Gibson LFO circuit

corliss1

Poster Extraordinaire
Gold Supporter
Joined
Sep 13, 2008
Posts
6,159
Location
Lansing, MI
Hey crew! I'm looking at a vintage Gibson GA-20RVT. Yes, from the garbage era of Gibson amps. I can't figure out what is supposed to trigger the LFO. In a Fender, such as the Deluxe Reverb, there is a bit of voltage taken from the bias supply to the grid of the trem tube, that kicks off the oscillation. But in the Gibson...I'm not seeing how any voltage gets to the grid. Any ideas are welcome. Images of the specific areas under question attached.

DRRI:

drri_lfo.png


Gibson GA-20RVT (it's not a 12ax7, so don't panic about the tube pin numbers :D Also, it's "v5" on the left and "v3" on the right - the writing is just sloppy):

gibson_trem.png
 

corliss1

Poster Extraordinaire
Gold Supporter
Joined
Sep 13, 2008
Posts
6,159
Location
Lansing, MI
Found a different version of the schematic, and it only shows half a volt on pin 5, so there likely shouldn't be as much there like a Fender, so I guess my lack of voltage is expected:

Screenshot 2023-01-23 at 23-50-42 20210810_231827.jpg (JPEG Image 2576 × 1932 pixels).png
 

corliss1

Poster Extraordinaire
Gold Supporter
Joined
Sep 13, 2008
Posts
6,159
Location
Lansing, MI
Replaced the three caps in the LFO and now we're good to go :D So all that for that.
 
  • Like
Reactions: BB

Jon Snell

Tele-Afflicted
Joined
Aug 31, 2015
Posts
1,563
Location
Jurassic Coast, Dorset. Great Britian.
Explanation of a simple R C Oscillator.

Anode on a cold valve is HT potential.
Valve warms up and voltage tries to go as low as it can, due to the bias configuration.
Pin 6 goes low, C19 has HT stored and travels south with the anode BUT 90° later due to phase shift. The 90° lag then moves C20 through its 90° lag in turn moving C21 through a 90° lag causing the grid Pin 5 to go negative. That is 3 X 90° which is 270° so far.
This "switches" the valve off, with an inverting action and that becomes the final 90° lag, returning the Anode to HT and the process starts again.
4 X 90° = 360°.
Simple R C or phase shift oscillator.

The -.59 volts measured on pin 5 is due to grid leak.
In the Fender design, the grid is at zero volts due to the cathode being raised through the cathode bias resistor.
Both work in exactly the same way.

All that is required is a Mu of at least 15 and it will work or for a transistor, a gain or Hfe of at least 50. A transistor of course, requires much less voltage and no heater.

If one capacitor is leaky or open circuit, one of the 90° phase shifting stops as does the oscillator.
The speed control is normally fitted between the 2nd and 3rd phase shifters, this varies the R C point and therefore the frequency of the oscillator.
 
Top