Can a MOSFET or transistor be used to switch on a 9V supply using a 3.3V signal?

Question

***Technical question, not a wiring problem.

I am building a bicycle alarm system and I'm a little stuck on the alarm output. I wonder if I am approaching this problem wrong.

My alarm piezo technically is a 12 volt device but it works well on 9V.

My device is battery operated with a large 9 volt battery pack. The system is driven by a PIC32 microcontroller.

The kind it is is not important, but its a 3.3V device, so its GPIO pins can only output 3.3V and don't supply much current, so obviously you would not want to power the device using a gpio. HOWEVER I was thinking I could use the gpio to signal a transistor or a MOSFET to switch on the 9V supply. The alarm just needs a 9V power supply to emit a really loud sound.

I have worked very hard trying to achieve this.

First I tried some 2N3904 transistors and also some 2N3906s. The current requirements are just too high and they both got burned up.

Next I tried a MOSFET, the IRFZ44N specifically. I selected this MOSFET from DigiKey thinking it would be good because it can handle so much current.

HOWEVER I can't seem to get it to give more voltage than my gpio! My gpio is 3.3V and the MOSFET won't output more than 3.3V.

To be clear this is not a wiring issue! That is why I am not posting schematics. I am not concerned about wiring. What I really want to know though is whether this approach is flawed. Is it even possible to use a MOSFET or transistor to switch on a higher voltage using a smaller voltage as a signal? If so, then what is a good mosfet for this type of thing?

This is the speaker I'm using.

I don't know exactly what the current requirements are since there is no documentation on that but I would assume between 100 and 500 mA.

Also if you really want to know, my microcontroller is the PIC32MX795F512l all though I don't think that's really too important here.

If you guys need me to post any schematics, I will but I'm honestly just looking for general information here, not really a step by step guide on how to construct my circuit since I can definitely handle that part and most people wouldn't find that relevant to their own questions on this topic anyway.

Answer 1

Yes, a BJT (bipolar junction transistor) or a MOSFET (metal oxide semiconductor field effect transistor) can be used to switch 9V from a 3.3V signal. It is done every day in approxiamtely 5 gazzilion devices. You just have to do it right.

From the descrption, you are doing something like this:

^{simulate this circuit – Schematic created using CircuitLab}

That circuit is known as a source follower. As you have found, the voltage at the source can never be higher than the voltage on the gate. That's why you can only get 3.3V to the buzzer.

What you need to do is this:

^{simulate this circuit}

That is called a low side switch. It will work properly, and deliver nearly the full 9V to the buzzer - if the MOSFET is a logic level MOSFET that can work with 3.3V on the gate.

There are similar problems when using bipolar junction transistors (BJTs) - the things you are calling just "transistors."

If you have an NPN connected this way, then it is called an emitter follower, and like the source follower will not deliver more than the base voltage to the buzzer:

^{simulate this circuit}

For the NPN transistor to work properly, you have to move it to the low side as well:

^{simulate this circuit}

Answer 2

^{simulate this circuit – Schematic created using CircuitLab}

Why not a 2N3904?

• beginners think hFE the linear current gain will work also as a switch but this is false. So degrade the hFE 80 to 90% or use 20% or preferably 10% of hFE and the 2N2904 has good speed but not as much hFE as the PN2222A
• too large Rb is the biggest failure and Vce becomes too large then Ic*Vce=Pd burns up at 220'C/W the thermal resistance value for Rja
• too large a load current for Rce = Vce(sat)/Ic (@ Ic= 10x Ib)
  • 2N2904 = 0.4V / 50mA = 8 Ohms
  • PN2222A is 4 Ohms? = (let me double-check datasheet) Nope it's 2 Ohms when driven properly.
    • Vce(sat)/Ic=Rce= 0.3V/150mA= 2.0 Ohms
      • and Rce = 1.0V/500mA = 2.0 Ohms at Ib=50 mA
      • from Fig 4 in onsemi specs if Ic/Ib= 100, Vce(sat)[nom]= 300mV/150mA = 2 Ohms [nom] but 4 Ohms worst case
      • also Rce= 1.0V/500mA = 2 @ Ib= 10 mA or Ic/Ib=50 and Pd = 500 mW burning up but if Ic/Ib=10 then at 500mA ,
        • Rce= 210 mV/500mA = 420 mOhms [nom] and Pd= 105 mW runs OK but warm at at 220'C/W x 105 mW= 23 'C rise above ambient.

The bottom line for small TO92 transistors

• For a 9V battery always choose the PN2222A with the A suffix for a 2 Ohms switch that drops to 420 mOhms ([nom] not worst-case or use a 1 Ohm FET and not the lowly 2N7000 which might work but is 1.9 Ohm nom and 9 Ohms max. - Or choose a lower RdsOn FET with logic Level thresholds ~ 1V if using a 3.3V uC to drive it.
• why is the PN2222A better? Its a slightly bigger chip with higher hFE and when used as a switch ALL transistors work best when using about 10% of the maximum hFE for Ic/Ib.
• SOME more expensive transistors have superbeta or hFE > 500 and thus can be used with 10% or hFE(sat)=50 or Ic/Ib=50.

- so for high(ish) switching currents on BJT's always use 10% of the maximum hFE for Ic/Ib

I am pretty sure Jim Williams would agree.

This becomes a trivial switching problem for “logic-level power FETs”.
However, the ones you chose are not, nor are the NPN's and your base R was probably too high.

Vgs must be at least 2x Vgs(th)or Vt max to enable low RdsOn.

Yours is the standard FET Vtv 2 to 4 V thus needs 8V min. So choose one with Vgs(th) << 1.5V

Consult with DK tech support if unsure how to pick in SMT for better options.

Regarding Vbat resistance, this depends on quality and condition. A new alkaline 9V battery may have a pulsed short circuit current of > 1A or < 9 Ohms but this quickly rises as the state of charge drops and ends up being 10 x this amount while still useful but almost depleted at 8V.