Quote:
Originally Posted by fltyrap
1. How do these radios step up the voltage (B+) that is used for the tubes? The radio I have uses a selenium rectifier. I have the schematic, but I can't tell what the output voltage should be (the first voltage shown after the rectifier and a subsequent resistor is 120v, but I don't know if this should be my target, or for that matter how 115v mains on the schematic becomes 120v after rectification).
|
First of all, take the voltages as listed on the schematic with a certain grain of salt. Your actual readings will probably be higher and will vary to a degree with variations in the supply voltage. Most mains voltage is higher now than it was when these radios were made. My AC power here runs between 120 volts but occasionally dips as low as 106!
How those power supplies output a
higher voltage than was seemingly fed into them was a mystery to me for many years until someone finally explained it in a way I could understand. I'll try to paraphrase their explanation here.
The actual value of an AC voltage is constantly changing. For example, 110 volt AC actually means 110 volts RMS. That means that the
average voltage of the constantly changing AC waveform is equivalent to 110 volts DC. Sometimes it's less, sometimes it's more. Sometimes it's zero volts and sometimes, it's at its peak voltage, at the top of the sine wave. The RMS voltage of a sine wave is equal to 0.707 it's peak voltage. A little math (or a peak reading meter) tells you that the highest voltage reached by the waveform of a 110 volt RMS sine wave (and only for an instant in time) is about 165 volts. The waveform spend it's time varying between 0 volts and 165 volts and it's average voltage equal about 110 volts.
When you convert that voltage to DC with a rectifier and feed it to a filter network, the capacitor in the filter network begins to charge from the DC which is pulsating between zero volts and 165 volts. If you drew no current from the power supply, the filter capacitor would quiclky charge to the peak value of the fluctuating DC feeding into it. When you operate the circuit, you load the power supply by drawing current. The less current you draw, the less the filter capacitor "gives up" to the load (and the better it filters out the ups and downs of the pulsating DC!) The more current you draw, the more the capacitor has to empty to feed the load. If the circuit is designed so that the load is fairly light compared to the storage capacity of the filter network and the ability of the rectifier to resupply it, then the capacitors will spend more time charging than discharging and will be able to charge to - and maintain - a higher voltage than if the load is drawing more heavily. If the load is greater, the output voltage will be lower since the filter capacitors must discharge to a lower level to feed current into the load. Since they are being fed with only a certain amount of power they don't get a chance to charge back up past a certain voltage level. This charge/discharge situation almost instantly reaches equilibrium at whatever DC output voltage the supply vs. load balances at. The circuit will be designed to achieve the desired DC B+ level with the intended load. That can be up to 165 volts for very light loads, but will never actually reach that level even with no load due to circuit losses.
Clear as mud? There are tutorials on the web that probably explain this better than I can.
Good luck,
--Dave