Circuit explanation.

[slash1986]

Hi, I have a question. Can you explain to me what happens to the signal after it has been amplified and has passed the decoupling capacitor C1. What type of route does it take and why? what type of RC filters are generated? and Why the Gain is reduced to millivolt at the output from the potentiometer.
Thank you.

[shooter]

read these til the lightbulb lites 
http://www.aikenamps.com/index.php/white-papers

[WimWalther]

Where is the output taken? It's not indicated

[slash1986]

Where is the output taken? It's not indicated

Hi, the output is on the centre tap of the potentiometer.


After my readings I think this: the 470k resistor and the potentiometer make a voltage divider. The 470p is a bypass cap for the high frequency??? The 1n is a potentiometer bypass and allows more frequency to go out without go to ground right?

[HotBluePlates]

... Can you explain to me what happens to the signal after it has been amplified and has passed the decoupling capacitor C1. What type of route does it take and why? ...

NEETS Modules 1, 2 and 6.

AC Voltage is present at the output of the coupling cap, and that voltage is impressed across the path through R6, the Pot and Ground.

The impressed voltage causes a current through these parts, and that current (and the resistance of the parts) results in "voltage drops" across the components in a manner that "divides the voltage."

"Current" goes somewhere around the circuit; "Voltage" mostly is/isn't, as a result of current & circuit resistance.

Our desired "signal" is the resulting Voltage, that doesn't really "flow through" the circuit.  Wherever the desired signal-volts are, we take a sample and pass that to the next bit of circuitry.

Mostly because we don't need all of the voltage present at the output of the coupling cap, we choose a lesser amount somewhere along the resistive track of the pot, and pass that sampling-point as output to the next circuit. (You have "L" for low and "H" for high drawn backwards on your pot; "H" is at the zero-output point)

The caps change the "apparent resistance" as frequency changes.  High frequency sees less resistance, experiences less voltage division, and is boosted in the output picked off the Pot-wiper.

[slash1986]

Thanks, the explanation is very clear. does the gain loss at the output of the circuit therefore depend on the frequency of the alternating current when we have capacitors? high frequencies bypass the voltage divider right? How can I calculate from what point in the spectrum the 470p capacitor starts to work?
Thanks.

[acheld]

You can calculate all that with tools for passive RC filters. 

Here is a godawful looking URL to Digikey's online calculator.   

Doesn't SPICE do this for you?

https://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-low-pass-and-high-pass-filter?utm_adgroup=Filters%C2%A0&utm_adgroup=&utm_source=google&utm_medium=cpc&utm_campaign=PMax_DK%2BProduct_Product%20Categories%20-%20Top%2015&utm_term=&utm_content=&utm_id=go_cmp-19646629144_adg-_ad-__dev-c_ext-_prd-_sig-CjwKCAiAkp6tBhB5EiwANTCx1HquchH9SPtrUnGBjHngny4Ohw0Lpaca3Lfzg1-UgdEbwrcxPFk0eRoCefcQAvD_BwE&gad_source=1&gclid=CjwKCAiAkp6tBhB5EiwANTCx1HquchH9SPtrUnGBjHngny4Ohw0Lpaca3Lfzg1-UgdEbwrcxPFk0eRoCefcQAvD_BwE

[WimWalther]

Try not to get into the habit of using an RC calculator. The basic formula is very simple and you ought to learn it:

Fc = 1/(2Pi*R*C)

Fc is the corner frequency, aka -3dB point (half of initial level).

2*Pi = ~6.28

R = Resistance in Ohms

C = Capacitance in FARADS.. not uF, nF or pF. If you wish to use those units, then you must multiply the result by 10^-6, -9 or -12, respectively.

Until you're comfortable with the math, definitely do use an online calculator to check your work. As your experience and confidence with the math grow, you can and will abandon the calculator apps.

[shooter]

it's usually not the math that hangs one up, it's that reducing a complex circuit down to simple, R & C & L, the rules to make that happen, so when you're to the point of math, there is no schematic left, just lines n circle things with arrows

[PRR]

> usually not the math that hangs one up

A LOT of people today can't math a Walmart check-out.

I had to force a Five back at the clerk at Mabel's because he over-changed me.

[WimWalther]

Where is the output taken? It's not indicated

Hi, the output is on the centre tap of the potentiometer.

The wiper, you mean to say?

