Couple of questions about the Fender 5F1 schematic - Page 1
Forum section: electronics / beginners

author: dcbrown73
author_rank: Regular Contributor
timestamp: March 06, 2021, 10:43:42 pm
content: Hello,I intend to build both the 5E3 and the 5F1 Fender Tweed Champ / Custom Deluxe amp kits, but before I do. I was hoping to ask a few question about the schematic so that I can better understand what is happening.Here is the full 5F1 schematic for reference.At the signal input, there are arrows that throw me off.What exactly are the arrows (circled in red) implying? I've always seen arrows like this as specifying inputs like for a chip or something vs outputs.On input 1, we have what appears to be a direct path to ground (unless the arrow means something else) and a second path with a 1m resistor to ground. (circled in blue) before hitting the 68k resistor. Does this just attenuate the signal and if so. Why not just use a resistor like the 68k already in place?Finally, I would like to understand the the intent of the capacitor in parallel with the 470 resistor that is biasing the 6V6 tube shown below. (ie what / why / how it works)

author: Andy Watson
author_rank: Super Contributor
timestamp: March 06, 2021, 11:20:32 pm
content: The arrows represent contacts (switches if you like). Insertion of the jack-plug causes the contacts to open and the signal passes through the jack on the contact represented by the kinky line!If you plug into jack 2, the arrowed contact is disconnected and the signal passes through to the potential divider formed by the series 68k resistors - note the lower end of this chain is grounded by the contacts of jack 1. So, your guitar or other instrument "sees" an input impedance of 132k - quite low for a guitar.If you plug into jack 1, the signal also passes through the contacts of jack 2 such that both 68k appear in parallel. In this case the instrument sees only the 1meg to ground - gives the input a "hotter" sound (so I'm told).The capacitor in the cathode is there to increaseACgain. To a first approximation the gain of the stage is set by the anode resistance (impedance) divided by the cathode resistance. More gain can be obtained by reducing the value of the cathode resistance, however, this would mess-up the DC biasing. Putting a capacitor across the cathode resistor presents a lower impedance to AC signals - therefore more gain.

author: Circlotron
author_rank: Super Contributor
timestamp: March 06, 2021, 11:38:21 pm
content: If you remove that capacitor the circuit will still work, but with lower gain, as has been mentioned. If the signal voltage on the control grid of that valve is positive-going at any particular instant then the cathode current increases, as you would expect. If the capacitor is removed, this increase in cathode current causes the cathode voltage to rise also and consequently the grid to cathode signal voltage is not as great as it would have been if the cathode voltage was held steady by the capacitor. This is actually a form of negative feedback. You would need somewhat more grid drive to get the same output to the speaker but presumably distortion would be lower.

author: Renate
author_rank: Super Contributor
timestamp: March 08, 2021, 06:48:04 pm
content: I consider that input jack stuff a bit overblown.The idea is that if you have one guitar you use jack #1.If you use two guitars you use jack #1 & #2.Jack #2 (with the goofy short on jack #1) was never intended to be used by itself.I'd have moved the 1 M to the right (to the grid) and skipped all the separate contacts.

author: Urshurak776
author_rank: Contributor
timestamp: March 08, 2021, 07:59:57 pm
content: Highly recommend this article specifically on how tube amps work using the 5F1 amp.https://robrobinette.com/How_Amps_Work.htm