I've been teaching myself electronics for a while now and am having a hard time understanding something. Voltage drop. When you want to drop voltage from one value down to a lesser value and are choosing the resistor to use, I've often heard it said something like this: "If we know the current and the voltage we want across, then it's just ohms law, IR=V". What I don't understand is how can you know the current because as soon as you place a resistor in the circuit it changes the current. V/R= I. The only thing I can think of is if the current source has such high internal resistance that adding any resistance would have little effect concerning the V/R=I formula. I hope i'm making sense. thanks A good example would be using screen resistors to drop some plate voltage. If I have say 450 B+ and want to drop it to 420, how would I go about sourcing a resistor? Or perhaps a better example would be choosing the right sized cathode resistor on a preamp section to make the cathode some positive value with respect to the grid, thereby making the grid negative with respect to the cathode.

I suspect the missing concept here is the "voltage divider." Yes, if you just strap lower value resistors across a supply it will drop the voltage, by "pulling down" the supply. What is happening is that the internal resistance of the supply is acting with the load resistor to form a voltage divider. Visualize a perfect supply with 0R internal, then picture its internal resistance as a seperate R following it, Now when you add your load resistor you can see that the perfect V supply is now divided between the load and internal Rs. The two examples you give are fairly dynamic and more complex than this visualization. Let's take the easier one, biasing up a 12AX7. When you look at the data sheet you will see a set of curves with Plate V on one axis and Plate I on another with each curve for a different bias voltage. Get familiar with these guys. The old military training books and the old ARRL Handbooks are gold for learning to design tube circuits. http://www.tubebooks.org/technical_books_online.htm You'll see it's a bit of a moving target. You'll find instruction on drawing a load line and developing the circuit you want. You can also just take what is known to work (Plate R 82K-330K and Cathode R between 820-5K) and just tweak from there. In your other example, the screen draws so much less current than is available to the plate circuit that you can use whatever combinations of Rs and zeners you like to get exactly your desired supply. Then again, doing some math is handy. Enjoy the journey.

The voltage divider concept makes sense. Within a circuit though, how would you ever know the voltage supply's internal resistance? Or just take any random point in an amps circuit. One can measure voltages and currents but how do you find out on paper what varying any resistor values will do? It seems you would need to know the dynamics of that circuit, whether internal voltage supply resistances or whether the current thru the circuit is constant by way of an ideal current source. I think maybe what i'm bumping into is the difference between a passive current source and an ideal current source. Ie...Some circuit's current is effected by varying the load resistance and some are not. I'm assuming that tubes are close to ideal current sources?

I would suggest picking up at least Merlin's preamp book if you're edging toward design aspects in your builds. The first chapter the common cathode triode gain stage (a book in itself) covers in detail using cathode and anode load lines to select a bias point and the potential outcomes of the selection and subsequent stages. You can download that chapter for free on his site. http://www.freewebs.com/valvewizard1/gainstage.html I haven't picked up his power supply design book yet, but will definitely buy it down the road.

I actually own both his books and know the chapter you're talking about. That is actually what caused my confusion. This part in particular. [FONT=TimesNewRoman,Bold][FONT=TimesNewRoman,Bold]Choosing the cathode bias resistor:[/FONT][/FONT]For now, we will use the traditional 100kΩ anode load. After drawing a load line we can then choose a bias voltage based on the degree of headroom and clipping we want from the stage, as discussed earlier. If this is the input stage of an amplifier we might choose to centre bias, or go a little warmer. In this case a bias of 1.5V is used. The cathode resistor may now be chosen. Looking at the load line in fig. 1.13 it can be seen that a bias of −1.5V causes 0.9mA of quiescent anode current. Of course, this current flows right through the valve, out of the cathode and down through the cathode resistor. The grid voltage is fixed at zero, but we want it to be 1.5V more negative than the cathode, so we can place the cathode at +1.5V instead. We now know the desired voltage across, and current through, the cathode resistor so we can apply Ohms law to find its value: R = V / I = 1.5 / 0.0009 = 1667 Where my confusion started and the crux of my question is; He says that .9ma flows thru the tube with a -1.5 bias. He wants +1.5v across the cathode resistor so simply divides V/I = R. My confusion was thinking that as soon as you found R and placed it in the circuit it would throw off the value of I. A sort of endlessly chasing one's tail due to not having 2 constants in the equation. I think where I went wrong was treating the tube as a passive current source where R's value would effect the current. However if the tube is a ideal current source and keeps the current constant I can see how ohms law can be applied without having 2 varying values or "one effecting the other".

Actually to get the 0.9 mA he used the plate characteristics graph (intersection of -1.5V curve and anode load line), I believe. On the next page you see that the bias is neither exactly -1.5 or 0.9mA because of the resistor value used. These values are used to solve for the cathode resistor because it is connected to ground. I don't understand your thinking behind an "ideal current source." Changing the cathode resistor definitely will have an effect on the quiescent current. Hope this helps.

Ideal current source meaning it doesn't change with respect to the load. Doesn't get "bogged down" if you will. I've heard it also described as an active current source (transistor, tube, etc) vs. passive current source (voltage across a resistor). While obviously I'm no expert on this stuff, I don't believe the cathode resistor is effecting the current in the way it would in a passive current source setup. My take on it is he is simply looking for a resistor to match what he needs to get -1.5v on the grid. That's where he chooses to bias on his 100k load line and knows with the grid -1.5v with respect to the cathode resistor that he'll get his intended .9ma cathode current. The cathode resistor doesn't seem to being "bogging down" the current as it would in a passive current source. It's job is simply to present 1.5v potential above the grid. For example. Let's suppose the grid wasn't at ground potential but instead at say 4v. That means the voltage across the cathode resistor would need to be 4+1.5 or 5.5v. The grid would still be -1.5v with respect to the cathode and the current would still be .9ma however the cathode resistors value would need to be 6111 ohms to get 5.5v across it. That's certainly different than our original value of 1667 ohms. However, nothing has changed as far as the tube knows. That's why I don't think the cathode resistor is acting like it would with a passive current source. The ACTUAL resistance doesn't seem to matter. It's job seems just to be make sure the cathode voltage stays +1.5v above the grid. Other than doing that, I don't see how it's affecting the current purely due to it's direct resistance. It seems rather to just control the grid and it's the grid that is determining the amount of current that flows. By the way, I'm not saying I'm right. I'm just trying to explain how I see it in relation to your response. I hope someone will either confim or correct this as I'm certainly no authority on it.

The voltage divider is formed by the plate R, tube and the cathode R. Yes, the I changes, but the curves show you what to expect. So once you've decided on any one value, you can go to the curves, know what the tube will do (more or less) and then calculate the other. Start with the Plate R, get the tube where you want it and then use the new I to select the K R.