[HotBluePlates]

... does the gain loss at the output of the circuit therefore depend on the frequency of the alternating current when we have capacitors? ...

It is wise to avoid calling this "gain loss."

   - No gain was lost: all the gain is at the tube's output, same as without the caps.

   - The caps C2, C3 bypass parts of a voltage divider.  It's as though "Z1" in the link was made smaller when frequency goes up.  Less voltage-division, output increases; tube gain has not changed.

   - Normally, R6 is like "Z1" while the Pot is like "Z2."  Except the pot is its own voltage-divider, and when you turn it down Z2 gets smaller while Z1 gets larger.

... How can I calculate from what point in the spectrum the 470p capacitor starts to work? ...

WimWalther provided the formula.

You would also be wise to read those NEETS Modules I linked, because you will need a basic understanding of how R and C (at least) behave in DC and AC circuits to make sense of anyone's reply.

[separateness]

I could probably not add better explanations than those given but I will say that I have found this online tool useful for understanding some filters in amplifier stages: https://falstad.com/afilter/

I drew up the relevant part of your schemo (ballparking the output impedance of the tube as 40kΩ) and below is what it looks like with the pot halfway. Tools like this are no substitute for understanding the underlying math and principles at work but they can be useful aids.

[acheld]

Thing is, OP is already using a high end tool (SPICE) to describe his circuit.

SPICE will calculate the frequency response, phase shifts, etc w/o any additional add-ins or tools.

[Lectroid]

> usually not the math that hangs one up

I had to force a Five back at the clerk at Mabel's because he over-changed me.

And forget giving a clerk enough extra change so that you can get back a flat amount, like a $5 bill,  instead of say, $4.39.  (Invariably handed over in a crumpled wad, mixed up with change in the same hand.)  One guy got angry with me for doing it once, I think maybe because I exposed his innumeracy. 

Arghh, get off my lawn!

[shooter]

I had to force a Five back at the clerk

same, almost; told the clerk "you gave me an extra 20" clerk argued they didn't, I went to the bullet store 
the "dark ages" has always baffled me, from "enlightened" to completely illiterate in all things, now I see how it's done

[WimWalther]

What can one say? Good help is hard to find..

[tubeswell]

...what happens to the signal after it has been amplified and has passed the decoupling capacitor C1. What type of route does it take and why?

After the signal has been generated in V1, it immediately faces several loads (signal impedance). Think of signal impedance as having the 'lowest impedance' at the signal source (plate of V1), and everything around that causes a load. The further the signal is away from the plate of v1, the more 'loaded down' it gets. The circuit is like the strands of a spider's web where the source of the signal (plate of V1) is the centre of the web, and the absence of signal occurs at the edge of the web where each web strand is anchored to a fixed point (these 'fixed points' occur at the signal ground return, and at the High Tension supply (which is hypothetically immoveable). If you push your finger on the centre of the spiders web, the web moves back and forth more easily than if you try to push your finger on the edge of the web. The plate resistor itself is one of these spider web strands - the further away you are along the resistor from the plate of v1, the smaller the relative signal becomes, until you get to the HT supply, where there is no voltage swing at all. Similarly, the coupling cap and the other resistors and capacitors following it, form a series of signal voltage dividers that load the signal. The further you are away from the plate of V1 along these various filter, the smaller the voltage swing is, until you get to the ground return, where there is no voltage swing. Also, the tube itself presents a load to the signal at V1. This is through the plate resistance of V1 and any unbypassed cathode resistance at the cathode of V1. All these various loads are in parallel when viewed from the PoV of the signal source at the plate of V1, and this determines the output impedance of the tube. And the output impedance of V1 itself forms the 'upper leg' of a voltage divider with the input impedance of whatever stage follows V1. Say the output impedance of V1 added up to 220k at frequency zHz, and the input impedance of the following stage was 1M; then Vout at frequency zHz = 1,000,000/(220,000 + 1,000,000) = 81.96% of Vin. What's more, the input capacitance of the following stage forms a filter with the input impedance of that stage, which determines the bandwidth of the signal seen at the input of the following stage - and so on it goes.

[d95err]

Comment about the SPICE simulation settings:

You are only simulating 10 ms. Sometimes, it takes a while before the circuit stabilises.

I recommend running the simulation longer before analysing the signal. I typically do at least 1 second, unless that takes forever (which can happen sometimes if you push things into clipping)