coupling gain stages?

[flemingmras]

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?

Voltage IS potential. Voltage is the unit of measurement for potential. A voltage (i.e. "charge differential" or "potential") must exist between two points (i.e. the two terminals of your filter cap) in order for electrons to potentially flow through a circuit connected between said two points (i.e. again your filter caps two terminals). No charge differential = no voltage = no potential for electrons to flow.

But voltage is relative to something. In the case of the filter cap, the positive voltage is relative to its negative side. You can't just connect one meter probe to the positive terminal only and see a voltage. The other meter lead has to be connected to the reference that said voltage is relative to in order to complete the circuit and see the voltage.

Kinda like when you measure something with a measuring tape...your measurement is relative to wherever "zero" on your measuring tape is. Same thing with electronics. Picture your multimeter as your "electrical measuring tape"...the "COM" meter lead (i.e. the black one) would be your electrical measuring tape's "zero".

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

Because capacitors are charged/discharged via alternating charges, whether they're "alternating DC" (i.e. DC that alternates above and below a fixed voltage) or pure AC (current/voltage that alternates above and below zero).

The key is to have them be alternating charges. With a static DC charge, the capacitor only draws current during it's charge up. Once charged up, no more current flows through RL. No current flow through RL = no voltage drop across RL.

But as soon as the DC plate voltage on the other side alternates below the coupling cap's charged voltage, the coupling cap will discharge through RL until the plate voltage alternates upward with input signal and recharges the coupling cap. It is this charge/discharge action that becomes your final output signal.

And yes...small value coupling caps will not fully charge/discharge below certain frequencies. This is why lower value coupling caps have a low frequency "roll off"....in other words, the low frequencies they pass will be at a lower amplitude than higher frequencies because the cap cannot fully charge/discharge at those frequencies. The value of RL plays a part in this as it draws current from the coupling cap, and the coupling cap cannot maintain its voltage long enough for the low frequency alternating DC to recharge it back up fully.

[flemingmras]

I did up a basic preamp stage in SPICE, which is an analog circuit simulation program used by engineers.

This first one is your input signal. If you look at the scale on the left of the graph, you will see that this signal measures 280mVpk-pk (milliVolts peak to peak), which is about the typical average output of a passive humbucker pickup.



This next signal is the signal taken directly off of the plate. You can see going by the scale on the left graph that the signal directly on the plate is alternating DC as it is alternating above and below the 200VDC plate voltage. It measures roughtly 14Vpk-pk (7 volts above and 7 volts below the static DC plate voltage).



And this third image is the signal taken right off of the coupling cap. It is now pure AC as you can see from the left graph scale that it is varying above and below zero volts. Its amplitude is also 14Vpk-pk just like the alternating DC on the plate. It swings above and below zero by a factor of 7 volts.



And here is a graph showing all 3 -



The bright blue line is your input signal. Due to the graph scaling and the small amplitude of the input signal it's hard to see the alternations in it but they're there.

The green line up at the top is your signal directly off the plate. You can see that it's actually AC, but the plate voltage provides a DC offset to it (alternating DC).

The red signal is what comes out of the coupling cap. The coupling cap passes the alternations in DC plate voltage, but blocks the DC offset of the static plate voltage so that you end up with pure AC at the output of the coupling cap.

Hopefully this will aid you in gaining the understanding that -

The output signal IS NOT the same signal as what came in at the input, but merely an amplified copy of it.

The preamp stage creates this copy via the input signal modulating the tube's grid voltage at input signal frequency, which also modulates electron flow from cathode to plate at input signal frequency, which in turn causes the plate voltage to alternate above and below its static zero signal value at input signal frequency. This allows the input signal to create an alternating DC copy of itself (alternating DC = AC with a DC offset).

The coupling cap and load resistor for the next stage convert this alternating DC into pure AC via blocking the static DC from the plate. The plate voltage alternations charge/discharge the coupling cap through RL at input signal frequency, and the charge/discharge voltage of the coupling cap becomes your final output signal.

[flemingmras]

And here's one more sim plot showing the AC with DC offset at the plate (alternating DC) along with the pure AC that gets reflected at the other side of the cap through it charging/discharging -



The top wave is the alternating plate voltage. As you can see, plate voltage appears to be somewhere around 200-210VDC, and this voltage is swinging above and below that static plate voltage.

The bottom wave is what ends up at the other side of the coupling cap. You can see that this sine wave is swinging above and below zero and is now pure AC since the coupling cap blocked out the DC offset of the positive plate voltage.

[noworrybeefcurry]

Thats cleared up alot, thank you.

