Voltage regulator goes to 0 with 100 ohm resistor

I'm a beginner on voltage regulators so I have started experimenting with them. I bought a LP2950 3.3V 0.1A voltage regulator (https://www.onsemi.com/pub/Collateral/LP2950-D.PDF). I have a real circuit hooked up like this:

I expect to see 3.3 V on my voltage meter but I see 0. If I remove the resistor, I see 3.3 V on my voltage meter, so the regulator seems okay. Moreover, the current through the resistor should be 3.3 / 100 = 0.033 A, which is below 0.1 A.

Do I need capacitors to get this simple setup working? Aren't they just for reducing problems with noise in a proper circuit?

• Where is the 5VDC from? An unregulated supply? A battery? Why don't you place a capacitor (at least 1uF) across the output? The datasheet recommends that. Requires Only a 1.0uF Output Capacitor for Stability May 7, 2020 at 5:47
• What is the input voltage of the regulator (the actual voltage at the regulator, not the claimed voltage of the source) when the 100 Ohm resistor is connected? May 7, 2020 at 6:06
• Double check, then triple check, the value of the resistor. Also, tell us exactly how you are measuring the voltage because your numbers are suspiciously exact. May 7, 2020 at 11:34
• I wasn't connecting any capacitor because I didn't think it was needed for just seeing if it works, but now I know better. The input voltage source was a regular bench supply. I triple checked the resistor. Yes these numbers are exact because I rounded them. Thanks for all help. I will be sure to use a capacitor. May 7, 2020 at 13:30

1 Answer

See page 9 of the datasheet:

These regulators are not internally compensated and thus require a 1.0 uF (or greater) capacitance between the LP2950/LP2951 output terminal and ground for stability....

In other words, the output capacitor is required, it will not work without an output capacitor. There's more evidence of this in Figure 17, which shows a "stable region" based on output capacitance between 0.1uF and 100uF as a function of the output capacitor's Effective Series Resistance (ESR). Basically it should be between 0.1uF up to 100uF.

Your calculation of 33mA load current (3.3V / 100ohm) is correct, and would be OK if everything was stable and the regulator was working correctly. But without the output capacitor, the circuit is not stable, and your DVM is not able to accurately measure what's really happening.

It's hard to explain at a beginner level why input and output capacitors are required (without going beyond Ohm's law and algebra into AC circuit theory, calculus, and control loops), but basically the regulator adapts to variations in the input voltage and load current, trying to maintain a nearly constant output voltage. Capacitors help prevent sudden changes in voltage. Without them, the regulator may be oscillating -- you won't be able to see that without an oscilloscope, may be too fast for DMM to measure, but one symptom is the voltage measurements out of regulation.

DC circuits theory alone (Ohm's law, KVL, KCL) only models how things work when everything is stable, you need AC circuits theory to model what happens when things are changing. But you can get some intuitive idea of how the quickly the regulator reacts to changes by examining figure 10 Line Transient Response and 12 Load Transient Response, which show typical response times during which time the regulator is adapting to a change in the input voltage or output load current. These oscilloscope graphs show events that happen in a fraction of a second, whereas your DVM measures only 2-3 readings per second, so this detail cannot be seen with a DVM.

Normally in PCB design we use a large bulk aluminum-electrolytic capacitor (like 10uF to 1000uF) where power enters the board, and then smaller individual bypass capacitors (like 0.1uF ceramic) close to each individual IC. This is a pattern you'll see on nearly every successful design.

• If the output capacitor is 0.1uF, then it really cannot be a ceramic as the stable region requires an ESR of between 5$\Omega$ and about 25$\Omega$ which is not typical for a ceramic. The minimum output capacitance across all loads is recommended to be at least 1uF. May 7, 2020 at 9:25
• Great answer, I will make sure I use a capacitor then. May 7, 2020 at 13:33