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Thread: Question about voltage dop in a power supply

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    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?

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    It's just ohms law. The voltages look about right to me. What currents are you using thru each resistor?

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    +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.

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    Last edited by Mick Bailey; 01-12-2018 at 02:40 PM.

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    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.

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    Last edited by tpaairman; 01-12-2018 at 06:44 PM.

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    Lifetime Member Enzo's Avatar
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    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.

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    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?

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    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.

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    Lifetime Member Enzo's Avatar
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    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.

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    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?

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    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?

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    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?

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    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

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    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

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    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.

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    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.

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    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?

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    "
    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"


    This is where I get dropped. How do we know the current of the screen grid? Is it another rule of thumb?

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    It's weird, because it WAS working fine.....

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    Look at the schematic voltages.

    The 22k resistor drops 320 to 280vDC. 40v dropped. Ohm's LAw tells us then 40v/22,000 ohms = 2ma. The only thing feeding off that 280v is the two triodes of the 12AX7. So 1ma each for 2ma.

    Now look at the screen node, the 320v. To get there we dropped also 40vDC (360v - 320v = 40v), and that was through the 10k resistor. Ohm's says 40v/10,000 ohms = 4ma.

    Now recall the 12AX7 had 2ma, and those milliamps flow through the 10k along with the screen current. SO of the 4ma through the 10k, 2ma of them were for the preamp tube. And that leaves the remaining 2ma for the screen. Screen is the only other thing drawing current through the 10k.

    SO not a rule of thumb, but calculated from the voltages provided on the schematic.


    But having said that, 2ma is a good average number to use in the absence of better numbers. So in this case, the power tube cathode current includes that 2ma along with whatever the plate is handling. In larger amps like a Twin Reverb, I usually just assume about 5ma per screen just to have a number. In that amp I could easily measure voltage drop across the screen resistor, but I don't care enough.

    MY way of looking at it is this: IF I calculate some dissipation for my power tubes using the cathode current, I know it includes some screen current. But if I ignore that, the fact that the true plate current is really a couple milliamps lower, it just gives me a little pad, my amp runs a hair cooler. This is like how my wife sets her car clock ten minutes fast to give herself a pad against being late.

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    I understand that everyone is trying to help, and I'm really not trying to be rude here, but this has become very frustrating because it's almost as if nobody has read the question all the way through.

    I KNOW it's ohms law. I KNOW it's volts divided by amps. What I can't seem to get anyone to see is that in order to do the ohms law calculation, aside from the voltage value, I need a current value. We have the voltage values all day long on this schematic, which was only an example so everyone could see what resistors I was talking about. And I know that on this schematic, we can work backwards to get the current value. But I needed to know the other way around. I'm designing an amp. I know I need a dropping resistor. What value resistor do I use?

    That aside, I have actually been able to piece together the answer to my question which was if I'm starting with a blank sheet of paper, how would I calculate the value of those resistors. I know that Fender, Marshall, and all the others didn't just throw darts and voltage and current values, they started from somewhere. Well, turns out the answer is called a load line calculation. I'm still getting my head around it, but you have to go to the data sheet, look at the graph that shows supply voltage vs plate current based on the grid input, decide what voltage and current values you want to use (that's where I'm still a bit hazy) and draw a line between the two. Then from that you can decide what plate voltage you want to use, which then gives you the difference in the supply voltage vs the plate voltage, and since you already know the current, you can figure the load resistor value. And since we now know the supply voltage we want, we know how much we want to drop it from the last voltage on the B+ rail, and using one thing that has been stated, add up all the current from that and all other downline stages.

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    tp, listen, I am glad you are where you are on this, but right above my last post was someone who still didn;t quite KNOW those things. So I want him to be as up to speed as the rest of us. My post was aiming to help Randall get it too.

    Fender took their circuits right out of the RCA book. RCA provided those basic circuits along with the typical resistors to use to make designing their tubes into things easier for engineers. If RCA used load lines, so be it, but you can grab that 100k plate and 1.5k cathode resistor right from the chart and have an excellent starting point. I would not be so sure Fender ever ran a load line for a 12AX7.

    To calculate anything you need two of the variables. You have a target current you want to design around? What power supply do you intend to use? Then the resistor value is simple. And that is why the tubes have families of curves instead of just one curve. I have to consider that the input stage sees up to a volt, but the next stage might see 10-20v or more of signal. That affects the bias point I want for my tube.

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    Quote Originally Posted by tpaairman View Post
    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.
    If you were designing the amp from scratch you would know the currents and voltages required by each section of the amp.

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    Quote Originally Posted by the fatch View Post
    If you were designing the amp from scratch you would know the currents and voltages required by each section of the amp.
    (posted incomplete comment, I don't know how) As I was saying, the current in each dropper resistor is the sum of the current required by the stage it is supplying and the sum of the currents in the previous stages. Knowing the current flowing in each dropper and the voltage to be dropped across it, finding the required resistance is a simple application of Ohm's Law.

