Plate resistor voltage drop

[ninjadave7]

Hey guys,

I've been tinkering with amps and effects pedals for a while, but haven't put as much time into understanding some of the theory behind it all. My question is: how does a plate resistor drop voltage? I know how to calculate voltage drop via voltage divider using Ohm's law, but I don't see a voltage divider on a plate resistor, if that makes sense.

On my Goodsell, or a Matchless Spitfire, I see 329v from the PS, thru a 220k resistor, then ~150v at the plate. How?

Thanks for indulging my ignorance and helping my understanding of electronics!

[jmaf]

The tube acts as a water faucet, when you drive the grid positive it's like opening a faucet, when you drive it negative it's like closing it. When you open it, electrons flow. The plate resistor is in series with the plate. When those electrons you've just let pass go out the tube, they must go through the plate resistor. Whenever there is current through a resistor, there is voltage.

When you increase the voltage at the grid, the voltage at the plate drops by an amount determined by the stage gain. Without the plate resistor, you wouldn't have a signal at the plate, it is precisely that resistor which makes a voltage appear there when a current goes through. That voltage is your signal, inverted. Hope this helps.

[Jazz P Bass]

Another way to describe it is the plate resistor is a "load resistor".
The plate must have a load to perform meaningful work.

[ninjadave7]

Ok, the faucet analogy makes sense, and that voltage is created because those electrons (current) have to cross resistance, but I'm still not clear why/how the voltage drops 179v from one side of the 220k resistor to the other. Does that make sense? The answer might be in your previous response, but I've read it several times carefully and I'm still not clear.

[g1]

"I know how to calculate voltage drop via voltage divider using Ohm's law". You understand that the voltage drop is due to current flow through the resistors? Think of the tube as the other resistor in the voltage divider. There is a voltage drop across the plate resistor, and across the tube. (I will ignore the cathode resistor to keep it simple). In your example you have 329V supply. You have 150V at the plate so 329-150=179V across the plate resistor. I=E/R so the current is 179/220K = approx. .8mA. The tube is in series with the plate resistor so the current through it will be the same. Therefore, the resistance of the tube at the time of measurement was R=E/I, 150/.0008=187500.
Hope that helps?

[Chuck H]

Great explaination. I was just going to say that the voltage drop is due to current through the tube. And site that if the tube were removed there would be no drop in voltage across the resistor. And conclude with the voltage drop across the resistor being analogous to the current through the resistor as a result of tube operation. Your response is more educational. I hope the OP is taking notes.

[jmaf]

Great explaination. I was just going to say that the voltage drop is due to current through the tube. And site that if the tube were removed there would be no drop in voltage across the resistor. And conclude with the voltage drop across the resistor being analogous to the current through the resistor as a result of tube operation. Your response is more educational. I hope the OP is taking notes.

Removing the tube and not having a voltage drop there is very illustrative, I think the OP will appreciate that.

In fact, taking examples to the extremes is what often helps me understand stuff. resistance A does this, resistance B does that. What if I remove resistance A(a infinite resistor) or what happens if I jumper resistor B(no resistance)...that is often a useful mind tool to solve stuff. Tube load lines are done just like that. With a jumper instead of the tube, current would be A, with an open tube current would be zero, from there draw a line between the two points.

[jmaf]

Ok, the faucet analogy makes sense, and that voltage is created because those electrons (current) have to cross resistance, but I'm still not clear why/how the voltage drops 179v from one side of the 220k resistor to the other. Does that make sense? The answer might be in your previous response, but I've read it several times carefully and I'm still not clear.

BTW: When I first started with tubes, it took me a while to grasp the plate load thing. I went to professors at my former university, they gave the talk. I went to engineers, they gave me the fancy talk.... Then I went to an uncle of mine, PhD, retired antenna engineer and teacher, sadly he was defeated by cancer a few years ago.

He taught me the water faucet analogy, then the water tower analogy for capacitors, transmitting information by tapping the water coming out a water hose, etc, etc, etc, it all became clear to me by just thinking about water.

If you resist the flow of water, pressure forms. Pressure is voltage in electricity. If there's a clog in a hose and water is trying to pass, pressure will form in that spot where there's the clog, it'll swell, pressure will form. When the tube is trying to send out electrons and there's a clog up in the plate(the resistor), the electrons can't pass freely, so pressure forms = voltage.

[MWJB]

"On my Goodsell, or a Matchless Spitfire, I see 329v from the PS, thru a 220k resistor, then ~150v at the plate. How?"