But their are several other things that still need clarifying if you dont mind. As far as controling the headroom within a givin stage or the plate voltage, it is simply a tap off of a filter cap via a plate resistor. But their is no set voltage, it more or less depends on the output of the tube and the given headroom from the load resistor.

so for example you could have say thousands of volts on a given plate and this would just provide more or less infinate headroom.

[flemingmras]

Thats cleared up alot, thank you.

But their are several other things that still need clarifying if you dont mind. As far as controling the headroom within a givin stage or the plate voltage, it is simply a tap off of a filter cap via a plate resistor. But their is no set voltage, it more or less depends on the output of the tube and the given headroom from the load resistor.

so for example you could have say thousands of volts on a given plate and this would just provide more or less infinate headroom.

Well first off a 12AX7 can only handle a max of 330 Volts on its plate. Since we're dealing primarily with guitar amps we will use a 12AX7 for all our examples.

Plate voltage can only swing as high as B+ (cutoff) and as low as whatever plate voltage the load line crosses the "Vg1=0" line (i.e. the point at which the input signal pushes the grid voltage to zero volts). The load resistor will set how far away plate voltage is from B+ for a given bias current. This sets your "swing margin" (i.e. "headroom").

If it were set up as a 100% symmetrical pure Class A voltage amplifier, the tube would be biased in such a way that would set your plate voltage at 1/2 the total swing margin. But they never are...they're almost always biased closer to cutoff than they are saturation. The higher the cathode resistor value, the closer to cutoff the tube is biased. This means that the positive swing of the signal will clip long before the negative swing ever will...going off the load lines.

The below example is a load line showing a typical 12AX7 stage with a 100K plate resistor and a 2.7K cathode bias resistor -


The black dot where the 2.7K bias lines meet the load line is your grid bias. The input signal basically "pushes" that dot up and down the load line when it modulates the grid voltage. With a B+ voltage of 250, our 2.7K cathode resistor sets our bias current to give a 65 volt drop from the plate resistor, which puts our static plate voltage at 185 volts.

In our example stage with a B+ voltage of 250 and a static plate voltage of 185, the positive swing of your signal is limited to 65 volts (185 + 65 = 270V...right at our B+). However, your negative swing can swing much lower than the positive swing can swing high. In the below example load line, the blue line right where the load line crosses Vg1=0 indicates that the negative swing of the signal can swing the plate voltage down to about 80 volts. That's about a 105V swing on the negative swing (185 - 80 = 105)! Quite a big difference from the positive swing of 65 volts.

So as you can see, most preamp stages aren't set up as a stage with a 100% symmetrical output. Nor would you want them to be for tonal reasons. Also, the grid curves themselves aren't even symmetrical...as you can see they're not evenly spaced apart from each other by default, which makes the tube itself asymmetrical by default.

[Tone Junkie]

Now this is some good stuff, I might have to read it a bunch of times to understand it. But thats because Im slow < LOL > Bill

[noworrybeefcurry]

So basically you have an ever changing plate voltage except for when your tube is idleing. Your filter caps will provide a potential for current to flow from your cathode to the plate, and this potential (voltage) can be varied via a load resistor. But as a signal is introduced onto the grid of the tube there will either be a voltage build up or drop on the plate depending on the swing of your grid signal. This pulsating DC will then feed through your coupling caps to the next stage as pure AC.

But basically you can provide a high enough voltage to the plate with respect to the tube to give your amplified signal swing enough headroom before it saturates, or you can lower the plate voltage to get your signal to clip.

[flemingmras]

So basically you have an ever changing plate voltage except for when your tube is idleing. Your filter caps will provide a potential for current to flow from your cathode to the plate, and this potential (voltage) can be varied via a load resistor.

The voltage at the plate itself can be varied yes. But B+ (i.e. the voltage across the filter cap itself) stays the same no matter what your plate load resistor is.

Typically though, your cathode resistor is used to bias the tube, and depending on what value you use will determine how hot or cold the tube gets biased and this will have an effect on static plate voltage as well.

But as a signal is introduced onto the grid of the tube there will either be a voltage build up or drop on the plate depending on the swing of your grid signal. This pulsating DC will then feed through your coupling caps to the next stage as pure AC.

Yep...you're getting it.

But basically you can provide a high enough voltage to the plate with respect to the tube to give your amplified signal swing enough headroom before it saturates, or you can lower the plate voltage to get your signal to clip.

Not exactly.

The difference between B+ and plate voltage (i.e. the voltage drop across the plate resistor) will dictate what the maximum possible swing will be on the positive swing of the output signal. The lower the voltage at the plate, the more available "swing" (i.e. "headroom") available for the positive swing of the output signal.

The trade off is that if you increase headroom for the output signal's positive swing, you decrease headroom for the negative swing.

