Two LM317 Based Voltage and Constant Current Regulator Circuit

Question

I have read some articles on on the following circuit. I was wondering if I can create such a Voltage and Constant Current Regulator for my Power Supply or not.

My Question is - is this circuit Practical? And will 1 Watt power rating's Variable Resistances and 2 Watt Ratings Resistances work?

Here the first LM317 is used as Current Regulator and the Second one as Voltage Regulator. I was also wondering if this ordering is correct, or if this has to be reversed - i.e. first voltage regulator and then Current Regulator to make it work better. Please help.

PS: I need this circuit to regulate Voltage and keep the output current to a configurable Constant level.

Answer 1

This really isn't a good solution. You have two regulators, each with their own voltage drop and power loss, full load current running through a potentiometer and inability to reduce output voltage to zero. It would be much better to get a proper design using one output stage with voltage and current limiting.

Figure 1. A two LM317 solution for those who insist. Source: ON-Semi datasheet.

See Smartest way to use current limit using LM317? for a full description of a working solution to some of these problems if you wish to continue with LM317s for this application. I give a detailed explanation of the circuit operation in that answer.

Answer 2

is this circuit Practical?

It's nearly practical - the main problem is that you haven't got a decoupler capacitor on the input power to the 2nd regulator in the chain. This will probably cause instability under certain load conditions.

Put another 1uF capacitor at the input pin to ground/0V.

Regarding the power ratings - the 100 ohm resistor is only going to "see" 1.25V between Vout and ref therefore the power is only about 16 mW.

The potentiometer (VR2) is subject to the same current as the 100 ohm i.e. 12.5 mA. Maximum power is when the potk is fully extended to 1k i.e. 160 mW.

VR1 is going to be a problem as per Wouter's answer.

Answer 3

There is a problem with VR1. At maximum current, the circuit can deliver 1A. The wiper of a normal potentiometer is not rated for such a current.

Calculation: regulate down to 0.5A. That will use 1.2 Ohm of VR1's resistance, in which 0.3 Watt will be dissipated. The full VR1 can easiliy handle this, but the very small part of VR1 you are using in this (0.5 A current-limited) setting lilkely won't. (1.2 Ohm is 0.1% of 1k)

Also note that the scale of VR1 will be far from linear: the above calculation shows that the 1A is fully right, 0.5A is 0.1% back from full.

The 100uF capacitor feels a bit low for my liking. IIR the rul;e of thumb more like 1000uF per 1A.

When ser for 1A and shorted, your output will be 0V, and the first LM317 will take the full 1A current. The input voltage can be sqrt(2)*12 ~ 17V, so the chip will dissipate 17W. That requires some good cooling. (It can be a bit better because the diodes have some drop, but it can get worse due to tolerances in the trafo and the mains voltage).