coupling gain stages?

[noworrybeefcurry]

Why are the plates of your preamp tubes grounded in most amps? From my knowledge that is the point were you would couple gain stages, but if the plates are grounded would you not just be shorting everything to ground?

[flemingmras]

Why are the plates of your preamp tubes grounded in most amps? From my knowledge that is the point were you would couple gain stages, but if the plates are grounded would you not just be shorting everything to ground?

In what amps are the PLATES grounded? The CATHODES are connected to a cathode resistor, which gets tied to ground but the plates get positive voltage via the plate load resistor, and an output coupling cap is used to filter out the static DC component of the fluctuating DC plate voltage while the fluctuations themselves appear at the other side of the coupling cap in the form of AC.

[noworrybeefcurry]

im sorry these are novice questions, but can you have both AC and DC voltages within the same line? You supply a constant positive DC voltage to the plates of tubes for electron flow to even occur, to my knowledge this is then collected by a coupling cap and fed to the next stage. But this would require both voltages in the same line.

[flemingmras]

im sorry these are novice questions, but can you have both AC and DC voltages within the same line? You supply a constant positive DC voltage to the plates of tubes for electron flow to even occur, to my knowledge this is then collected by a coupling cap and fed to the next stage. But this would require both voltages in the same line.

Under "zero signal" conditions, you have a static DC current flowing from cathode to plate. The AC copy of the input signal is created via the input signal modulating the static DC current that flows from cathode to plate at input signal frequency (i.e. causing the DC current to rise/fall above/below its "static" value with input signal). This causes the DC plate voltage to also rise and fall at input signal frequency. It is the rising/falling voltage that the coupling cap passes through while the static DC plate voltage gets blocked, and you end up with pure AC at the other side of the coupling cap.

When the input signal at the grid swings positive, it offsets the negative voltage at the grid and drives the grid voltage less negative. This allows cathode/plate current to increase above its static zero signal value.

When the input signal swings negative, this offsets the grid's idle negative voltage and pushes the grid more negative than its idle voltage, which causes cathode/plate current to decrease below its static zero signal value.

This in turn causes plate voltage to do the opposite...fall, then rise (plate voltage and plate current are out of phase with each other). This is what inverts the output signal and causes the output to be out of phase with the input.

The static plate voltage is blocked by the coupling cap, but the rises and falls in plate voltage pass through it and end up as a pure inverted AC copy of the input signal at the other side of the coupling cap.

Essentially you could say that you have AC superimposed onto the DC plate voltage when input signal is applied. The coupling cap blocks the DC component, and you now have AC superimposed onto zero volts on the other side of the coupling cap (since caps block DC).

[axeman]

Are fucking a wake rite now.

[noworrybeefcurry]

When you say "zero signal conditions" is that the tube's idle current once the cathode is heated? Also, if you simple have a static DC on the plate of the tube then how do you modify the plate voltage or headroom in that stage?

[flemingmras]

When you say "zero signal conditions" is that the tube's idle current once the cathode is heated?

Yes...the idle cathode/plate current that is flowing from cathode to plate when no signal exists on the tube's grid.

Also, if you simple have a static DC on the plate of the tube then how do you modify the plate voltage or headroom in that stage?

"Headroom" is controlled via the B+ voltage where the plate resistor connects to the power supply. The peak-peak swing of the output signal (i.e. the voltage measurement from the positive peak to the negative peak) cannot exceed the B+ supply voltage at this point.

Plate voltage is determined via the B+ voltage, the plate resistor value along with where the tube is biased at via the cathode resistor. The plate resistor also controls the gain of the stage as well. The cathode resistor will also have SOME effect on the stage's gain but not to the extent that the plate resistor does, mainly because the cathode resistor only affects gain by shifting the operating point to a different set of curves that are spaced slightly different.

[noworrybeefcurry]

I apologize i am still learning and my references may be wrong, but when i need to reference to a circuit board layout i generally use the layouts on "www.ceriatone.com"

But back to the topic, so their is a coupling cap and a plate resistor in parallel with each other and that is in series with a filter cap and the plate of your tube?