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    Tpaairman said " I need a current value. We have the voltage values all day long on this schematic, which was only an example so everyone could see what resistors I was talking about. And I know that on this schematic, we can work backwards to get the current value. But I needed to know the other way around. I'm designing an amp. I know I need a dropping resistor. What value resistor do I use?"

    I may not get your question, but when I think about designing an amp or stage I think about what tube(s) to use, considering gain and impedance and linearity, and that choice tells me how much current I'll need. Choose plate resistors for the gain and distortion you want, with higher values less distortion and more gain. I then decide what voltage I want to run the tube at, considering voltage swing needed and the linearity of the plate curves, and that sets the B+. Adjusting the cathode resistor size gets the tubes into the bias I want. The filter effect of each RC stage lowers the power supply noise, too, and that's a consideration towards big caps, but big caps sag less, if that's important. In the push/pull part of the amp, the effect of PS ripple is somewhat lessened, but the preamp triodes want it quiet, so I try to find a balance of high enough voltage and little noise. Sorry if this is too trivial.

    Dan

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    Quote Originally Posted by tpaairman View Post
    I understand that everyone is trying to help, and I'm really not trying to be rude here, but this has become very frustrating because it's almost as if nobody has read the question all the way through.
    We HAVE read your question and answered it, including "you should study a little more":
    Now you are frustrated because you donīt "get it" .... but you didnīt show due diligence studying.

    This included actually downloading and reading 12AX7 datasheets, where the typical design examples are shown.

    Hereīs the GE datasheet "example chart".

    The typical 100k - 1k5 - 250V combo is in the RCA one but this is what I found and will use:
    Click image for larger version. 

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    I marked one choice, the one closest to typical Fender use, and which you might have picked for your design.

    I chose 180V +V .
    Why?
    Because I liked it, all 3 values work but in general itīs safer to choose a "midrange" option and not go to extremes, unless you have a very definite reason to do otherwise.

    This example uses 100k plate resistor, a 1100 ohms cathode one, and is happy driving a 240k load.

    It uses an easy to achieve +180V rail and can put out some 20V RMS "clean" .
    Actually 5% distortion but by Tube standards that passes as clean.

    I KNOW it's ohms law. I KNOW it's volts divided by amps. What I can't seem to get anyone to see is that in order to do the ohms law calculation, aside from the voltage value, I need a current value.
    *ITīS-SHOWN-IN-THE-DATASHEETS*

    ...... where? .......... I donīt see it!!!

    You need to apply Ohmīs Law:

    1) rail voltage is +180V

    2) usually plate voltage is about half that, so signal has the same "room" going upwards than going downwards, for symmetrical clipping and maximum Vout.
    No need to spell that out, itīs a common sense choice.

    So plate voltage, even if they donīt *specifically* show it, must be around +90V .

    So you are dropping some 90V across the plate resistor.

    Applying Ohmīs Law: 90V/100000 ohms=0.9mA ...... which is very close to 1mA estimated by Enzo, simply based on his ample experience.
    But now you know where it comes from.

    As of dropping resistors: suppose you have a, say, 300V supply, and want to tame that down to 180V .
    R=V/I so (300-180)V/0.9mA=133k .

    This is just an example, in a full design you choose every single gain stage from the chart as you wish, calculate each triode current needs, add them up as needed in different nodes (you donīt feed them all straight from main supply but in general make a string of RC filters to reduce hum and possible instability) and calculate each section dropping resistor based on available voltage - desired voltage and current needs.

    In fact I suggest as "homework" you choose any Amplifier schematic you like (Iīd start with a Champ just to make it quicker, then it applies to any other) , download the RCA datasheet which is what Leo actually used, calculate all currents at different spots, repeat Fender Voltage choices and calculate the dropping resistors, which was your original doubt.

    I would NOT be surprised if what you calculate roughly matches what Leo used


    We have the voltage values all day long on this schematic, which was only an example so everyone could see what resistors I was talking about. And I know that on this schematic, we can work backwards to get the current value. But I needed to know the other way around. I'm designing an amp. I know I need a dropping resistor. What value resistor do I use?

    That aside, I have actually been able to piece together the answer to my question which was if I'm starting with a blank sheet of paper, how would I calculate the value of those resistors. I know that Fender, Marshall, and all the others didn't just throw darts and voltage and current values, they started from somewhere. Well, turns out the answer is called a load line calculation. I'm still getting my head around it, but you have to go to the data sheet, look at the graph that shows supply voltage vs plate current based on the grid input, decide what voltage and current values you want to use (that's where I'm still a bit hazy) and draw a line between the two. Then from that you can decide what plate voltage you want to use, which then gives you the difference in the supply voltage vs the plate voltage, and since you already know the current, you can figure the load resistor value. And since we now know the supply voltage we want, we know how much we want to drop it from the last voltage on the B+ rail, and using one thing that has been stated, add up all the current from that and all other downline stages.
    See above explanation

    And yes, the next higher stage is to dismiss the example chart and cook your own, straight from the graphic curves ... but as-is the "simpler" method has served well to tons of Designers.

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    Last edited by J M Fahey; 05-20-2018 at 12:00 PM.
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