How to calculate voltage drop

[MBD115]

Hi guys. This is my first post.

My question is how can I calculate the voltage drop across a simple series resistor. I've read the answer a bunch of times but I just can't seem to understand how to do it.

Lets say I have a 12AX7 tube that has 121V on it. The main supply voltage is 250V. The inline series resistor is a 200k. The voltage drop across that resistor is 129V (250-121=129). How do I figure what the voltage would be if I were to change that 200k resistor to say a 100k, or a 150k, etc. This should be possible to calculate but I don't seem to be able to understand how. lol

I see the Trainwreck amp in this same place has 267V on B+5 and a 100k resistor to the preamp tube. Is there a way to figure how much the voltage would drop across that 100k resistor?

Thanks for the help

[Zippy]

V = RI

[stubbyfex]

Hello! You need to google "Ohm's Law". Mr. Ohm figured out a whole bunch of math formulas that if you know two things, like Voltage, Current, Watts or Ohms, you can find a third. Here is a wheel. You plug in what you know, and can find the third "piece". Then you can plug in the third piece and find the fourth. Using your example: I=E/R, so 250 volts /200000 ohms = .00125 amps. Now from those two pieces you know the third.
Oh, here is a site that will show you how to calculate the voltage drop across resistors.
http://www.bcae1.com/resistrs.htm

[martin manning]

That's fine for the example case where the voltage drop across the plate resistor and the resistance are both known.

The second part of the question was: what happens if you change the value of the resistor? You know the current for the example case, but with a different value for the plate load resistor, the idle current will change. To solve this you need to plot the load line for the new resistor value on the plate characteristics for the tube, and find the new bias point. The new plate voltage and idle current are then known.

MPM

[MBD115]

Thanks guys for the replys

I still can't get it. lol

I know if I replace that 200k with a 100k that the voltage will be 165volts because I've did that. 140k gives me about 140volts on my tube because I've tried that one too.

But I can't seem to get it through my head how to calculate all that without doing resistor swapping....

Ive read that web page 3 times over the past couple weeks and it ain't helping. lol I still don't get it.

[MBD115]

Sorry Martin, I was typing when you posted.lol

I pretty noob ok

All I know is
200k = 121 V
140k = 140 V
100k = 165 V

I've got a 100k and a 100k variable in there right now but I wish I could do this in a better fashion like with my calculator then know what will happen.

Thanks for posting

[M Fowler]

MBD115,

I don't know how to calculate that what I use is a chart. The Trainwreck has three gain stages so depending on the gain, B+180 volts on 12AX7:

Plate resistor/Cathode/Gain
100k 1.8k 40
100k 2k 47
100k 2.2k 52
220k 3k 53
220k 3.5k 59
220k 3.9k 63
470k 5.8k 62
470k 6.7k 66
470k 7.4k 68

The relationship between the plate and the cathode resistors has to be kept constant to provide the bias to the tube stage.

I use my books and references. Some one smarter than I can help both of us here.

Mark

[Ears]

Your question is a little more involved than ohms law on its own. Changing the load resistor changes the conditions on the tube and the tube will behave by changing the current through it. You need to understand the concept of the load line to find your answer.
There are many books articles etc on this . The NEETS material is as good as any http://www.tpub.com/neets/ or http://www.tpub.com/content/neets/14178/

[MBD115]

Thank you all again for trying to help me. I need it. lol

Tell me if this is right....

I need to make a graph. Up one side would be the voltage and across the bottom would be the resistance then plot on my graph the known values I have measured.

I would mark my graph at 250V on the 0k line because 0k equals no-resistance, or a solid wire, and I know if I put a solid wire in there the voltage will be 250k, my line voltage.

I would mark my graph at 121V on the 200k line because I know a 200k resistor will give me that amount.

I would then mark my graph at 165V on the 100k line, then again at 140V on the 140k line because I know this is what will happen.

