7
$\begingroup$

I noticed that the pinout for cheap Baofeng connectors has +V available. I'm interested in powering some electronics directly from this connector.

When unloaded, this seems to have about 3.3V on it, but when I try to power my load (an ESP32), the voltage drops to 0.8V.

How much current can this pin actually provide without significant voltage drop? Or, what does the voltage/current curve look like?

$\endgroup$
  • 1
    $\begingroup$ Have you measured the current draw of your ESP32 to rule out a fault? $\endgroup$ – Phil Frost - W8II May 29 at 17:42
  • 3
    $\begingroup$ The voltage is probably meant to bias an electret microphone, so I wouldn't be surprised to hear that that terminal can't source much current. $\endgroup$ – rclocher3 May 29 at 19:48
  • $\begingroup$ 3.3v is a common level for logic signals in electronics. This page linked in the other question seems to indicate V+ can be used for PTT 2 - is that not the case? $\endgroup$ – JPhi1618 May 30 at 18:16
7
$\begingroup$

I would first measure the current draw of your ESP32 to rule out a fault there.

The +V pin on the Baofeng microphone connection is probably intended to bias an electret microphone. These are essentially a capacitor, with sound pressure changing the spacing between the plates and thus the voltage. They contain a FET buffer since the capacitor could not drive a cable directly. The buffer requires power, but very little: the bias current is probably far less than 1 mA.

Thus it's quite likely the +V pin is unable to supply even a small current. Adding a series resistance on such a pin would have no impact to the intended use of biasing a microphone, but is a cheap and effective way to protect the radio from all kinds of faults that could happen on such a connector.

$\endgroup$
  • $\begingroup$ Almost certainly an electret mic. The only other possibility would be for the PTT; pressing it might feed 3.3 volts back to the main board. $\endgroup$ – Mike Waters May 30 at 15:39
7
$\begingroup$
  • With no load, I measure 3.17V
  • With 100kΩ load, 2.91V
  • With 10kΩ load, 1.59V
  • With 1kΩ load, 0.29V

The math works out to a 10kΩ series resistance between +V and ground.

As such the maximum power that can be drawn is at 1.59V/10kΩ load, or 0.25mW max power. Short circuit current would be about 0.3mA.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.