In an attempt to repair a presumably faulty preamp I realised the construction deals with two voltages; +14 and -14 (as well as ground). I don't have access to the transformer unit, so thought I could construct my own.
Am I correct in my assumption that what I need is a 28 volt power source, and to connect two resistors in series across the two leads, and I'll end up with +14 volts on the positive lead, -14 on the negative, and a 0 volt "ground" where the two resistors join? Kinda like this (hoping the ASCII-schematic looks like it's meant to)..
Using a resistor pair is not a particularly good idea for any significant current (significant being more than a few milliamps)
A better solution might be to use about 36V in total across a centre tapped transformer. i.e. each side relative to the centre tap has 18V above the ripple after rectification and *under load*. Use the centre tap as your local common (ground).
Then use Linear regulators (if you are less than 1A and you stay close to the drop out - about 2.5V each - then the heat sinking won't be too bad).
I know the drop out is a little highly spec'd here, but that's standard for the old style 78xx beasts (and various others). A low dropout regulator has it's own very special problems that will bite you in the behind.
To get the zero volts, you'd need R1=R2. The thevenin equivalent would then be a 0V source through an R of (R1)/2. To keep the variation of your "ground" to, let's say, 1/2V about ground at up to one amp you'd need to have (R1)/2*(1A) = .5V. This means R1 of 1 ohm.
So, R1 and R2 would each need to be at least able to handle 200 watts. Just in case, make them 1 kW resistors. Your 28V supply should need to support about 15 amps. About 500 watts worth. This would then supply your up-to-1A pre-amplifier, assuming it could tolerate the .5V variations in the ground.
You will probably want to avoid that solution, though.
If you do this, it'll keep GND pretty much at 0V, even if the impedances of L1 and L2 are not equal.
Note that the current through R1 and R2 should be about 1/10 of what the maximum current through either transistor is. This will depend on how different L1 and L2 are. If L1 is not there, for example (an infinite impedance) and L2 is 14 ohms, then it'll draw 1A through the NPN transistor, and no current through the PNP transistor. So, you'll need about 100mA through R1/R2, meaning R1 and R2 should be
140 ohm 2W resistors. The Transistors should be beefy, since they may be dissipating 14W. Big TO-3 cases and heatsinks are probably a good idea.
If you use an opamp to drive the transistors, with feedback from the virtual ground, it'll be far more precise. Something like this:
With this circuit, your virtual ground would be far more accurate, and the resistors R1 and R2 can be large, like 100k. Make R1 = R2, and Rf = R1/2. That way, your opamp inputs will see similar impedance.
An even better way would be to just generate the right voltages to begin with. You can get 'center tapped' transformers, which provide AC voltages which are mirror images of each other, and a ground. If you use one of these, you can generate +-14V fairly easily.
For example, assuming a 1A current draw on either or both, you'll need a 28VAC transformer that is rated for something like 35 VA. Using this, use the following circuit:
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