I'm building a small circuit to power a heater at 380mA using a 3.7V battery. I'm using an AMC7135 to regulate the current to 380mA. Without the AMC7135 the heater would typically draw ~500mA, so the AMC7135 should have no trouble regulating it. This is the circuit I currently have:


simulate this circuit – Schematic created using CircuitLab

The problem is that it always regulates the current to just 30mA, instead of the expected 380mA. I assumed this was some sort of inrush current problem, so I tried putting capacitors between GND and OUT, and GND and VDD, and OUT and VDD as well. But that didn't work either.

Interestingly, if I hold the switch down then connect the battery, it goes to 380mA as expected. But pushing the switch alone just gives 30mA.

In the final product, the switch will be replaced with a microcontroller. So it can't be attached directly between the battery and load (otherwise it would be fried by the 380mA).

How can this circuit be modified to get the 380mA regulation as expected?

My inspiration for this circuit comes from this webpage:
I'm assuming that if PWM works, then turning it on and off manually should work too.

  • 2
    \$\begingroup\$ your schematic is upside down ... also horizontally inside out ... please follow schematic drawing convention ... +V at top, GND at bottom, input at left, output at right \$\endgroup\$
    – jsotola
    Apr 7 '21 at 0:50
  • \$\begingroup\$ Why do you have a cap across OUT and ground? \$\endgroup\$
    – Passerby
    Apr 7 '21 at 1:38
  • \$\begingroup\$ @Passerby I've seen the cap across OUT and GND on various schematics. It doesn't work at all without it. I can't explain why, but it might have something to do with a feedback loop. \$\endgroup\$ Apr 7 '21 at 2:12
  • \$\begingroup\$ According to the datasheet it should be across vdd and ground. And likely just a 0.1uF not 22 uF. The link you provided also is using a 0.1uF though across out and ground. \$\endgroup\$
    – Passerby
    Apr 7 '21 at 3:27
  • \$\begingroup\$ What kind of heater is it? The way you have it wired, in the off state the capacitor gets charged to battery voltage through the load. Is there any reason you can't switch both the AMC7135 and the load VDD at once? \$\endgroup\$
    – K H
    Apr 7 '21 at 6:43

Attempting to apply voltage to OUT while switching VDD is extremely likely to put the internal control electronics into either a nonsensical or lockout state, and the switch ought to go between the load and the regulator.

You have two options:

  1. Use a MOSFET as the switch (e.g. an nFET on the low side as shown below, with a microcontroller driving the gate). The microcontroller will not get "fried"; this is a standard way of switching a load whose current or voltage are too large for the microcontroller's I/O pins.

enter image description here

  1. select a different current regulator with an enable input.

Note further that I've redrawn your schematic right-side-up. Putting ground on top and positive supply on the bottom is confusing.

  • \$\begingroup\$ In the past, I thought I had seen this done in flashlights while only using an AMC7135, a microcontroller, and a switch not directly connected to the battery. I'm starting to doubt myself though. \$\endgroup\$ Apr 7 '21 at 1:03
  • \$\begingroup\$ I remember now: Many flashlights use PWM directly on the VDD pin of the 7135. The 380mA definitely doesn't go through the microcontroller. That seems like a solution without a mosfet. How is that possible? \$\endgroup\$ Apr 7 '21 at 2:18
  • \$\begingroup\$ @DanielWilliams You'll need to obtain one of those devices and measure its operation when it's misconfigured in a way that works. It's very likely that it's regulating to a lower voltage applied to load which is enough to turn on your load weakly, but below the Vf of the LED. \$\endgroup\$
    – nanofarad
    Apr 7 '21 at 2:38
  • \$\begingroup\$ Do you have a recommendation for a similar regulator with an enable input? \$\endgroup\$ Apr 22 '21 at 23:37

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