Results 1 to 16 of 16
Like Tree3Likes
  • 1 Post By Mick Bailey
  • 1 Post By Enzo
  • 1 Post By J M Fahey

Thread: Question about voltage dop in a power supply

  1. #1
    Junior Member
    Join Date
    Jan 2018
    Posts
    9

    Question about voltage dop in a power supply

    Greetings everyone. First of all I didn't see an area to introduce yourself (forgive me if I missed it). My name is Michael, and I've worked on antique radios for quite some time, but am moving to the area of amplifiers.

    For my question, I'm using the schematic below as an example. The B+ starts out at 360 volts. As you go down the line, there is a 10K resistor, and it shows the voltage drops to 320. Then a 22K and we're at 280 volts and so on. What I can't figure out is how does it come out to those voltages? When I try to calculate the voltage drops, I get completely different figures, and that's accounting for the plate and cathode resistors.

    Or to turn he question around, if I was designing an amp, how would I calculate what resistors I would need in the B+ to get to the plate voltage I need?

    fender6g10.gif

  2. #2
    "Thermionic Apocalypse" -JT nickb's Avatar
    Join Date
    Dec 2009
    Location
    Devon, UK
    Posts
    2,478
    It's just ohms law. The voltages look about right to me. What currents are you using thru each resistor?
    Experience is something you get, just after you really needed it.

  3. #3
    Supporting Member
    Join Date
    Dec 2009
    Location
    UK
    Posts
    2,975
    +1. The voltages are pretty close.

    For example, take a look at the preamp sections - they both show 1.4v dropped across a 1500 Ohm resistor. All of the current in a tube flows through the cathode, so it's a simple calculation to determine how much idle current is draw through the 12AX7; 1.4/1500*2= 1.867 mA. That is the current flowing through the 22K dropper resistor, so the voltage drop across that resistor is .0018*22,000=39.6v. The schematic shows a drop of 320-280=40v.

    Similarly the voltage drops across the 100k plate load resistors are OK; 1.4/1500*100,000=93v. The schematic shows 90v.

    It could be that the schematic shows actual voltages measured in an amp, rather than calculated voltages. Then you have component tolerances to take into account. That's why there's a note to say all voltages +/-20%.

    A tube amp is an imprecise instrument.

    Duncan's PSU calculator is a useful tool - pretty bang-on to real-world voltages.
    Last edited by Mick Bailey; 01-12-2018 at 03:40 PM.
    Tom Phillips likes this.

  4. #4
    Junior Member
    Join Date
    Jan 2018
    Posts
    9
    That answers my question. I was looking at this more from a voltage divider standpoint, and using that formula, so I was coming up way off.
    Last edited by tpaairman; 01-12-2018 at 07:44 PM.

  5. #5
    Lifetime Member Enzo's Avatar
    Join Date
    May 2006
    Location
    Lansing, Michigan, USA
    Posts
    28,926
    Voltage division would require a return to ground somewhere, but the B+ never connects to ground.

    For 12AX7s, in general, for a very crude estimate, I use 1ma per triode. In this case, we get 1.8ma for two of them, or 0.9ma. Which is close enough to my 1ma for rock and roll.
    Education is what you're left with after you have forgotten what you have learned.

  6. #6
    Old Timer
    Join Date
    May 2006
    Location
    Cheshire, UK
    Posts
    1,633
    Quote Originally Posted by tpaairman View Post
    That answers my question. I was looking at this more from a voltage divider standpoint, and using that formula, so I was coming up way off.
    The voltage divider equation only works if there's no current taken from the junction of the two resistors and there's always current taken from each node of the power supply chain so you have to use Ohms law. For example the junction of the 10k and 22k has to supply the 6V6 screen current. Can you calculate the screen current?

  7. #7
    Junior Member
    Join Date
    Jan 2018
    Posts
    9
    Quote Originally Posted by Dave H View Post
    Can you calculate the screen current?
    I'd say 4 mA based on the 40 volt drop across the 10K resistor.

    But here's where my hangup is. In this example, we're going with the cathode current to work back from, which makes sense. And as Enzo said, we can use 1mA as a rule of thumb for the 12AX7. However, if I was designing my own amp from scratch, where would I start with a value to work with? Or put another way, when the folks at Fender designed this amp, how would they have come up with those dropping resistor values?

    I apologize if I'm coming off as a pain in the rear on this one. It's just that I really want to learn so I can pick up the data sheet and know how to do it.

