> > First, a few quick "whys": > > 1) The 7805 is actually a Motorola MC7805CP, date > code 7308 with gold leads. Very hard to exactly > replace.
Any reason why it would have to be an exact replacement? In any case, the behaviour of the 7805 if you apply a voltage to the output with the input floating may well depend on the manufacturer and even the date (some devices were improved over the years). Unless you have a 1973-or-so data sheet from Motorola, I don't think you know whether it will be damaged or not. [...] > 1) I have a 12 volt DC supply. 12 volts seems to be > within the VIN range for the 7805s whose data > sheets I've now read. Can I simply apply 12 volts? Yes, but... The power disipated in the 7805 will increase, in fact it will be more than doubled. To put it crudely, a linear regulator acts like an automatic variable resistor. I have no idea what current the load takes, let's call it I. If you supply 8V, then the power disipated in the 7805 is 3*I watts, if you supply 12V it's 7*I. This may or may not be a problem. > 2) Could I place a resistor in series between the 12V > supply and the 7805 to drop the voltage at the 7805 > to somewhere around 8? Yes. You need to know the maximum current the load will draw, which will be much the same as the current drawn from the PSU. Then just calculate the resistor to drop 4V at that current. If you can find one, you could probably use a 7808 to supply 8V to the unit from a 12V supply. Or a 7805 'jacked up' with a 3.3V zener diode (in series with the common lead to the extra 7805 only). My guess is that giving it 12V will be fine though. What is the device, and do you have any idea how much current it is going to draw? -tony
