0
\$\begingroup\$

I am trying to design a circuit that will completely turn off the Adafruit Powerboost 500 (by connecting the ground and EN pins) automatically. As part of this, I have come up with the attached circuit. I have very little experience with electronics, but my reasoning is as follows:

When the battery is first connected, the EN pin will be grounded and the powerboost will be off. Upon pressing the button, current will begin to flow and the transistor will be inhibited, so the device will continue to operate.

If this circuit works as I expect, I will add a microcontroller between the transistor and 5V pins, allowing me to cut power automatically when a certain event is triggered. This will require pressing the button again to turn on the circuit.

Powerboost manual: https://learn.adafruit.com/downloads/pdf/adafruit-powerboost.pdf

Thank you.enter image description here

\$\endgroup\$
  • \$\begingroup\$ I... don't think that works the way you describe?? \$\endgroup\$ – Passerby Apr 3 '16 at 18:30
  • \$\begingroup\$ I suspected that might be the case! Any idea how I could achieve what I intended? \$\endgroup\$ – Reckless Apr 3 '16 at 18:33
  • \$\begingroup\$ Is the Powerboost powering the micro that's going to switch the Powerboost off? If so, then as soon as the Powerboost is disabled the micro will power down, the transistor that's pulling the EN signal low will turn off and the Powerboost will be enabled again, starting up the micro. Anyway, your circuit as it stands doesn't work. \$\endgroup\$ – Transistor Apr 3 '16 at 19:10
1
\$\begingroup\$

The EN pin is pulled up through R13 to UBAT. Transistor (see comments above) correctly points out that as soon as the unit is disabled you lose the 5V, if anything the whole circuit is liable to oscillate.

You could try this circuit (I haven't tried it so it would be an interesting experiment).

enter image description here

On power up Q1 is switched ON (due to the small current into its base) taking EN to ground. This keeps the output OFF.

Closing SW1 turns Q1 OFF (Vbe less than 0.6V) allowing EN to be pulled up through R13 (an 'on board component'). The 5V output is now available.

SW1 can be replaced by a second transistor (Q2) which can be controlled by a logic signal. When the input to Q2 is HIGH (Q1 = ON) it turns Q1 OFF, when the input to Q2 is LOW (Q2 = OFF) Q1 is turned ON.

Both transistors are any common small signal NPN types (silicon) e.g. BC548

\$\endgroup\$
0
\$\begingroup\$

If the Powerboost is powering the micro that's going to switch the Powerboost off then you're going to have to do something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Hold-on circuit.

Rather than make a circuit to switch the system off, have it default to off and create a circuit to hold it on.

  • R1 pulls the EN input low so the Powerboost is normally disabled. You might need to experiment here. The higher the resistance the less current will be wasted during hold-on but at some point it will stop working. Check the data sheet for the V-in LOW threshold and check with your multimeter with various values of R1.
  • Pressing SW1 will pull R1 high turning on the EN and allowing the Powerboost to work.
  • You now need to get the micro to turn on Q1 to keep the EN input high. If GPIO pin is pulled low a base current will flow from Q1 through R2 and D1 to the GPIO. This will turn on Q1 and pull R1 high. (SW1 needs to be held long enough for the micro to perform its power-up reset and start running the code.)
  • If the GPIO is switched high the base current will stop, Q1 will go open-circuit, R1 will pull EN low and the Powerboost will shut down. D1 was intended to prevent backfeed from the micro to the battery circuit but I don't think it's required.

I haven't tested this. As I say, some experimentation will be required.

\$\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.