Confusion about max current for this TI TPS613221A boost converter

Question

I would like to power a circuit with 2 AA batteries for an input voltage range of about 1V-3V and an output of 3.3V to the circuit. I don't yet know what the current consumption would be but at this stage, I'm just trying to figure out how this particular IC works.

According to the datasheet,

9.1 Application Information

The TPS61322xx is designed to operates at a wide input voltage range from 0.9-V to 5.5-V. The minimum peak switch current limit is 0.5 A for TPS61322, with 0.75 A for TPS613221A and 1.1 A for TPS613222A.

Specifically for TPS613221A:

5 Device Comparison Table

PART NUMBER	OUTPUT VOLTAGE	TYPICAL CURRENT LIMIT
TPS613221A	3.3V	1.2A

and

7.5 Electrical Characteristics

PARAMETER	TEST CONDITIONS	MIN	TYP	MAX	UNIT
ILIM Peak switch current limit	TPS613221A	0.75	1.20	1.60	A

At this point, I'm still unclear what the maximum current that can be drawn by the circuit is and I don't see anything else in the datasheet related to maximum output current. I'm assuming it's the max peak switch current of 1.6A.

NOTE: There is also some information about high-output current with an external Schottky diode but this is specific to the TPS613222A (5V output) IC.

Now, if you head on over to the WEBENCH POWER DESIGNER website tool to configure the TPS613221A:

You can see that I_OUT Max has a range from 0 to 0.7A. This range changes depending on the part number entered but doesn't correspond to any current limit on the datasheet. Moreover, if I increase the value to, say 0.2 for example, I get this error when I select the "view design" button:

Design cannot be created!

Your preferred device does not work with the inputs you selected.

If I select the default values and then try to change I_OUT Max afterwards, I get this error:

Design update failed. The current limit of the part has been exceeded.

And just for reference, this is the circuit that was designed. It suggests using a 2.2uH inductor.

So here's what I'm trying to figure out:

1. How can I figure out from the datasheet what the maximum current I can draw from a boost converter is? It doesn't seem to be as straightforward as reading off a value.
2. Is peak switch current limit the same as this maximum current? Seems like this relates to the instantaneous switching currents.
3. How does this differ from I_OUT Max?

Thanks in advance!

Answer 1

How can I figure out from the datasheet what the maximum current I can draw from a boost converter is? It doesn't seem to be as straightforward as reading off a value.

The only figure you know is that the peak current current may be limited to 0.5 amps for the TPS61322. I'm not going to triplicate my answer for the other two devices in the data sheet. I'll leave that to you.

This current is drawn from your batteries and, it isn't the peak current that can be drawn by the load at 3.3 volts. Load current will be less and, how much less depends on how much voltage boosting the chip has to do.

So, if your input is as low as 1 volt and, you need 3.3 volts out, the average output current has to be proportionately less than the input current to avoid breaking the laws of physics. In other words, you can't get more power out than what you are putting in.

Here's a simulation from my basic website: -

• I've set the input voltage to be 1 volts and the output voltage to be 3.3 volts
• I've set the operating frequency to be 2 MHz
• I've then tweaked the load resistor to cause the input peak current to be just below 0.5 amps (0.498 amps)
• Basically, the peak input current is the inductor current

So, this is telling me that your output current is limited to about 127 mA. Notice also that the average inductor current (0.419 amps) is precisely 3.3 times the average output current (0.127 amps). This is the output to input voltage ratio. You can't get better than this.

But, my calculator is for an ideal boost converter (without losses in the output diode). The chip you use may implement a synchronous MOSFET instead of an output diode hence, my simulation would be fairly accurate. However, if the output uses a diode it will "drop" circa 0.6 volts in supplying energy to the output and, this also needs to be taken into account.