[flemingmras]

I apologize i am still learning and my references may be wrong, but when i need to reference to a circuit board layout i generally use the layouts on "www.ceriatone.com"

But back to the topic, so their is a coupling cap and a plate resistor in parallel with each other and that is in series with a filter cap and the plate of your tube?

No need to apologize. We all gotta start somewhere. I'm always willing to teach those who are willing to learn.

Below is a schematic of a typical common cathode triode stage -



Arrows indicate direction of static DC electron flow. The filter cap powers the stage while the power supply continuously recharges it. Think of Cf as your car battery while the power transformer/power supply is your alternator.

When no signal is present at the grid, this DC current is constant. Current flowing through a resistance creates a voltage drop across said resistance. The current flowing through the tube is exactly the same as the current flowing through Rk and Rp because they are in series with each other (current flow is the same at all points within a series circuit).

Using Ohm's Law and assuming 100K (100,000) ohms for Rp and a static plate current of 1.2mA, we can calculate the voltage drop across Rp -

Voltage = Current x Resistance

We know that the tube is drawing 1.2mA of current through Rp and we know that Rp = 100K ohms (100,000 ohms) -

1.2mA (0.0012 Amps) x 100K Ohms (100,000 Ohms) = 120 Volts

Now, assuming a B+ voltage of 270 volts, we can subtract Rp's voltage drop from our B+ voltage to find our static plate voltage -

270 Volts - 120 Volts = 150 Volts plate voltage

Now remember we are still under "zero signal" conditions so our plate is seeing a static voltage of 180 Volts and our plate current is a constant 1.2mA.

1.2mA of current ALSO exists through Rk at the cathode because it is part of the series branch. Assuming that Rk's value is 820 Ohms, we can calculate our cathode voltage -

1.2mA (0.0012 Amps) x 820 Ohms = 1 Volt

This places our cathode at 1 Volt.

Now...Rg-k references the grid to ground, which is our zero volt reference so this places the grid at zero volts relative to ground, meaning they're at the same potential. Since the grid is at zero volts and the cathode is at +1 volt, the cathode is said to be elevated above ground by +1 volt. This makes the cathode 1 volt more positive than the grid, which makes the grid more negative than the cathode. Relative to the cathode, the grid is sitting at -1 volt.

Now let's apply a 1 volt peak-peak input signal -



"Volts peak to peak" (Vpk-pk) means voltage measured from the positive peak of the AC sine wave to the negative peak of the AC sine wave. Peak voltage (Vpk) means voltage measured from zero to one or the other peak and is equal to 1/2 the Vpk-pk. This means that our 1Vpk-pk signal = 0.5Vpk.

When the signal swings to the positive peak, this offsets our -1 volt at the grid by the value of the signal swing. This means that it makes our grid less negative by a value of 0.5V. When the grid is made less negative, this "opens the valve" and allows more electrons to flow from the cathode to the plate. Electrons being negatively charged, they offset the plate voltage by making it less positive and causing the plate voltage to drop. In our stage, plate voltage will drop from its static value of 150V down to about 118V. This is a drop of about 32 volts.

Now when our signal swings back toward zero, electron flow from cathode to plate starts to decrease back toward its static value. This causes plate voltage to rise back up toward its static value since less and less negative electrons are flowing to the plate. Once the signal is back at zero, electron current and plate voltage will be returned to their static value.

On the negative swing, this pushes our -1V at the grid MORE negative. This causes electron flow from cathode to plate to DECREASE BELOW its static zero signal value. Now even less and less electrons are flowing to the plate, which causes the plate voltage to rise more positive. At the peak of the negative input signal, plate voltage will have increase to about 182 volts. This is a plate voltage INCREASE of 32 volts.

So we now have a fluctuating plate voltage that is rising above and below its static value of 150 volts by a margin of 32 volts above and below the static plate voltage.

