# Undervoltage Testing of a LDO or Switching DC-DC Converter

I am using a Regulator section in - UJA1076 this IC

Input Voltage = 16V

Output Voltage = 5V

My question is, I want to test the undervoltage condition 90% and 70% as mentioned in the datasheet. What are the ways to bring the output voltage below the undervoltage limit of the regulator? Like, I want to check at what conditions does my output voltage drop from the the nominal 5V to some thing like below 4.5V and 3.5V.

My answers - We provide a step load at the output to see the voltage dip. But how much should the load current be and for how long?

And other solution is that we reduce the input voltage till we achieve the required output low voltage of 4.5V and 3.5V.

Are there any other methods to bring the output voltage low?

• Input Voltage = 16V
• Output Voltage = 5V

That means you want to convert (16-5) V · 0.1 A = 1.1 W to heat inside the IC.

The thermal data says that on a two layer board, you'll be heating up your IC by 1.1 W · 78 K/W = 85 K. I don't think you want that.

So, reduce the supply voltage with an external voltage regulator. To avoid producing the same heat elsewhere (and needing heatsinks and other expensive design constraints), use a simple buck converter.

Alternatively, use the external PNP transistor to get the heat production out of the IC, but honestly, that's more useful for having higher currents at a not-so-high input voltage than for absurdly high 16 V input voltage if you're still drawing 0.1 A. I mean, that's 1.6 W of power from the supply, just so you have 0.5 W at 5 V! What a wasteful design.

What are the ways to bring the output voltage of the under the undervoltage limit of the regulator?

Not quite sure what you mean, but undervoltage protection refers to input. So, you need an adjustable voltage supply.

We provide a step load at the output to see the voltage dip. But how much should the load current be and for how long?

You can't use the internal regulator alone because of the thermal considerations above, so you need to use an external PNP transistor as in Figure 6 of the datasheet.

Now, the U/V curve of the output will mostly depend on that, and your question can only be answered after you've chosen a PNP transistor, and knowledge on the control loop in the IC. I don't see any information on the control loop, only vague terms like "fast load ramping". You'll find a bit of information on how quick an undervoltage situation is detected in the "dynamic properties" table, but again, that's an input thing, not an output thing.

• Thank you for the answer. Thank you for the additional information. I edited the error in my question regarding the undervoltage. I just want to understand, what are the ways in which we can bring the output voltage of a linear regulator or a switching regulator down from its nominal voltage other than the step load current and decreasing the input voltage?
– user220456
Commented Jul 26, 2020 at 9:22
• none. It's a regulator. it will keep the voltage regulated, until you draw more current and it has to follow up on regulating. Commented Jul 26, 2020 at 9:37
• There is no other way. Either load the regulator with extra load until it cannot provide enough current so the output voltage will drop, or lower the input voltage until it is too low for the regulator to stay in regulation so the output voltage will drop. Commented Jul 26, 2020 at 9:40
• Thank you for the answer. Any thumb rule on how much load current I need to provide for the step load response?
– user220456
Commented Jul 26, 2020 at 9:47
• no, there can't be. As said, it depends on the supply-to-output-voltage difference, the supply internal resistance, it depends on the control loop and the dropout transistor. Commented Jul 26, 2020 at 9:50