# Linear Voltage Regulator Maximum Input Current

I am trying to understand why the maximum rated input current is not listed as a parameter on linear voltage regulators. For example, the LM1117 lists nowhere on the datasheet the maximum input current. The maximum input voltage is listed but not the input current. At first I thought it may have just been the particular regulator I was looking at; however, if you go on Mouser maximum input current is not even a selectable parameter. I find it hard to imagine the input current doesn't matter especially when the maximum output current is often listed.

Excluding the bias (quiescent) current consumed internally by the linear regulator, the input current and the output current of the linear regulator are equal. Linear regulators are not power converters — they use a pass transistor that acts sort of like a variable resistor that is regulated to maintain a fixed output voltage or current, regardless of what the load is doing.

Furthermore, linear regulators take a constant voltage input — you do not supply them with a controlled current, so it would be unusual to have a specification for the current you ought to supply to the part, since it's irrelevant.

Instead, you have to make sure the maximum output current of the linear regulator is less than or equal to what you need to deliver to your load. In the case of an 1117 regulator like the LM1117, that means keeping the load at or less than 800 mA.

Furthermore (and often overlooked by beginners), you need to make sure the junction temperature of the linear regulator stays below the maximum rating. In this case, that means 125°C. Operating in an ambient temperature of 25°C, you'd have 100°C of latitude. The thermal information in the datasheet lists a junction-to-ambient resistance of 23.8 °C/W for the TO-220 package (under the conditions specified in the datasheet), which means the regulator can only dissipate 100 / 23.8 = 4.2 W of power.

What does this spec mean? It means there's a limited amount of voltage drop the regulator can handle (since the power dissipated by the linear regulator is equal to the voltage drop across the regulator multiplied by the current going through it).

If you're operating with the maximum 800 mA load specified for the part, 4.2 Watt / 800 mA = 5.25 V maximum voltage drop. So even though the device is specified for 20V maximum input voltage, you cannot drop 20V to, say, 3.3V with an 800 mA load without exceeding the thermal specifications for the device (but you could drop 20V to 15V without exceeding this spec).

These calculations may seem bizarre because they're being done backward. Typically, you would start by calculating your voltage drop and current requirements and then select a regulator based on that specification.

For example, if you wanted to regulate a 12V source to 3.3V to power a system that consumes 150 mA of current, you would calculate 12V - 3.3V = 8.7V drop. 150 mA * 8.7V = 1.3W, which is within specification for this regulator, so it would work. Note, however, there are many other specifications to consider when selecting a linear regulator, and an 1117 is a poor choice for many applications (it's rather cheap, though!)

• thank you for your very thorough answer. This is probably a really simple follow-up question, but something I have never thought of before until your answer. If a power supply is rated for 12V 10A, the current listed is just the maximum current not the produced current that current is decided by the load on the supply? Commented Apr 20, 2020 at 13:22
• Correct! A power supply cannot guarantee to simultaneously produce both 12V and 10A at the same time. It will only happen to work out that way if it's connected to a 1.2-ohm equivalent load. Commented Apr 22, 2020 at 5:28

The maximum input current will be determined by the power dissipation of the device.

Note input current effectively is going to be same as output current, with some small amount used internally for device function.

Maximum OUTPUT current is usually specified, but it's always at some voltage differential (VIN to VOUT)

It's all about the devices temperature (i.e. power dissipation) The power the device dissipates is simply P=(Vin-Vout)*Iout. Your job is to make sure you don't try to dissipate too much power (heat). If your Vin to Vout is very large, you won't be able to pass as much current. If it's very small, you can pass more current. You can increase the allowed power by heatsinking.

So yes, for such a simple device, there's a fair amount you have to consider. But it's pretty simple once it's explained.