On the other side of the coupling cap, the 150V DC plate voltage cannot pass through it so the voltage on that side of the cap is zero. However, the 32 volt fluctuations CAN pass through it since they're fluctuating. Since our DC voltage on the other side of the cap is zero and our 32 volt fluctuations are passing through Cc, we end up with a +/-32V AC signal on the RL side of the coupling cap across RL!

Now...AC sees ground the same as it sees B+, so to AC signals, Rp and RL DO in fact appear to be in parallel with each other. But DC only sees Rp since RL is on the AC side of the coupling cap.

So...we have a +/- 0.5VAC input signal on the grid, yet we have a +/-32VAC signal at the output coupling cap. Since our output is much larger than our input, we now have amplification.

This particular stage has a voltage gain of 64. Gain is figured out by dividing the output signal voltage by the input signal voltage -

32V / 0.5V = 64

This means that the voltage output of the stage will be 64 times what the input signal to the stage is.

One more thing to note...as you noticed, when the input signal swings positive, the plate becomes LESS positive and causes plate voltage to DROP. On the negative swing of the input, plate voltage INCREASED. This means that our output signal is said to be "inverted", or 180* out of phase with the input signal (i.e. the output signal is swinging negative when the input signal is swinging positive and vice versa).

[axeman]

[noworrybeefcurry]

You said that the filter cap was powering that particular stage and your power supply was recharging the cap, i was aware that this was how it works. But i was under the impression that your filter caps were fed by the rectifier circuit and then provided a positive high voltage current to the tubes plates. In the schematic that you provided it shows the static DC flow from going out of the plate through the filter cap back into the cathode of the same triode, could you clarify on this. If their is no plate voltage then how can their be electron flow within the tube?

Also your explanation of how coupling caps and plate load resistors work as far as filtering static DC and reflected AC really explained the whole concept perfectly to me, and i can understand it within the illustration that you provided. The coupling cap and plate resistor are in parallel going to separate circuits, but in every schematic i have seen the cap and resistor are in parallel but in series with the plate of the tube and the next stage; and because they are in parallel/series i think thats why i am having trouble understanding how you can have static DC and reflected AC on the same wire. Could you clarify?

[flemingmras]

You said that the filter cap was powering that particular stage and your power supply was recharging the cap, i was aware that this was how it works. But i was under the impression that your filter caps were fed by the rectifier circuit and then provided a positive high voltage current to the tubes plates. In the schematic that you provided it shows the static DC flow from going out of the plate through the filter cap back into the cathode of the same triode, could you clarify on this. If their is no plate voltage then how can their be electron flow within the tube?

They're two different circuits...the preamp stage circuit and the power supply circuit. At the same time you have current being drawn by the stage from the filter cap as shown in the schematic, you also have "recharge current" coming from the power supply recharging the cap. The power supply was just omitted from the schematic for the sake of clarity.

There is no such thing as "positive current". The positive side of the filter cap positively charges the plate while the negative side of the filter cap negatively charges the cathode. The heater heats up the cathode, which causes electrons to boil off of the cathode. Electrons being negatively charged, the plate's positive charge draws the boiled off electrons to it (opposite charges attract) in an attempt to discharge the cap. If there were no power supply present, it would discharge the cap completely, but the power supply acts as a charge pump to continuously recharge the capacitor.

Also your explanation of how coupling caps and plate load resistors work as far as filtering static DC and reflected AC really explained the whole concept perfectly to me, and i can understand it within the illustration that you provided. The coupling cap and plate resistor are in parallel going to separate circuits, but in every schematic i have seen the cap and resistor are in parallel but in series with the plate of the tube and the next stage; and because they are in parallel/series i think thats why i am having trouble understanding how you can have static DC and reflected AC on the same wire. Could you clarify?