  8. #8
    Lifetime Member Enzo's Avatar
    Join Date
    May 2006
    Location
    Lansing, Michigan, USA
    Posts
    28,926
    fender made a zillion models of amps, and none were designed in a vacuum. The front end of most Fender amps looks pretty much the same. They already know what sort of current draw the stages will have. They might decide for whatever reason that they want a B+ of 260v in one amp, but 300v makes sense in another. An amp with a pair of 6V6 will likely have a lower B+ at the power stage than a 6L6 amp. SO we might see 380v in one and 480v in the other. If we want 280v for the input stages, we calculate a dropping resistor based on what voltage was there to start with. In other words the identical preamp circuit would need different dropping resistor values in a Deluxe type than a Twin Reverb type.

    Get yourself a copy of the RCA tube manual, the RC-30 edition is offered in reprint. And it can be downloaded from online somewhere too. Look at a typical Fender input stage like on your schematic. Note the cathode resistors of the 12AX7 are 1500 ohms, and the plate loads are 100k? The numbers are right out of the RCA manual. In the rear sections of the book are suggested circuit values for different tubes at different supply voltages.
    J M Fahey likes this.
    Education is what you're left with after you have forgotten what you have learned.

  9. #9
    Junior Member
    Join Date
    Jan 2018
    Posts
    9
    Quote Originally Posted by Enzo View Post
    Note the cathode resistors of the 12AX7 are 1500 ohms, and the plate loads are 100k?
    For those two, yes, it's in the book. How about the 22K or the 10K, and how would I calculate that those values drop the voltage to 320 volts and 280 volts respectively? And then looking the other way, while 100K is listed in the book, how did they calculate that it then drops the voltage to 109 volts?

  10. #10
    Lifetime Member Enzo's Avatar
    Join Date
    May 2006
    Location
    Lansing, Michigan, USA
    Posts
    28,926
    The only current through the 22k resistor is the two 12AX7 triodes. My 1ma each rule means 2ma through the 22k. They had 320v and 280v? So 40v. What resistor will drop 40v with 2ma running through it? Ohms Law says 20k. 20k is not a standard resistor value, while 22k is, so use 22k, close enough. As it turns out my 2ma is really 1.8ma , in which case Mr. Ohm says 22.222k for 40v. Again, 22k is the closest standard value.. They COULD have used precision resistors and gotten values like 20.248k or whatever, but that would be expensive, so 22k is more than close enough. In fact, they used 20% tolerance resistors back then.

    The 10k? Same deal. The same 2ma from the 12AX7 goes through it too, plus the power tube screen grid current - about 2ma. That makes 4ma. What resistor will drop 40v at 4ma current? Ohm's Law says 10k.

    I hope that was a typo, you mean 190v on the plate? Ohm's Law again. You have 280v of B+ running through a 100k resistor making 190v there. 90v dropped. What current drops 90v across 100k of resistance? 0.9ma Very close to my rule of thumb 1ma, and exactly what the schematic shows for this circuit.

    Your power tube has a 470 ohm cathode resistor dropping 20v. Solving for current - 42ma. Subtract a couple for screen current and we have 40ma. Note the primary of the output transformer drops 10v. 40ma dropping 10v, means the transformer must have a winding resistance of 250 ohms. How close is that to reality?
    Education is what you're left with after you have forgotten what you have learned.

  11. #11
    Junior Member
    Join Date
    Jan 2018
    Posts
    9
    Yes, the 109 volts was a typo.

    On to the rest of this. I get every thing you just posted. The problem is you are basing all of that on the fact that there should be 1 mA (or .9 mA) at the cathode, which is great, but neither the manual nor the data sheets show that, at least anywhere that I can find. That's the part I'm missing. If I pick a given tube, and I crack open the manual, how do I determine what current value I should be looking for at the cathode, so I can then use that current value in my ohms law equation to determine what resistor I need to use?

  12. #12
    Supporting Member
    Join Date
    Dec 2009
    Location
    UK
    Posts
    2,975
    You're not looking in the right place. Just look for 'static anode characteristics' or 'average plate characteristics' in the datasheets and that gives the Va against Ia. Most tube manufacturers give sample circuits and these show tables for typical resistor values against plate voltage.

    1mA is an approximate figure based on the typical voltages found in tube amps, rule-of-thumb, but not based on nothing. The actual value may be a little more, or a little less. Ultimately it doesn't matter too much and you can see the latitude from the characteristics charts.

