
Bon, j'ai sur Toulon un magasin qui fait des tube NOS. Dans le stock il a des trucs genre telefunken. Je vais voir suivant le prix.
si ton ampli sonne encore brillant et péchu les lampes sont en pleine formetbo a écrit : Moralité, court jus en sortie et lampes qui n'aiment pas du tout, genre de couleur orange ....
- J'ai une couleur bleu qui apparait dans les lampes ?? Est ce un rinçage des lampes dû à ma précédente erreur de manip ?
Alors ça va, les lampes sont encore bonnes.si ton ampli sonne encore brillant et péchu
Oui, il s'agit bien d'une grille. Je penche pour la grille 2 (j'ai du mal à les différencier dans la lampe).la mise au rouge des filaments ??? tu parles du filament qui rougit normalement ou des grilles (les spirales entre le filament de chauffage et la plaque).
A voir à tête reposée (faut quand même faire passer quelques dizaines de mA)Q: I understand you can put a 50V reverse-biased zener diode in the center-tap of a power transformer and lower the output voltage by 50V. Exactly how does this work?
A. A zener diode is simply a diode that has a designed-in "breakdown" reverse voltage that is fairly tightly controlled. In the forward direction, it is a "normal" diode, with a voltage drop of around 0.7V. If you drive current through it backwards, it will drop a voltage equal to the zener voltage. This makes it useful as a level-shifter. Note that "regular" diodes also have a breakdown voltage, and will conduct in the reverse direction if you exceed that voltage. If you put a 50V reverse zener in the center-tap <>of the power transformer, it will conduct on the filter capacitor charging pulses and make the center-tap <>more negative than ground by 50V. Since the main rectifiers can't conduct until the voltage is higher than the voltage stored on the filter capacitor, conduction will be delayed until the voltage is effectively 50V higher than it was before you put the zener in, which in turn reduces the output voltage by 50V.
Another way to look at it is this: it makes the center-tap negative by 50V during the capacitor charging pulse. Since this is negative with respect to ground (and the output voltage), it "subtracts" 50V from the output voltage. Either way, the "real" total voltage differential between B+ and the center-tap remains the same, however, since we have level-shifted the center-tap negatively by 50V, the output voltage is lowered by 50V. The drawback is that you don't get something for nothing. In this case, the price you pay is in power dissipation. The voltage dropped across the zener multiplied by the current through the zener results in a power loss that is dissipated in the zener as heat.