Basically the higher the plate voltage, the less positive swing you'll have available and will be able to hit "cutoff" sooner (i.e. the point at which current flow through the tube is "cut off"). The less plate voltage you have, the later you'll hit cutoff, but you'll hit saturation (the point at which the negative swing clips) sooner. It's a trade off either way. You can "center bias" it so that you have an equal amount of swing for both the positive and negative swings (i.e. symmetrical output).

[noworrybeefcurry]

Makes sense. So is that why lower plate voltages tend to filter out higher frequencies?

[noworrybeefcurry]

This is somewhat varying off topic, but why are certain filter caps only grounded. I cant find any taps to any other part of the circuit, they are just fed by the power supply then grounded?

[flemingmras]

This is somewhat varying off topic, but why are certain filter caps only grounded. I cant find any taps to any other part of the circuit, they are just fed by the power supply then grounded?

Not sure what you mean by "lower plate voltages tend to filter out high frequencies".

As to why filter caps are grounded...look at a power supply circuit. The negative side of the supply is grounded, so essentially ground is the "electron path" TO the filter cap. Electrons flow from the negative side of the supply to ground...through ground....then to the negative side of the filter cap.

In reality, the way older amps were built the negative side of the supply goes straight to the chassis and all the filter caps go straight to the chassis. Electrons flow from the negative side of the supply to the chassis, then use the chassis as "one big phat wire" to flow to the negative side of all the filter caps.

"Ground" is just a reference...but you have to have one side of the circuit connected to ground in order for ground to work as your reference. In the case of HT supplies it's always the negative side of the supply that goes to ground. Remember, voltage is relative to a reference, and the other side of the circuit is your reference for said voltage. By referencing the negative side of the circuit to "ground", ground then becomes your reference.

If you recall our original example circuit, drawing it up this way -



is the exact same as drawing it the original way -



Notice everything that goes to the negative side of the supply is grounded...because the negative side of the supply itself is also grounded.

[HTH]

did someone say STICKY? - this info is too good to slide off into old-thread obscurity. Mods!!!

[noworrybeefcurry]

So to the best of my understanding filter caps are polarized and when you introduce a high voltage positive current onto the positive side of the cap from your rectifier circuit the filter cap pulls electrons from ground into the cap towards the positive terminal so it can charge? But how does this work, electrons flow negative to positive and is ground not more positive in relation to the negative terminal of the cap?

Secondly what do you mean by the negative side of your power supply? Are the filte caps not running off of a tap from your rectifier circuit which is positive?

[flemingmras]

So to the best of my understanding filter caps are polarized and when you introduce a high voltage positive current onto the positive side of the cap from your rectifier circuit the filter cap pulls electrons from ground into the cap towards the positive terminal so it can charge? But how does this work, electrons flow negative to positive and is ground not more positive in relation to the negative terminal of the cap?

Everything is relative. The positive side of the supply is POSITIVE relative to the negative side of the supply. Since the negative supply is at zero relative to ground (they're at the same potential because the negative side of the supply is connected to ground), this makes ground more negative than the positive side.

Remember...ground is nothing more than a reference point. When the negative side of the supply is connected to ground it is said to be "referenced to ground". When you measure voltages in an amp, you most often reference the meter to ground and all positive voltages are relative to the ground reference. Since the negative side of the supply is connected to ground, you will see the full positive voltage of the supply when referencing your meter to ground.

Again the electrical measuring tape concept...all measurements are relative to zero, which is ground and to make the negative side of your supply your "zero", you connect it to ground.

The positive charge on the positive side of the cap is felt on the negative side, which pulls electrons from ground because ground is more negative than the positive side of the supply (zero is -450V less than +450V because (+)450V + (-)450V = 0 right?). The positive terminal and the negative terminal of the cap are internally isolated via a dielectric insulator between the positive and negative plates in the cap so current can't flow through the cap, but the effects of the positive charge that exist on the positive side are felt on the negative side just enough to pull electrons into the negative side.

In short...a capacitor is just like a "temporary rechargeable battery" that is constantly being recharged as current is drawn from it.

Secondly what do you mean by the negative side of your power supply? Are the filte caps not running off of a tap from your rectifier circuit which is positive?

You have two sides to a filter cap...the negative side and the positive side. On a full wave rectifier that uses a grounded center tap, the center tap is the negative side of your supply. On a full wave bridge rectifier, the two backwards facing diodes on the BR circuit is the negative side of the supply. This is the side of the supply that goes to ground.

Through ground, the negative side of the supply connects to the negative side of the filter cap, while the positive side of the filter cap connects to the positive side of the supply via the rec diodes.

[noworrybeefcurry]

Thank you, that makes so much sense. But when your filter cap is fully charged is the negative terminal showing zero volts or -450?