Essentially the signal is "superimposed" onto the static plate DC. Think of it like this -



That is what your plate voltage is doing when under signal...fluctuating above and below its static value. The plate side of the cap sees this fluctuating DC. However, on the other side of the coupling cap (the side opposite the plate), the static DC gets filtered out, but the fluctuations above and below it get passed through the cap as the fluctuations on the other side of it charge and discharge the cap. Thus, you're left with this on the RL side of the cap -



As the DC plate voltage fluctuates above/below its static value when under signal, it charges/discharges the coupling cap. It is the charging/discharging of the cap that becomes the AC copy of your input signal. Because the plate voltage flucuations are much larger than the input signal induced grid voltage fluctuations, you end up with a bigger signal at the output of the stage than at the input, hence the amplification.

So you don't really have AC on the same wires as DC. You're converting DC into AC by making the plate voltage fluctuate above and below its static value, then filtering out the static DC plate voltage with the coupling cap. It's Newton's 2nd law at work -

"Energy can neither be created nor destroyed. It can only change from one form to another"

The tube uses the input signal to convert static DC current into fluctuating DC current. The coupling cap then converts the flucuating DC voltage at the plate generated by the fluctuating DC current through the tube/plate resistor into pure AC via filtering out the static DC plate voltage from the fluctuations. Through all of this, the AC at the output ends up being an amplified copy of your input signal.

[noworrybeefcurry]

I was under the impression that filter caps were polarized and would only provide rectified positive DC out of one side of the cap onto the plate of a given tube. But a filter cap will discharge both ways?

To my best understanding you have a constant static DC voltage that is provided to the plate by your filter cap, but as signal is introduced into the triode then only the reflected AC will pass through the coupling cap into your next stage?

also, at the beginning of this topic i had asked why the coupling caps and load resistors after the tubes plate were grounded. Every time i look at a schematic the resistor and cap are always going to ground potential, but from my understanding they should be fed into the next stage. Could you clarify?

-Thank You

[flemingmras]

I was under the impression that filter caps were polarized and would only provide rectified positive DC out of one side of the cap onto the plate of a given tube. But a filter cap will discharge both ways?

No. You're misinterpreting.

You have TWO charges on a filter cap. The negative side has a negative charge (a surplus of electrons,which are negatively charged) and the positive side possesses a positive charge (a SHORTAGE of electrons). Just like a battery, when the positive side of your filter cap is connected to the plate, the positive charge from the filter cap is pulling negative electrons from the cathode to the plate, which are being sourced from the negative side of the filter cap's electron surplus. Think of the negative side of the filter cap as your "send" while the positive side is your "return". The grid is right smack dab in the middle of this electron stream. As such, grid voltage has direct control of this electron flow from cathode to plate.

If there's no power supply recharging the cap, this flow will continue until the charges on both sides of the filter cap are equalized. When both sides of the cap possess an equal amount of electrons, there is no charge differential (i.e. "potential"). There MUST be a charge differential on both sides of the cap for electrons to flow from it. When a charge differential exists between both sides of the filter cap (one side possessing more electrons than the other), there is a potential for current to flow, and we measure this potential as "voltage".

However, in amplifiers we have a power supply continuously recharging the filter cap. As electrons leave the negative side of it, more electrons are pulled into the negative side of the cap from the power supply. As electrons are collected by the positive side of the cap, the power supply pulls them away from the positive side of the filter cap. It is this pulling force on the positive side of the filter cap that pulls negative electrons into the negative side of the filter cap.

To my best understanding you have a constant static DC voltage that is provided to the plate by your filter cap, but as signal is introduced into the triode then only the reflected AC will pass through the coupling cap into your next stage?

That is correct but you're missing something. How is that AC created? That is what I explained in my above post. The AC that passes through the coupling cap IS NOT the same signal as what came in from your guitar. It is a mirrored copy of it, but at a much higher voltage than what it came in as.