    More information here - see Fig. 1.3/p.5

    www.valvewizard.co.uk/Common_Gain_Stage.pdf

  13. #13
    Old Timer
    Join Date
    May 2006
    Location
    Cheshire, UK
    Posts
    1,633
    Quote Originally Posted by tpaairman View Post
    I'd say 4 mA based on the 40 volt drop across the 10K resistor
    You need to subtract the current through the 22k from that to calculate the screen current. It's 4-1.8 mA

  14. #14
    Junior Member
    Join Date
    Jan 2018
    Posts
    9
    Quote Originally Posted by Mick Bailey View Post
    You're not looking in the right place. Just look for 'static anode characteristics' or 'average plate characteristics' in the datasheets and that gives the Va against Ia.
    Thanks You. That's what I needed to know.

  15. #15
    Old Timer J M Fahey's Avatar
    Join Date
    Oct 2007
    Location
    Buenos Aires, Argentina
    Posts
    9,443
    Quote Originally Posted by tpaairman View Post
    how do I determine what current value I should be looking for at the cathode, so I can then use that current value in my ohms law equation to determine what resistor I need to use?
    Trivial answer: "if you had studied Electronic Engineering they would have taught you that" , which is not an impolite answer as it looks but simply states that such things *can* be calculated by users, no Magic involved at all and that is the main point.

    That said, Tube Manbufacturers want to sell their stuff , as much of it as possible, so they make it easy for you, as easy as possible.

    Basically, they split possible users in 3 areas:

    a) fully trained Techs and Engineers, who want to custom design something very specific , starting with a blank page, pencil and a rubber (yes, design is interactive, you start it and then correct it as much as you wish so the rubber is essential )
    Those use the various curves and graphs abbundant in any tube datasheet.

    b) trained Techs and Engineers who don't want or need to start from ZERO , just need a properly working and trusty "generic/cookie cutter" gain stage, they will use their brain or ideas in other stuff, such as EQ, general amp layout, features, etc. .
    For those Manufacturers offer a wide assortment of gain stages, for different supply voltages, gain value, drive capability, ets.
    As wide as often showing stages fed from meager 90 to 160V DC to hair rising 250/300/350V .
    Usually one of those general purpose ones will be fine, they are guaranteed bto work, and in any case you are free to tweak them later.
    Not surprisingly, that is the most popular option, including none less than Leo Fender, go figura.

    The typical "Fender Gain Stage": a 12AX7 triode with about 50/60X gain, 100k plate resistor, 1k5 cathode bias one, fed around 250V +V which you see repeated ad nauseam all over the place, and by most other designers/manufacturers, can be found exactly like that in the RCA/GE/Philips/Tungsram/etc. datasheets from the 50īs

    c) for those hobbyists (and maybe a couple Manufacturers) who "just want a finished design that works" , some datasheet, and definitely Factory "Application Notes" and even full books , suggest fully made and tested projects, with all values fully worked out.

    Very common even today, and with SS stuff
    Just check 99% of 15W entry level Guitar amps, the class individually most sold in the World, most/all use TDA2030 chipamp, exactly the datasheet application example, some with a minor feedback tweak (involves adding just 2 resistors) to make them more Guitar friendly.
    Since it works so well, hardly anybody would want to design his own from ground up, go figure.
    g1 likes this.
    Juan Manuel Fahey

  16. #16
    Senior Member SoulFetish's Avatar
    Join Date
    Jan 2016
    Location
    Massachusetts
    Posts
    492
    Quote Originally Posted by Dave H View Post
    The voltage divider equation only works if there's no current taken from the junction of the two resistors and there's always current taken from each node of the power supply chain so you have to use Ohms law. For example the junction of the 10k and 22k has to supply the 6V6 screen current. Can you calculate the screen current?
    This would essentially equate to a parallel Load across R2 in a voltage divider, right?
    If I have a 50% chance of guessing the right answer, I guess wrong 80% of the time.

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Similar Threads

  1. Power Supply Question
    By Paleo Pete in forum Guitar Effects
    Replies: 8
    Last Post: 11-19-2016, 04:58 PM
  2. Dual positive voltage supply question
    By Austin in forum Theory & Design
    Replies: 8
    Last Post: 06-21-2012, 07:18 AM
  3. Replies: 3
    Last Post: 06-01-2010, 12:09 PM
  4. Reducing Power Supply voltage
    By graphitiac in forum Music Electronics
    Replies: 6
    Last Post: 06-20-2008, 02:11 PM
  5. high power supply voltage
    By Hubie in forum Debugging Your Build
    Replies: 11
    Last Post: 05-04-2007, 12:30 AM

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •