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I'm building a small circuit that needs to be powered by another device. I will power my own circuit from the already existing devices battery. I need to be able to shutdown/power-up my circuit through a trigger signal from the already existing device. My circuit needs ~200mA and at least 3.3V.

The only "trigger" signal available is a 3.3V signal, capable of providing something like 5mA.

When the existing device is powered off, the "trigger" signal will still have a small voltage, around 0.8V, so my idea is to use an opamp as a comparator, so when the "trigger" signal is higher than ~1.5V it will activate the power to my circuit, and when it´s lower than 1.5V it will shutdown my circuit.

I´ve attached a simulation of my idea and it seems to work, but i don´t know if this is the best approach to solve this.

Do you guys have any suggestions/changes, or maybe I should scrap my whole idea and do something else?

I´ve used transistors in this schematic, maybe it´s better to use mosfets? Any suggestions?

  • The switch in the image is there so simulate the trigger signal, I have no control over it, it´s controlled by the device.

  • I use a 20ohm resistor to simulate my load. Schematic idea.

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  • \$\begingroup\$ Welcome to EE SE! Nicely animated schematic. Do you have any specific power efficiency and budget requirements? Does the 1.5V reference voltage needs to be precise and accurate? \$\endgroup\$
    – Unknown123
    Commented Apr 2, 2019 at 19:52
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    \$\begingroup\$ you say "I've used transistors [...] maybe it's better to use mosfets"--I feel like I should point out that a mosfet is a type of transistor. By "transistor" you seem to mean BJTs. \$\endgroup\$
    – Hearth
    Commented Apr 2, 2019 at 19:59
  • \$\begingroup\$ Maybe a relay could work? The only issues would be finding one rated to switch reliably at 3V (preferably with the freewheeling diode included) and rated for contact wetting current (minimum required current needed by the contacts) of 200mA or less. \$\endgroup\$
    – MarkU
    Commented Apr 2, 2019 at 20:22
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    \$\begingroup\$ Is the reason you've selected \$4.2\:\text{V}\$ as the voltage rail due to the fact that the system power source is a Lithium ion? If so, when you write "at least \$3.3\:\text{V}\$ do you mean that more is okay? If so, how much more is okay? \$\endgroup\$
    – jonk
    Commented Apr 2, 2019 at 20:58
  • \$\begingroup\$ i would simply use an optocoupler. the 0.8 won't do anything, and the 3.3v is plenty. it probably needs about 1.8v to fire up, which is pretty close to your dialed-in opamp config... add a fet for more current handling if needed. \$\endgroup\$
    – dandavis
    Commented Apr 2, 2019 at 21:03

1 Answer 1

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Here is a classic high-side switch circuit that should work for you.

enter image description here

A couple of points about the transistor choices. You can't just use any P-channel MOSFET for Q1. Because of the low-ish voltages involved you need a logic-level MOSFET which simply means that the process is tweaked for lower VGS. Q1 (IRF3708PBF) specifies a max RDSon (resistance) of 0.029 ohms with a gate voltage of -2.8V. Available at Digikey for a couple of bucks.

So... with a load current of 200mA, Ohms law tells us our IR voltage drop will then be 0.2A * 0.029 ohms = 0.006V of drop. Very little voltage loss - so far so good.

To drive the gate of Q1 low, we use a common NPN transistor at Q2. To avoid turning on Q2 (and Q1) when the CTRL voltage is 0.8, we put a diode in series with the gate. This raises the turn-on threshold of Q2 to about 1.2V. When Q2 is fully on, you should see about 0.2V at the collector which drives the Gate of Q1 nice and low turning it on fully.

The purpose of R1 is to keep the VGS of Q1 solidly at 0V (OFF) even in the presence of any leakage current at Q2.

R2 limits current into the base of Q2 when the CTRL input is ON.

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  • \$\begingroup\$ Randy, when you use the CircuitLab button on the editor toolbar your schematic gets saved inline with the post. No need for an account, screengrab, upload or background grid. You can edit it again easily. It also means we can copy it into other answers and edit it. \$\endgroup\$
    – Transistor
    Commented Apr 6, 2019 at 18:51
  • \$\begingroup\$ Yah, I found it confusing to go back and forth from the schematic editor to the text and the tool wouldn't let me save the file without buying. I'll try to make it work next time. \$\endgroup\$
    – Randy Nuss
    Commented Apr 7, 2019 at 14:17
  • \$\begingroup\$ Just use the Save and Insert button. It will save an image in your post with a link to edit the schematic. Open a second tab with the question in it for reference to the text. \$\endgroup\$
    – Transistor
    Commented Apr 7, 2019 at 14:21
  • \$\begingroup\$ Thanks guys for all the answers ! I think i will try to find a suitable p-mos and go with Randys suggestion. seems like an easy solution. \$\endgroup\$
    – yellowtape
    Commented Apr 8, 2019 at 6:56
  • \$\begingroup\$ I think i found a suitible p-mos for this. "PMN42XPEA" PMN42XPEA DATASHEET what i got when calculating was that @ 600mA at -3V Vgs it will disapate 0.018W, that should be fine. and that 600mA at -3v vgs should not be any problem ? what do you guys think? \$\endgroup\$
    – yellowtape
    Commented Apr 8, 2019 at 8:13

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