How does it do this? It uses the guitar's pickup signal to modulate the voltage at the grid. As mentioned earlier, the grid has direct control of electron flow from cathode to plate so any alternations in grid voltage will cause the same alternations in electron flow from cathode to plate, but in much bigger amounts (a small change in grid voltage causes a much larger change in cathode/plate current). This causes the static DC electron flow through the tube to alternate above and below its static value at input signal frequency, which in turn causes the plate voltage to alternate at the same frequency as the input signal. Those alternations in DC plate voltage are what we call "alternating DC". It is the alternations in the DC plate voltage that pass through the coupling cap. But the static DC gets blocked so you end up with an alternating voltage that alternates above and below zero, which is pure AC.

In short, the input signal first converts the static DC electron flow from cathode to plate into an alternating DC copy of itself. It is called "alternating DC" because it's still DC in the regard that it's still flowing in the same direction, it's just alternating above and below a static value at input signal frequency. The coupling cap converts the alternating DC plate voltage into pure AC by blocking the static DC on the plate. Coupling caps will block STATIC DC, but they will pass ALTERNATING DC (i.e. Alternating DC = DC voltage that alternates above and below a static DC voltage). It is the alternations (i.e. "changes") in DC voltage that pass through the coupling cap while the static DC voltage is blocked. It does this via charging/discharging the coupling cap as the DC voltage alternates above and below its static value.

also, at the beginning of this topic i had asked why the coupling caps and load resistors after the tubes plate were grounded. Every time i look at a schematic the resistor and cap are always going to ground potential, but from my understanding they should be fed into the next stage. Could you clarify?

-Thank You

The resistor you see going to ground in my schematic labeled "RL"...that is actually the grid leak resistor of the next stage. The resistor is there for two reasons -

Reason 1 is to complete the coupling cap circuit. In order for a cap to pass alternating DC in the form of pure AC, the alternating DC at the plate has to be able to charge/discharge the coupling cap. Since one side of the cap is connected to the positive source (i.e. the plate), the other side must be connected to a negative source that the positive charge on the other side of it can pull electrons into the cap from, which is ground (ground is seen as being negative relative to the plate since the plate is positive relative to ground..it's all relative).

Reason 2 is that the next stage's grid needs a ground reference. RL also functioning as the grid leak resistor, it gives the next stage's grid the ground reference. Since it is referenced to ground via the leak resistor and the cathode is lifted above ground via the cathode resistor's voltage drop, the cathode is placed more positive than the grid (which is sitting at ground potential...i.e. "zero"), which makes the grid more negative than the cathode and biases the tube. The static negative voltage at the grid (negative relative to the cathode, not ground) repels electrons from the cathode (electrons being negative and like charges repel), which limits static electron flow from the cathode to its static value. The input signal modulates this negative grid voltage, making it more/less negative as the input signal swings positive/negative, which causes the grid to repel more/less electrons from the cathode, which causes electron flow from the cathode to alternate above/below its static value.



RL can also be a volume pot that controls how much of stage 1's signal gets to proceed into the input of stage 2, which is how it's done in most amps.

Now in reality, if there's no pot in place of RL (or any other signal limiting circuitry) there will usually be a 2nd resistor between Cc and RL, which makes a voltage divider, or a "fixed pot" per se. This is because the grid of the next stage cannot handle the full output signal of the driving stage and you'll get "grid limiting" where the grid is driven positive and electrons will flow to the grid. Electrons being negative, they will offset the positive swing of the input signal and "clamp" the positive swing. We call this "blocking distortion" and it's not a very musical form of distortion.

This added limiting resistor is shown below labeled as "RLM" (for Limiting Resistor). You can also bypass this resistor with a low value signal cap to allow more highs to come through -



Of course another way to do it would be to omit RLM altogether and use a low value resistor for RL (like say 47K - 100K instead of the more common 1M like you commonly see), which will load the output of the driving stage and reduce its output gain some.

[noworrybeefcurry]

So the filter caps in your power section dont specifically provide a voltage to your plate, but they are more a source of potential for current to flow, is that correct?

I do understand the concept of gain within a tube, but their is another thing that has me slightly confused. How is it that the coupling caps will pass the alternating DC? In the end it is still DC, so is it just that the cap itself does not have a chance to fully charge?

-Thank You