Adjustable Voltage Regulator Resistor Values

Question

I'm working with an adjustable voltage regulator and am having a hard time understanding my equation for determining resistor values.

Datasheet: http://www.chinaeds.com/zl/20103612559506780_AS2850YU-X,AS2850YT-X,AS2850AUPDF.pdf

I'm able to get ~5V (my goal) on Vout when using the following values:

• Vin = 12V
• R1 = 22K
• R2 = 1.47K (1K and 470 in series)

As soon as I put a load on Vout it drops to ~3.7V

What am I doing wrong?

Edit: turns out I had multiple regulators that were bad. Once I found a good one everything works as expected (with the proper resistance values).

Answer 1

The current \$I_{ADJ}\$ is very small and constant. Therefore its contribution to the overall output voltage is very small and can generally be ignored.

\$I_{ADJ}\$ is 40-80\$\mu\$A.

So how are you getting 5V? Calc gives:

\$ V_{OUT} = V_{REF} \times \left({1 + \frac {R_2} {R_1}} \right) = 1.25V \times \left({1 + \frac {1.47k\Omega} {22k\Omega}} \right) = 1.333V \$

I'd replace \$R_1\$ with 120\$\Omega\$. Standard value from LM337 calcs.

\$ R_{2} = \left( {\frac {V_{OUT}} {V_{REF}} - 1} \right) \times {R_1} = \left( {\frac {5V} {1.25V} - 1} \right) \times {120\Omega} = 360\Omega\$

Answer 2

It isn't 100% clear from the datasheet, but based on the language below figure 2, it looks like this regulator requires a ~10mA reference current through R1 to maintain regulation. If this is true, then you should put a 500 ohm resistor in place of R1, and recalculate your other values.

Answer 3

This part is a linear LDO regulator. You cannot get more current out of it than you put into it, so your 12V 1A supply can provide only 5V at 1A (plus a lot of heat, and minus a few mA the regulator itself uses).

If you really need 1.5A at 5V you should use a switching regulator, in which case your 12W power supply will work fine, and the regulator should run cool. This regulator will not. Even if your 12V supply could deliver 1.5A it would be dissipating more than 10W which means a big heatsink and/or a fan.

Edit: If you are seeing the drop with a light load, you are probably experiencing oscillation. The output capacitor on virtually all LDO regulators is not optional (they really should show it in all the schematics on the datasheet). It should be 100uF electrolytic or 22uF tantalum in this case. Without the cap you will definitely see oscillation at the output for some load resistances and this could account for a multimeter reading of 3.7V