![2n3055 transistor ampeg 2n3055 transistor ampeg](https://320volt.com/wp-content/uploads/2008/10/amplifier-circuit-2n3055.jpg)
#2n3055 transistor ampeg series
The shunt voltage regulator sits across the load and acts as a bypass for current if the voltage gets too high (and generally has a small power resistor in series with the supply). This tends to be best suited to situations where the output current varies a lot (say a general bench supply), and has the characteristic of running almost cold when there is no current being drawn, and hottest when current is a maximum. (these also come in a negative voltage form as the 79xx series). These days the most common form would have to be the 78xx series, where "xx" is the output voltage, 7805 for 5V, 7812 for 12V and so on. This sits between the supply and the load and cuts the supply voltage down to the required load voltage. The most common form of voltage regulator is the series type. What is shunt regulation, and why is it well suited for a DC heater supply in a valve amplifier? If I was designing from scratch and using valves fron the 12Axx family, I'd specify 12 volt heater windings on the transformer which would allow rectification/regulation You could voltage double the 6.3 volts and then rectify and regulate that (but this will halve the supply current) so again you have compromises.
![2n3055 transistor ampeg 2n3055 transistor ampeg](https://i.ytimg.com/vi/IQ-qGAIT1o8/maxresdefault.jpg)
There are low dropout regulators available that have 10,000uf) and choice parts, but it may not work for everyday fluctuations in the AC mains. The regulator will shut down (ie not output volts, if the drop out voltage is not exceeded) Regulators require the input voltage to be higher than the output voltage to work.īasic regulators such as 78xx series (7806 for this case) require typically 2.5V so we have a problem. There are 2 diode forward voltage drops in a bridge rectifier to consider, (typically 0.7V per diode) so this brings your unregulated DC voltage down to 7.5V. When you bridge rectify, you multiply by 1.414 (square root of 2) to get your peak voltage, so 6.3 volts becomes 8.9 volts or so.