# An LM350, can I have both constant current, and specific voltage?

I have a regulator such as in this setup to provide 1.2 Amps to a load that requires constant current:

R1 is set to 1 Ohm, which nicely brings the current within range (1.25A according to a small calculator below the image), but what I do not understand is that it speaks of a 1.25V reference. Does it output 1.25V only, or does it output the input voltage (minus) voltage drop (vdrop maybe 3V) and the 1.25V is just specific to ADJ?

Can I output voltage to, say, 2V as well as 1.2A constant current with just one LM350 device?

The datasheet for the LM350 I will be using is the following: http://www.ti.com/litv/pdf/snvs772a

In constant current mode, it will output whatever voltage is necessary to push (in your case) 1.2A through the load.
This will be limited by the supply voltage (minus a bit for the regulator drop) as obviously it can't output a higher voltage than goes in. So you can't expect it to put 1.2A through an e.g. $1 M\Omega$ resistor unless you have a supply of 1.2 Megavolts handy :-)

Foe example, if you have a 1 ohm resistor as the load, the voltage at the load will be:
1.2V ($1.2V \div1 \Omega$ = 1.2A)
If you have a 5 ohm resistor as the load, the voltage at the top of the load will be 6V ($6V\div5\Omega$ = 1.2A)

You can't set it up for constant current and constant voltage, as to keep one constant requires varying the other.
However, with a static load in constant current mode you can set it to drop 2V - e.g. $2V\div 1.2A = 1.6666\Omega$ resistor needed. So if you have a load of $1.6666 \Omega$ on the output of a constant current of 1.2A, the voltage will be 2V.

• Hah! There are a lot of 5W+ 1.5-2.0 resistors at Digikey I could use. So maybe I can assume.. 5V input is required for the LM350 for this to work? 3 drop leaves 2V, I think I can get this on my own, not sure of what to calculate in my head at the moment though. Thank you. – Kenny Robinson Feb 10 '12 at 0:58
• The input supply doesn't matter as long as it's high enough to allow for the minimum drop for regulation (stated in the datasheet, say at least a couple of volts for a standard reg) All you need is the right load resistor, which in this case is 1.6666 ohms and you will get a 2V drop (with 1.2A). The input voltage can be anything > 5V to whatever the max voltage is for the regulator. Obviously you need to account for power dissipation as Photon mentions - the higher the input voltage above the output voltage the more power dissipated by the regulator. – Oli Glaser Feb 10 '12 at 1:04
• Thank you, I am very satisfied with my project planning now. – Kenny Robinson Feb 10 '12 at 1:06

It's a 1.25 V reference because the circuit will operate to keep the ADJ pin 1.25 V below the OUT pin. The OUT pin will be driven to whatever it takes (up to Vin - 3 or so) to force the ADJ pin to be 1.25 V below Vout.

To answer the question the way you phrased it in the title, you can't get the power supply circuit to force both a fixed current and fixed voltage at the same time. That wouldn't leave any freedom for the load to affect the circuit. For example, if you have a 1 Ohm load, you need 1 V across it to get 1 A. If you have a 2 Ohm load, you need 2 V across it to get 1 A. If the supply circuit tried to force both at the same time, you'd have a logical contradiction on your hands.

Note that to get 1 A output you will need to be sure to pick the right package type for your LM317 and also you will probably want to use an external heat sink (or a good thermal connection to your PCB) to avoid overheating the part.

• Absolutely, I am trying to get a metal case that I can mount this on, however heat sinks are on my shopping list and a spare 'LM if one of my attempts fail. Thank you. – Kenny Robinson Feb 10 '12 at 1:07