No! B+ connected to one of the leads of the output transformer. The other lead of the output transformer is connected to the 26 plate. Follow the schematic that Andy posted!
Have you got a 150v glow tube e.g. 0D3 from B+ to ground? As in my schematic? If you do and you have the DN2540 set for 32.5 to 35mA, then the 0D3 should light up pink and you can only have 150v on the top of the LL1660 and 145v on the plate. Follow the schematic and the voltages should be correct!
Your wiring diagram for the LL1660 is fine - doesn't really matter that the output is reversed or not. If you have wired the LL1660 as in your schematic it will be correct.
Your wiring diagram for the LL1660 is fine - doesn't really matter that the output is reversed or not. If you have wired the LL1660 as in your schematic it will be correct.
Hi Merlin, you need at least that dropping resistor from your B+. Look back at the values given by stajo. So from B+ into the resistor, other end of resistor decoupled to ground (with the 220uF capacitor you mentioned before), and this voltage goes to LL1660, and other end of LL1660 goes to plate.
Suerte hermano, pero no haga (o pense )cosas que nuestro hermano Argentino no aprobaria (good luck brotther, but dont do, or think, thinks that our Argentinian brother wouldn't aprove).
Erik
Suerte hermano, pero no haga (o pense )cosas que nuestro hermano Argentino no aprobaria (good luck brotther, but dont do, or think, thinks that our Argentinian brother wouldn't aprove).
Erik
Im not sure since Im not familiar with Rods board, but you could try replace it with a 1,2 V battery maybe. For functioncheck of the rest of the circuit.
So, back, if you have 307V B+, 5mA through the #26, and wish to have around 150V at the #26's plate, calculate value of dropper resistor as follows.
R = V/I, where V is value to be dropper (157V), R is value of resistor, and I is value through the #26. So, if V is 157V, and I is 0,005A, R is = 31k4. But, is the PS shared? If you are using the same 307V PS for both channels, total I is 0,01A, and R is therefore 15k7 (15K will do).
Good luck! Erik
R = V/I, where V is value to be dropper (157V), R is value of resistor, and I is value through the #26. So, if V is 157V, and I is 0,005A, R is = 31k4. But, is the PS shared? If you are using the same 307V PS for both channels, total I is 0,01A, and R is therefore 15k7 (15K will do).
Good luck! Erik