When I try to connect these points I must create an 'arc' because a straight line will not connect them all. This 'arc' I create on my graph should also tell me that it will never touch the bottom line because that line goes on to 'infinite k". Because any resistance at all in there will give me some voltage and to equal "0Volts" the connection must be broken, ie: infinite ohms.

After I have created my graph I could then know what a new resistor value would give me in voltage and the ohms law can be used to calculate the current.

I think this is what I ment to say.
That is hard to explain

Is my thinking right on this?????

[Ears]

server shutdown error

[Ears]

server shutdown error

[Ears]

For common cathode voltage amps:
You do need a graph yes.
But it is plate current vs plate voltage.

Plate current (y axis) vs plate voltage (x axis)
At Current =0 mark in your supply voltage. (i.e. on the x axis)
At Voltage = 0 Divide the supply voltage by the resistor value and mark the resultant current on the y axis (i.e when voltage is zero).
Draw a straight line between the points. This (load) line has a slope of -1/Rp where Rp is value of your plate resistor. Increase the value of plate resistor and the slope becomes less steep, lower the value of the resistor and the slope becomes steeper.

For a case wher plate load resistor is very much higher than cathode resistor the tube's voltage and current will basically settle somewhere on that line. Exactly where is determined by the bias on the tube's grid. So the load line is usually drawn on the tubes plate characteristics from the data sheets. Draw a few lines in, try using the same bias value and see how the plate current and voltage change for varying loads.
Actually, valve wizard has good article on the common cathode, see recent garage talk post for the link.

[martin manning]

This is the graph, with the three cases plotted. I picked a cathode resistor of 1.7K because that matched the Va (plate voltage) of the first case, then held it constant. The intersection of the cathode load line (purple) with the anode load lines defines the idle points. Note that the plate voltages are pretty close to the experimental values. Remember that the plot shows the voltage from anode to cathode, so the anode voltage to ground (as measured) will be higher than the plotted value by the cathode voltage Vk.

I used the Excel tool here:

https://tubeamparchive.com/viewtopic.php?t=7463

MPM

Edit: Sorry, My update attempt failed first time... the file is there now.

[MBD115]

Thanks again guys for all the replies. I do need all the help I can get. lol

But....... you all are talking WAY way way over my head!!!!! I'm a doob and a noob. You got to remember than. haha

I'm going to do some more reading because I have a lot to learn, you all just showed me that. lol

Thanks for the help and the replies. I can't absorb any of it because I am too dumb and I do not know enough about what you are talking about. I do know what grid and plate means. I'm a little lost when you mention bias on a 12AX7 tube, but I think I get the idea that the resistor on the line to ground, in relationship to the load voltage and the voltage(?) of the signal line establishes the bias of the tube. (?)

I'm not sure at this point in my life just how deep I want to get involved with understanding all this because I was just trying to crank up the gain on my tube amp a little and "kinda wing it" and hopefully get lucky. haha

You guys have so much knowledge and I am so much of a doob I just can't keep up. lol

Thanks to all. Take pride in your abilities and your know how.
I know I'm stupid and this is all above my brains capacity.
Thanks again to all. I need to find a big rock to crawl under and hide now. lol

[Stanz]

If I remember correctly, the way to calculate the voltages at each point on a string goes something like this:


Voltage In (Vs)
R1 R2 R3 R4 R5


@ Voltage In - Vs * (R1+R2+R3+R4+R5) / (R1+R2+R3+R4+R5)
(in other words, the full voltage)

@ R1/R2 node - Vs * (R2+R3+R4+R5) / (R1+R2+R3+R4+R5)

@ R2/R3 node - Vs * (R3+R4+R5) / (R1+R2+R3+R4+R5)

@ R3/R4 node - Vs * (R4+R5) / (R1+R2+R3+R4+R5)

@ R4/R5 node - Vs * (R5) / (R1+R2+R3+R4+R5)


Here is an excel sheet that you can throw numbers in to get a quick look at what happens by changing a resistor.