LM317 - Voltage regulator

Question

I want to drive a board with 5V and 3.3V but my input voltage is around 40VDC. The board will draw around 750mA. If I check the voltage on NODE1 without the 7805, I've got 28V. With the 7805 connected, the voltage drops to 1.6V on NODE1 R2 is a 10K POT to adjust the voltage. I used LTSpice to simulate the circuit and when the NODE1 draws:

800mA --> 28V 840mA -->

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

Issues:

1. The LM7805 ground should be to the voltage source ground
2. 40V - 5V out = 35V you want to drop, at 750mA this is 26.3W, way too much for these regulators even with heatsinking.

If you want 5V, I recommend you use something like a R-78C5.0-1.0/X9, then using a normal LDO to get 3.3V from that assuming that both the 5V and 3.3V load use less than 1A.

Answer 2

The function of LM317 is always maintain fixed V OUT-ADJ=Vref in order to design an internal negative feedback loop to raise the OUT voltage.

Vref is too low to drive 7805 and the xx_uA current is also just enough to drive the two series Resistors connected to ADJ and not any significant load.

Answer 3

It is not clear what your question is but there is something you can do to improve the design. I recommend adjusting R2 to lower the node 1 voltage below 24 V, because that is the limit on some 7805 datasheets.

For better reliability, adjust R32 such that the power dissipated in both regulators is equal at 750 mA output current. The current through each device is 0.75 A, the voltage is

((30 - 5) / 2) V

If the problem is that the node 1 voltage drops too much when the 7805 is connected, NOTE that the pinouts for the two regulators are significantly different.

And the equation is

P = E x I