 has a sensitivity of 11mA/ V but a 6V6 has a lower 4.1mA/V  I'll double the MV control, one for the EL84 and a separated other for the 6V6 tube to have more control and obtain different tones

K

[mwelch55]

Shooter,

One of those threads used the PSUD to design a power supply.  I downloaded the program and configured it like the power supply in my schematic.  Here is the screenshot of my results.  The dropping resistors are not even close to my estimations.  I'll have to see if those resistor values give me the voltages in the simulation.

[AlNewman]

I think your schematic needs work.

With the screens and preamp altogether you MAY draw 20 mA.

Over a 10k resistor that's a 200V voltage drop, not a 20V voltage drop.

I think you carried one too many zeroes.

[Merlin]

The dropping resistors are not even close to my estimations.

You're over-estimating the current for the preamp tubes in the simulation. 1mA per triode is more realistic (2mA per bottle)

[mwelch55]

The reason I had two 10K resistors and a 5K resistor in the power supply is that I was looking at the Fender Champ, Harvard and Princeton schematics and they had a 10K between the power tube plate and screen and a 22K before the preamp.  I didn't know how to calculate the resistor values and didn't know about PSUD.  If the PSUD utililty is correct, my values were way off, but as I said in the original post, it was only a wild guess.

I am working on the changes for my schematic.

In PSUD, I adjusted the current load for the different nodes and adjusted the resistor values to get the desired voltage for each node.
Here is a screenshot of PSUD with the results of those changes.

[jjasilli]

If sluckey were here he'd probably say to do it intuitively.  But his intuition is based on tons of hands-on experience.  What I do is use Ohm's Law.  Measure: the voltage drop across the dropping resistor in question; and the actual value of that resistor.  Then use Ohm's Law to find the current (in Amps, not mA, or your decimal will be in the wrong place).  Then use Ohm's Law again:  plug-in the voltage drop you want, and the current, to determine the R value you need.  Use the closest stock R value; or, place resistors in series, or in parallel, to get the exact value.

[tubeswell]

Look at a Fender/Marshall/Vox etc amp schematic for 2 x 6V6s or 2 x EL84s and see what they used.

[mwelch55]

The dropping resistors are not even close to my estimations.

You're over-estimating the current for the preamp tubes in the simulation. 1mA per triode is more realistic (2mA per bottle)

This information helped me get my simulation a little more accurate.  Thank you.

What I do is use Ohm's Law.  Measure: the voltage drop across the dropping resistor in question; and the actual value of that resistor.  Then use Ohm's Law to find the current (in Amps, not mA, or your decimal will be in the wrong place).  Then use Ohm's Law again:  plug-in the voltage drop you want, and the current, to determine the R value you need.  Use the closest stock R value; or, place resistors in series, or in parallel, to get the exact value.

I knew there was a way to do it using ohms law, but I couldn't figure out the logic.  Thank you for that.

Mike