It's also important to recognise the role of the cathode resistor, this decides the current that the tube will draw when a voltage is applied to the plate, increase the value of the cathode resistor and voltage drop accross that 220K plate resistor will drop & voltage at the plate will rise. Decrease the value of the cathode resistor and current drawn will rise, voltage drop accross the resistor will increase, plate voltage will drop.

In reality, you wouldn't normally mess with plate & cathode values unless you were specifically revoicing the amp (rather than fine tuning the current voicing).

Also different tubes draw different currents in the same circuit. If we take the common 100K plate resistor/1.5K cathode resistor set up, a 12AX7 will show ~2/3 of the voltage at the B+ rail at it's plate, after the 100K. Sub that 12AX7 for a 12AY7, which is a higher current tube, and you get half the B+ supply voltage dropped accross the plate resistor, rather than 1/3.

So the plate resistor does not drop a specific voltage, it just sits in the power supply, the cathode resistor largely dictates how much current is drawn, along with the B+ supply voltage & the plate resistor largely just "resists" and Ohms law does it's stuff, so if the conditions are right for a 12AX7 with 100K plate, 1.5K cathode & ~300vdc at the B+ rail, the tube will draw about 1mA, dropping 100v accross the plate resistor, leaving 200vdc at the plate itself. Increase the B+ supply to 400vdc and the same set up yeilds ~130vdc dropped accross the 100K plate resistor & 270vdc (@1.3mA) at the tube plate.

Just increasing the plate resistor value will increase resistance, dropping more voltage across the plate resistor, leaving less voltage & more current at the tube plate. But as noted before, this will change the character of the amp, revoicing it darker.

[Chuck H]

+++

I think we were just omitting the actual amount of current or it's control method for purposes of clarity. The Q was "why is there a voltage drop?" Too much info right off could blow a vein in someones head. But I can only speak for my perception of the issue. Perhaps the OP will be able to grasp the extra detail and broad strokes aren't in order.

That said... Yes. The actual amount of current through the tube will dictate the voltage drop across the plate load resistor. The tubes "bias" dictates the current that will flow through the tube. The bias of the tube can be controlled at the cathode or the grid. The bias is the voltage relationship between the grid and cathode and is adjusted relative to plate voltage and desired operation of the tube. And so, is analogous to the current through the tube and therefor the current through the resistor and therefor the voltage drop across that resistor.

[ninjadave7]

Wow, this has really turned into a good thread, it makes a lot more sense now! I might even print it for reference. Thanks guys! Keep talking if you want to... I'm just absorbing

[ninjadave7]

"Think of the tube as the other resistor in the voltage divider. There is a voltage drop across the plate resistor, and across the tube. (I will ignore the cathode resistor to keep it simple). In your example you have 329V supply. You have 150V at the plate so 329-150=179V across the plate resistor. I=E/R so the current is 179/220K = approx. .8mA. The tube is in series with the plate resistor so the current through it will be the same. Therefore, the resistance of the tube at the time of measurement was R=E/I, 150/.0008=187500.
Hope that helps?

OK, I wasn't thinking of the tube as a resistor, that helps. The thought crossed my mind, but because I wasn't sure how that would work I didn't pursue it more. So follow-up question on this response: You used the measured voltage drop (179v) in the equation to find the current through that part of the circuit, but used 150v (the voltage immediately before the tube) to calculate the resistance of the tube. Why didn't you use 329v then to calculate the current through the plate resistor? I'm sure the answer is right under my nose, but apparently I can't smell it...

[mooreamps]

I hope it does help.. because that's exactly how it works...

-g

[g1]

So follow-up question on this response: You used the measured voltage drop (179v) in the equation to find the current through that part of the circuit, but used 150v (the voltage immediately before the tube) to calculate the resistance of the tube. Why didn't you use 329v then to calculate the current through the plate resistor?

The tube and plate resistor are a series circuit. The plate resistor does not have 329V across it. One end (supply) has 329V. The other end (tube plate) has 150V. Therefore the voltage drop across the plate resistor is 179V. If you measured voltage with one meter probe to each end of the plate resistor you would measure 179V.
The 329V is shared between the plate resistor and the tube. In a series circuit all the voltage drops must add up to the supply voltage.

Edit: Once again I have omitted the cathode resistor for simplicity.

[ninjadave7]

How does the cathode resistor play into deciding the current the plate draws?