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I want to create a NC switch in my electrical circuit to take power away from the 3.7 V internal battery when 5 V is received on a micro-USB for charging. By 0 V I mean No Connection or a mV, not GND.

What I am thinking is: I need a transistor whose gate is looking for 5 V and when it receives the 5 V cuts the voltage from the circuit. When the voltage is not detected on the gate the voltage from the battery will stay engaged. Is that sort of thing achievable?

enter image description here

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  • \$\begingroup\$ It might help us to understand what you are hoping to achieve if you explain how you think this circuit should work. I suspect htat you have your emitters and collectors swapped. \$\endgroup\$
    – Transistor
    Apr 18, 2023 at 17:42
  • \$\begingroup\$ I do indeed have E & C swapped lol. I updated the post. Thx! \$\endgroup\$
    – GildedWasp
    Apr 18, 2023 at 17:46
  • \$\begingroup\$ I'm not too sure how your charger behaves when there is no voltage at its output (open circuit, short to ground, etc ). But in any case there should be a resistor between base and ground. \$\endgroup\$ Apr 18, 2023 at 18:36
  • \$\begingroup\$ I've seen that before I just didn't know what size resistor to use yet, so I didn't include it. \$\endgroup\$
    – GildedWasp
    Apr 18, 2023 at 19:10

2 Answers 2

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Here is a conceptual solution. You only need to replace the manually operated switch with an automatic one (I hope you will cope with this task)... or with a short connection if possible. You can just edit the schematics below. And of course I expect you to explain the circuit operation...

V1 = 0 V

schematic

simulate this circuit – Schematic created using CircuitLab

V1 = 5 V

schematic

simulate this circuit

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  • \$\begingroup\$ This may be a really dumb question what is SW? A relay? This is what I am struggling with. I have seen that SW symbol, but idk if it's just a relay, or am I just looking for the wrong kind of "switch" component when I searching? \$\endgroup\$
    – GildedWasp
    Apr 19, 2023 at 17:59
  • \$\begingroup\$ @GildedWasp, SW is an ordinary off-on (SPST) switch. Here it has to be open when the voltage from the charger is 0 V and closed when the voltage is 5 V. Note that the voltage across it has opposite polarities in the two cases. Do you know such an element to put it in the place of the switch? \$\endgroup\$ Apr 19, 2023 at 18:43
  • \$\begingroup\$ another transistor could replace the switch maybe? I don't think I am understanding the diagram well. You said the bottom of the switch would have negative polarity? What is happening when the switch is connected, that would send 5v into the gnd side of the battery right? \$\endgroup\$
    – GildedWasp
    Apr 19, 2023 at 20:03
  • \$\begingroup\$ @GildedWasp, Maybe they didn't present it to you that way, but besides transistors, diodes are also switches that close or open the circuit depending on the polarity. So they can play the role of the SW switches in the schematic above. I just want you to understand the idea and finish the circuit you started yourself... \$\endgroup\$ Apr 19, 2023 at 20:12
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If you insist on using BJTs, a lot depends on the current that can be drawn by your "circuit". If it's less than 100mA or so, a single BJT can work:

schematic

simulate this circuit – Schematic created using CircuitLab

For load currents exceeding 100mA, you'll need to buffer the disable signal:

schematic

simulate this circuit

The simplest solution will employ a single P-channel MOSFET, with \$V_{GS(TH)} << 3.7V\$:

schematic

simulate this circuit

In all the above circuits, R1 ensures the "normally closed" (NC) condition that you stipulated. If the input DIS is left unconnected, or is held low (0V), the output OUT will be +3.7V, or very close.

The use of a BJT for Q1 means that OUT is unlikely to get higher than 3.6V in practice, due to the constraint \$V_{CE(SAT)}>0.1V\$. The MOSFET solution will perform better in this respect.

In all cases, when the potential at DIS exceeds 3V, node OUT will effectively be disconnected from everything, but you must still ensure that there's nothing in the load "circuit" than might cause OUT to rise above +3.7V. That will cause OUT to lose isolation from the +3.7V supply, since things inside Q1 that should be reverse biased become forward biased.

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