# Quick question for techies

#### yankee

I have been experimenting with feedback options for an input triode tube in a simple se el84 amplifier. The input tube is 7dj8. The plate resistor is 47K and the cathode 1k. So on a lark I connected a 120k resistor directly from the plate to the cathode and the amp seems to sound better. Just wondering what is going on electrically. Any ideas.

#### PRR

Paid Member
The 1k may be too low for that tube. Did you do any calculation or plotting?

#### jjasniew

Well, from a DC perspective some current is going to flow through the 120k, pulling up on the 1K, making the voltage drop across it a little higher. (You might even be able to measure that change with a DMM...)

Typically, this would bias the tube so that the quiescent grid voltage is apparently (from the tube's perspective) a little bit more negative, perhaps putting the tubes operating point in a somewhat better spot, at the expense of amplification stage gain.

AC wise I couldnt begin to offer a descriptive analysis. FAIK, that's the change you might be hearing, vs from the change in DC operating point.

#### nigelwright7557

I have been experimenting with feedback options for an input triode tube in a simple se el84 amplifier. The input tube is 7dj8. The plate resistor is 47K and the cathode 1k. So on a lark I connected a 120k resistor directly from the plate to the cathode and the amp seems to sound better. Just wondering what is going on electrically. Any ideas.

You could try a 2k2 pot in place of cathode resistor and tune it to the sound you like.

#### rayma

Try a 0.22uF coupling capacitor in series with the 120k, and see if the improvement is still heard.
If so, it's not due to the shift in operating DC point.

#### shermanr

Interesting scenario.

If what you describe is accurate, if you connect a 120K resistor from the Plate to the Cathode on a 7DJ8, keeping your circuit the same, you are bypassing the Tube (a.k.a. Connecting a 120K resistor in parallel with your 7DJ8), but you still have the 1K resistor at the Bottom to hold the bias.

Without the DC voltage being applied I cannot calculate, but I can comment on the scenario.

This is not a typical scenario, so without the actual B+ Voltage, I can only speculate or extrapolate what happens.

The 7DJ8 tube will develop a voltage drop that can be calculated or determined from the Characteristic Curves only when we know the voltage being generated across across the Plate-Cathode when a specific current flows through the tube and we know the bias condition (1K).

Assuming that the tube makes music, and we connect the 120K resistor...

A 120K Resistor will definitely be in parallel with the Plate to Cathode and load-down the circuit, and will create a electrical current split between the current flowing though the 7DJ8 and the 120K Resistor.

Without the 120K Resistor, your tube will settle at an Operating Point, with the 120K the operating point will settle at a point that is different than without the 120K.

The behavior of a Resistor in Parallel with a Plate to Cathode is probably not unusual as it will interact with the current flow by providing a parallel path for the current to flow through.

How much current flows through the 120K resistor can be calculated by knowing the DC voltage being applied to the circuit.

Lets assume that we have 200 Volts into the 47K resistor. Then we have the node where the Plate and the 120K connect. Then the other node is the Cathode where we have the other lead of the 120K connected and the 1K, the other 1K lead to Ground.

So what you are doing by bypassing the tube with a 120K resistor is basically drawing extra current through the 120K resistor, that is dissipated as Heat. And introducing a damping effect across the 7DJ8 and attenuating effect. It will likely have a sonic impact, it will act as a signal attenuator and a disturbance that was not there previously. The only way to describe it is to say that we have a Pure Tube or a Bypassed Tube.

Resistors are non-reactive, so all they do is limit current. dissipate heat, set voltage levels, attenuate signals and drop voltages. They don't act on the AC component, except to dissipate this as RMS Heat (calculates just like DC), superposition.

The end sonic effect would require us to know if your Power Supply Voltage is, or is not Regulated.

If the Power Supply is regulated, then the placement of a 120K Resistor in Parallel would not have much of an effect on the sound as the regulated Power Supply will seek to maintain voltage.

I also may think that the 120K would act a a damper to soften transients.

Without the 120K resistor you only have the Tube Vacuum Gap. When you introduce a Resistor across this "air gap" you are altering the performance of this electrical circuit, just like having an extra amplification stage, more stuff that gets in the way.

The higher you make the Resistor in value, the less effect it will have, and beyond 10 to 20 Megohms, it literally disappears electrically.

But if the DC supply is un-regulated, the effect that the 120K resistor has is to draw extra DC current and load down the power supply a bit.

It may drop a few volts, maybe even 5-8 Volts, speculating that your DC supply is delivering 200 VDC. But not much of an effect as these are tens of milliamps on a large DC supply.

A 7DJ8 draws more filament Current than a 6DJ8 but electrically they are quite similar. And I know these tubes work best around 150V across the Plate to Cathode.

So the answer is that it will not kill the circuit, it just introduces a Resistive DC load across the Tube that will most likely alter the dynamic characteristics of the Tube while the DC quiescent voltages will settle but not the frequency response.

Note that there is a limit on how Low you can go with this Resistor.

The only way to really know hos this affects the music is to connect the before and after to measurement equipment and look for the differences.

Electrical example: Vdc (Plate-Cathode) = 150VDC

A 120K resistor at 150 Volts DC would draw 1.25 milliamps.

So the 1K Cathode Resistor will act as negative feedback and increase the DC voltage drop across the 1K Cathode Resistor.

How much, we need actual numbers to know.

Good idea, hummmm.

Interesting question

#### audiowize

From the basic AC perspective, you have cut the impedance loading the triode down, though we can't say by how much without having the schematic drawn out and posted. Connecting the resistor from plate to ground vs. plate to cathode will make a little difference, as you are injecting a little bit of what's going on at the plate back into the cathode. The only place I've seen this done is on the split load phase inverter in some of the old Fisher amps. The cathode is pretty tough to drive though, so I would expect a rather modest difference when comparing connecting the non plate end of this resistor to ground vs. cathode. Perhaps you could start by measuring the difference in gain between the 120K resistor connected to the cathode vs. connected to ground?

Since you are driving an EL84 with a 7DJ8, I would guess the EL84 is triode strapped, so we'll guess that your B+ is about 300V. In this case, the 47K plate load and 1K cathode resistor would give about 3.7V of bias and 3.7mA of plate current with 125V on the plate, so there's nothing wrong there.

#### yankee

It has around 250-260v on the input of the 47k plate resistor and maybe 3.6v on the cathode with the 120k resistor in the circuit. It does sound like there could be some negative feedback but likely a small amount because the overall gain of the amp isn't very diminished. Yes, the el84 is indeed triode strapped but I have it is running a little hot at 330v and 30-35ma on the plates.

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#### PRR

Paid Member
> It does sound like there could be some negative feedback

It's positive feedback. Just not enough to matter.

In a related scheme, Bogen used to run a resistor *from B+* to an under-size cathode resistor. This gave them the DC bias with a small cathode resistor and more gain without a cathode capacitor.

But in general I think "the amp seems to sound better" is essentially meaningless.

#### PRR

Paid Member
THD at 10V out is raised from 0.4% to 0.7%.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(basic)
TOTAL HARMONIC DISTORTION = 4.131270E-01 PERCENT
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(frill)
TOTAL HARMONIC DISTORTION = 7.277961E-01 PERCENT
Tube is driving 2/3 the load with 3/4 the current, so "double" makes sense.

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#### yankee

That's pretty close simulation and very cool. I know what you mean about subjective evaluations. I've been wrong before where I thought I liked something better only to find out I didn't. Thanks for that.