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So I can buy a Positive Linear Regulator rated for -60V input (LM317HV). But the Negative Linear Regulators are -35v (LM7909) range. Yes there are a few odd parts that are -40v or -50v, but none I've found are -60v.

My question is, can Linear Regulators be serial connected to increase the operating range? or is there are simpler way to increase the Input rang I've missed?

My thought was having the first regulator inline with a Zenor Diode (25V) to get it to handle the raw 60V and feed its output (under -35v) to a second to produce -9v. It would have to scale or the first NLR would drop out as the voltage sags to -12V so that the circuit and still produce the -9v.

Specifically I'm looking for -9v from an input of range -12 to -60. And in the 10mA range, which would dissipate 500mW

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  • \$\begingroup\$ You have to use a part that is specified for the input voltage. The voltage regulators share a common ground, so you cannot "stack" them to increase the input voltage range. It may be easier to use a positive regulator for the higher input voltage range, then use a small switching regulator (caps-only for example) to create a negative voltage from the positive one. \$\endgroup\$
    – Janka
    Jan 24, 2017 at 21:47

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You could use a transistor zener regulator to drop most of the voltage, and put it in series with a linear regulator. A basic transistor zener regulator looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The output voltage is set by the zener. Q1 must be rated for the full Vin voltage, and for dissipating a power of Vin * Imax, so it may need a heatsink. R1 is chosen by trading off the current and heating at high Vin against having sufficient current for effective regulation at low Vin. For a negative regulator, turn the zener around and swap the transistor for a PNP.

The design can be improved in several ways, such as by adding a capacitor in parallel with the diode to give better regulation at higher frequencies. Or using a MOSFET for larger currents and/or lower dropout voltage. Ultimately, this kind of circuit is not going to be as good at regulating as a linear regulator, but that doesn't matter if you have a regulator too. And it's pretty cheap.

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  • \$\begingroup\$ P = (V_In - V_Out) I_Max and V_Out = V_Z - 0.7V \$\endgroup\$ Jan 24, 2017 at 23:08

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