Plate loading question

[caps]

IF, I repeat, IF :grin: , I understand this correctly,

A varying potential on the grid will cause a variation of current flowing through the plate.

Ok, so we need to add a resistor, because for amplification, ( E = I X R ), we need a voltage drop...

and bingo, we have with the voltage drop across the plate resistor, so we have amplification.

It seems very obvious this is how we get amplification in the case of say, a power amplifier....

However, for a mic. PREamplifier, do we get amplification out of the tube in exactly the same way???

 

[pstamler]

In a word, yes.

Peace,
Paul

 

[PRR]

> obvious this is how we get amplification in the case of say, a power amplifier.... However, for a mic. PREamplifier, do we get amplification out of the tube in exactly the same way???

All amplifiers work the same. Just like a 0.1HP weed-trimmer engine and a 1,600HP prop fighter-plane engine work the same. The details are different, but the basics are the same.

 

[Larrchild]

it's all hydraulics, basically.

 

[caps]

At this site:

http://www.diyparadise.com/tubeloadline/tubeloadlines.html


He states selecting the value of a cathode resistor to set the grid bias point goes like this:

"What the operating point also tells you is that, you are now ready to build the circuit! You now have B+ of 260V, plate resistor of 10kohm, what's missing is cathode resistor. This is easy. At -6V grid, you are going to get ~12mA. Back to Mr Ohm... R = VDropAcrossResistor/Current = 6V/12mA = 500ohm. Choose the nearest 470ohm for simplicity's sake."

However, dosent the plate load resistor have a effect somehow on the cathode resistor value. If so, how ?

 

[alk509]

[quote author="caps"]dosent the plate load resistor have a effect somehow on the cathode resistor value?[/quote]

Nope.

Peace,
Al.

 

[PRR]

[quote author="caps"]dosent the plate load resistor have a effect somehow on the cathode resistor value.[/quote]
Yup.

> B+ of 260V, plate resistor of 10kohm

Change that 10K (an absurdly low value for many uses) to 100K (a fine value if you are feeding another tube 2 inches away). Now there is no way you can run the proposed 12mA: 100K*12mA= you need 1,200 Volts for the resistor, and some left for the tube, say ~1,300 Volts. Also 15 Watts, which may be more than you care to feed it.

With a "perfect" tube (that can suck infinite electrons at zero volts) the loadline for 260V B+, 100K, runs from 260V and 0mA to 0V and 2.6mA. As a wild first rough guess, idle the tube halfway between: 130V and 1.3mA. Can the tube pass 1.3mA at 130V? For that monster 5687, sure, easy, and it takes (squint) about -8V on the grid (or +8V on the cathode) to put it there. Now the cathode resistor must be 8V/1.3mA= 6K. Or 4.7K will work about as well: tube will settle toward 1.5mA with 110V across it. 1.5mA*4.7K= 7V, which (squint squint) is about the right bias for 110V and 1.5mA in a 5687.

All this curve stuff is for fools, or extreme cases. For an audio voltage amp (load resistance very high), you look up the tube's data-sheet plate resistance, pick a plate resistor 2 to 5 times higher, pick a load resistance 2 to 5 times higher than that. Then you build it, and fiddle the cathode resistor so that the plate voltage sits near 1/2 to 2/3 the B+. You really CAN do this on the breadboard/prototype. Don't drink a case of whiskey, as the "BOGUS circuit designer" did, to figure a simple tube circuit.

Example: 12AU7 book says Rp=7K. Pick 7K*5= 35K plate resistor. Pick 35K*5= 175K next stage load resistor. You can fudge this a lot: to 100K if you are driving big power tube grids, to 1Meg if you only have to drive a small-tube's grid.

Example: 12AX7 book says Rp=62K. 62K*5= 310K plate load resistor. 310K*5= 1.55 Meg next-stage grid resistor. That's above spec for some tubes. A likely set of values is 220K and 1Meg. For a little less gain and a little more distortion, try 100K and 250K: this is Fender preamp ballpark.

BTW: the cathode resistor will almost always be the plate resistor divided by the tube Mu. 12AX7 with 100K or 220K plate resistor usually uses a 1K or 2.2K cathode resistor. BCD picked 10K plate resistor and tube with Mu=17, 10K/17= 588Ω, BCD drew all those curves and arrived at 500Ω. Either answer (or 470Ω) will give a fully working amp. Dinking around with the cathode resistor may increase gain and reduce distortion but not enough to notice (making it way off will reduce gain). Dinking cathode resistor may increase maximum unclipped output, but most stages should never go near clipping and clip-action should be checked with ear and scope, not from the quite bogus data in tube books.

 

[Swedish Chef]

I love this place! :green:
PRR can make anything seem like falling off a log!

chef

 

[NewYorkDave]

An excellent "rule of thumb" approach!

The rules are a little different when you're driving an impedance that is not large compared to the plate resistance, though. In other words, when you have to deliver a measurable amount of power. In those cases, you'll usually find the cathode resistor/grid bias chosen to maximize the plate current (within the ratings of the tube) and transconductance. For example, in the "One Bottle Pre" design I posted on the Lab some time ago, the cathode resistor is a measly 330 ohms, which is the plate resistor divided by 66. The Mu of the tube is about 37.