I need to build a regulated 48VDC ~1amp power supply.

I have two transformers on hand: One with a 36VAC secondary which would give around 51 volts DC once rectified The other with a 44VAC secondary which would give around 62 volts DC once rectified

Now, feeding the LM317(HV) with 51 Volts DC doesn't give enough headroom to my taste if we consider the regulator voltage dropout and house voltage variations.

Feeding the LM317(HV) with 62 Volts DC would exceeding regulator's input-to-output max voltage.

While searching the internet for solutions, I found an application note on TI Website about "Maida High Voltage Regulator" (floating regulator).

Maida Regulator, Figure 2 of http://www.ti.com/lit/an/snoa648/snoa648.pdf

Now my questions:

  1. On the schematic, Vin is 170Volts. Can this schematic be used with much lower voltages (let's say 60VDC)
  2. According to the schematic, Output current is 25mA but I need around 1 Amp. Do I need to modify something to get 1 Amp output ?
  • \$\begingroup\$ The 170V input is required because the output is adjustable up to 160V. If you don't need that, you don't need such a high input. \$\endgroup\$
    – user16324
    Commented Jul 15, 2016 at 19:27
  • 1
    \$\begingroup\$ 1 ampere is a lot for an LM317 no matter what you do. Even with a 3 volt difference, you will dissipate 3 W. \$\endgroup\$
    – pipe
    Commented Jul 16, 2016 at 8:46

1 Answer 1


First, let's figure out how the circuit works.

enter image description here

Figure 1. The circuit works by dropping almost all the voltage across Q2.

This application is using the circuit built around Q2 as a pre-regulator.

  • R1 and D1 form a Zener voltage regulator which keeps the base of Q1 at 6.2 V higher than Vout.
  • Q1 and Q2 form a Darlington voltage follower. The voltage on the output of Q2 will follow that on the base of Q1 less two diode voltage drops.
  • The voltage between the LM317 input and output will be restricted to 4.8 V. As more current flows R3 will cause the voltage to drop even further. At 25 mA the drop will be 2.5 V giving the input voltage as Vout + 2.3 V. At this point we will be approaching the minimum headroom required for the LM317 to regulate.
  • The end effect is that the LM317 never has to dissipate much power and probably works without a heatsink. \$ P_{MAX} = VI = 2.3 \times 25m \approx 60~mW \$.
  • Note that Q2 is working much harder. If the output is set at 20 V and 25 mA is drawn then Q2 will dissipate about \$ VI \approx 150 \times 25m = 3.75~W\$. Heat sink required.

On the schematic, Vin is 170 volts. Can this schematic be used with much lower voltages (let's say 60 V DC).

Certainly. Recalculate the resistance (and power) required for R7 to achieve your maximum output voltage.

According to the schematic, output current is 25 mA but I need around 1 amp. Do I need to modify something to get 1 amp output?

Yes. As explained above R3 is designed to limit the current. For 2.5 V drop at 1 A R3 would be 2.5 Ω. The problem will be power dissipation in Q2. You would probably have to parallel several transistors and put resistors (maybe 0.1 Ω) in each emitter to balance out the currents.

Do the power calculations and you'll start to see the benefits of switched mode regulation.

  • \$\begingroup\$ In your final section, do you mix up R1 and R3? \$\endgroup\$ Commented Jul 15, 2016 at 23:26
  • \$\begingroup\$ Thanks. I must have lost concentration at the end. Fixed. \$\endgroup\$
    – Transistor
    Commented Jul 16, 2016 at 8:25

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