Hello, all!
This'll be my first post, so my apologies if this question has been asked/answered elsewhere already.
I'm an enthusiastic newb to DIY audio. I can solder and have upgraded household circuits without incident, but since I have only a beginner's knowledge of audio electronics, I've been spending time with the manual before Crack kit ships... as well as trying to better understand what those electrons are getting up to while they move through the circuit.
I think I correctly understand how a triode works: The heater causes the cathode to emit electrons; the B+ voltage attracts those electrons to the plate; current flows thru the tube and gets modulated by AC signal applied to the grid. But something I noticed in the Crack's diagram has me scratching my head:
On most triode schematics I've seen, the lead from the power tube to the output jack is connected between the B+ rail and the tube's anode. On the crack, that output lead is connected between the cathode and ground rail. Why is this? Does it not matter where that lead connects, as long as current is flowing in the tube and a coupling capacitor blocks DC at the output?
Or is there some other reason? Thanks for any insights!!
This'll be my first post, so my apologies if this question has been asked/answered elsewhere already.
I'm an enthusiastic newb to DIY audio. I can solder and have upgraded household circuits without incident, but since I have only a beginner's knowledge of audio electronics, I've been spending time with the manual before Crack kit ships... as well as trying to better understand what those electrons are getting up to while they move through the circuit.
I think I correctly understand how a triode works: The heater causes the cathode to emit electrons; the B+ voltage attracts those electrons to the plate; current flows thru the tube and gets modulated by AC signal applied to the grid. But something I noticed in the Crack's diagram has me scratching my head:
On most triode schematics I've seen, the lead from the power tube to the output jack is connected between the B+ rail and the tube's anode. On the crack, that output lead is connected between the cathode and ground rail. Why is this? Does it not matter where that lead connects, as long as current is flowing in the tube and a coupling capacitor blocks DC at the output?
Or is there some other reason? Thanks for any insights!!