I'm working on an application where I need to step down a 3.7V LiPo down to a 1.0-1.5V for an embedded device that takes AAs (effectively an AA-to-LiPo conversion kit). In order for "low battery" detection to still work on device, I need the stepped-down voltage to swing with the input. The embedded device is tolerant of high voltages up to 5V.

In an effort to be simple, in my head, the easiest option is just a voltage divider followed by a unity buffer in order to have the current to drive load. The device only really takes in 15-20mA, so most off the shelf op-amps should be able to handle it. enter image description here

With the circuit simulated and breadboarded, it works exactly as expected if the load is a simple LED. However, once it's rigged into the actual device, the voltage at output drops precipitously once connected but rises again when disconnected (as measured by a multimeter).

I was able to smooth this out to a certain extent with a 3900uF (it's all I had at this very second) capacitor at the output. However, the behavior would still come back if input voltage fell below a certain extent.

My suspicions are because my embedded device is presenting a signifcant capacitive load beyond the mere 100pF my op-amp is designed to handle. Therefore, it's causing major oscillations in the output that are too quick for my meter to pick up.

I imagine adding an RC snubber network at the output would fix the oscillations to an extent? Alternatively, should I add an emitter follower after the op-amp to decouple the capacitive load?

  • 1
    \$\begingroup\$ If the device is tolerant of 5 V, and the battery is only 3.7 V, then what is there to convert? \$\endgroup\$
    – AnalogKid
    Commented Jun 29 at 4:15

1 Answer 1


The op-amp is not able to source 20mA. At best you'll get 15mA from a typical device, but that's only with a severely diminished output voltage swing.

The second problem is that a typical op-amp is probably unable to get the output anywhere near its positive supply, with or without a load, so you'll have to use an op-amp which has a so-called "rail-to-rail output".

Using such a rail-to-rail output device will also enable you boost the output current capability using a single NPN bipolar junction transistor:


simulate this circuit – Schematic created using CircuitLab

This design requires that the op-amp output be able to reach at least 0.7V higher than your desired output, due to the \$V_{BE}=0.7V\$ drop from the transistor, so you must use a rail-to-rail output device.

You should be able to get at least 200mA through the load.

  • \$\begingroup\$ Thank you! I tested this solution in situ and it worked wonderfully. However, I opted to omit R4 and connect the output straight to the base. What is the purpose of R4 in this circuit? Simply to limit the base current of the BJT to reduce power consumption? Would it not cause a reduced voltage at the base? Thanks! \$\endgroup\$
    – Felix Jen
    Commented Jul 6 at 8:38
  • \$\begingroup\$ @FelixJen R4 protects the op-amp from a short circuit across the load, limiting current as you say. Yes, it would drop a small voltage. At 200mA load current, base current would be perhaps 1 or 2mA, for less than 0.5V drop. R4 also helps stabilise the loop, in case of a nasty reactive load, but if you don't expect one, then there's no harm in omitting R4. \$\endgroup\$ Commented Jul 6 at 9:41

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