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Context

Note: I am a beginner with electronics so please provide detailed explanations and add descriptions that are more beginner friendly. I really do want to learn how this all works.

I am trying to make a diffuser using a 113 kHz piezoelectric disc and a 9 V DC power supply from a battery. I was trying to follow GreatScott's video on creating a DIY mist maker, but ended up getting confused with some portions. I have successfully got the N555 timer to output a 113 kHz frequency powered by the 9 V battery, but I am quite stumped on how the MOSFET portion works. Also, I do not have a toroidal inductor, I have instead cylindrical inductors with bands.

I have tried just attaching the output from the N555 timer to the piezoelectric disc since I reached the 113 kHz frequency, but it didn't work, which I believe is because I didn't have a high enough amount of current/power.

Questions

  1. Is it possible to get this piezoelectric disc to work with a 9 V power supply instead of a 15 V power supply?
  2. What is the N-channel MOSFET and inductor used for in this schematic? Can I make this work without it since I already have a 113 kHz signal?
  3. How can I increase the amperage in my circuit so that I can power the piezoelectric disc from the N555 timer?

Components

GreatScott Mist Maker Circuit

winny
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Alex Thomas
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    You need to use an inductor with a high saturation current, about 300 mA. The cylindrical inductors are unlikely to have a saturation current anywhere near that. – Mattman944 Oct 09 '22 at 21:01
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    9V batteries are weak, it is unlikely that you can drive this piezo with it. Use 6 or more AAs instead. – Mattman944 Oct 09 '22 at 21:03
  • Would a toroidal inductor have a higher saturation current then? Also, I am confused as to why the schematic wants me to connect the VCC to the inductor and directly into the piezo. Wouldn't this drown out the 113khz frequency I made with the timer? – Alex Thomas Oct 10 '22 at 07:22
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    If you are inexperienced and want to change the design, you need to start with a working design, and change one parameter at a time. Built it as close as possible to the original. Even experienced designers sometimes follow this plan. – Mattman944 Oct 10 '22 at 08:44
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    9V batteries are almost always a wrong choice unless you're absolutely sure you need very little power (for a remote or something like that). They're comparable in capacity to coin cells, many times weaker than AA(A)s. This circuit will probably drain a 9V battery in an hour or so. – TooTea Oct 10 '22 at 11:17
  • Thank you very much for all the suggestions! I am buying AA batteries and AA battery packs with wires so I can prototype with that instead. I will try to follow this design as best as I can. I also bought a toroidal inductor to see if that works instead. – Alex Thomas Oct 11 '22 at 00:53

3 Answers3

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Is it possible to get this piezoelectric disc to work with a 9V power supply instead of a 15V power supply?

yes, change the inductor to 150uH

What is the N-Channel MOSFET and inductor used for in this schematic?

Any N-channel Mosfet that's good for 5A ore more should work well enough.

Can I make this work without it since I already have a 113khz signal?

no

How can I increase the amperage in my circuit so that I can power the piezoelectric disc from the N555 Timer?

you need to increase to voltage to increase amperage, this is what the inductor and MOSFET do.

Jasen Слава Україні
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  • The piezo voltage will raise quite high, so the MOSFET should have at least 50V Vds rating. Most do, but some are only rated 20-30V. – jpa Oct 10 '22 at 06:58
  • Thank you for answering my questions! I have some follow up questions regarding how I use the inductor and MOSFET. I Tried to connect the output of the N555 timer to the MOSFET's gate and connected the MOSFET's source to ground, but the discharge pin only has a fraction of a volt as it's output. I don't think my amperage increased either. Also, does this schematic want me to connect my 9V DC supply directly to the inductor and then connect that directly to the Piezo? That doesn't make sense to me. – Alex Thomas Oct 10 '22 at 07:14
  • The discharge pin is an open collector output, so without a pull-up there is nothing to raise the output above the base-collector diode forward voltage. The inductor is to be connected to your Vcc supply, and the action of the boost converter (as explained by @jpa ) creates a higher voltage across the transducer to GND through the capacitor. – PStechPaul Oct 10 '22 at 08:34
  • When you say "pull-up" do you mean a pull-up resistor? What do pull-up resistors do? If there are any good resources you suggest I look at please let me know as well :) – Alex Thomas Oct 11 '22 at 01:00
  • Changing the inductor to 150 uH will make no material difference. The inductors do not have the correct characterics. He needs ones with higher saturation current. || MOSFET should have a high enough Vds rating to survive the inductive ringing voltage. – Russell McMahon Oct 11 '22 at 23:01
  • My MOSFET has a VDS of 55V. I went and checked the Datasheet. So I should be good with the MOSFET – Alex Thomas Oct 13 '22 at 02:18
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Jasen's answer covers the practical points. For learning more, I'll add some background:

What is the N-Channel MOSFET and inductor used for in this schematic?

It implements a boost converter, which raises voltage level. In this application the boost converter output is not rectified with a diode, instead it is fed through a capacitor into the Piezo disc.

The inductor raises the voltage by converting energy into a magnetic field and back again. It is similar in function to transformers: current × voltage stays constant, but high current low voltage can be converted to low current high voltage. The MOSFET is used to switch the power to the inductor as it can handle higher current and voltage than NE555.

Piezo discs bend when voltage is applied over them. The higher voltage, the more the disc bends. Then when it bends back and forth quickly, if forms tiny droplets of water. The 9 volts supply voltage is not high enough to move the water fast enough, which is why the inductor is used to raise the voltage.

jpa
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  • Should this be viewed as a class E or F amplifier rather than a boost converter? Boost converters usually only allow current to flow in one direction but this one works both ways. – user253751 Oct 10 '22 at 10:12
  • @user253751 It does not have all the characteristics of a class E/F amplifier either. I guess it is just a circuit that has similarities to common building blocks but no exact match :) – jpa Oct 10 '22 at 10:39
  • Thank you very much JPA! I am understanding more why this was included in the original design now. So assuming I make this MOSFET circuit correct I should see a higher voltage but less current than if I didn't use it. This sounds to me that the net total of Power would be the same though which wouldn't solve my issue with lack of sufficient power. So does this come down to me using a more powerful power source in addition to all of this? Thank you again for your time and patience. (I will test this later. I have ordered a toroidal inductor and a battery pack with AA batteries that hold 8) – Alex Thomas Oct 11 '22 at 01:21
  • @AlexThomas Yes, power or energy cannot be created out of thin air. You may or may not be using the maximum power *available* from your current power source. In general piezos do not require much current, but they need a high voltage to work. – jpa Oct 11 '22 at 04:57
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The other answers have good information, here is more on the inductor.

Basic inductor equation:

\$V = L \frac{di}{dt}\$

Re-arranging and assuming voltage is constant:

\$\Delta I = \frac{V \Delta t}{L}\$

The inductor will "charge" when the MOSFET is ON:

\$\Delta t = DutyCycle * \frac{1}{f}\$

For 50% at 115 kHz, \$\Delta t = 4.35 \mu S \$. You can substitute the duty cycle of your 555 circuit.

\$\Delta I = (9V)(4.35\mu s)/150\mu H = 0.26 A\$

The saturation current of the inductor must be at least this high. The inductor assortment that you have is useful for filters, but not for most power circuits.

The inductor "charges" when the MOSFET is ON, then "discharges" through the piezo when the MOSFET is OFF. If the inductor saturates, it won't hold any more energy, so you don't want it to saturate. The circuit can be simulated if you know the equivalent circuit for the piezo.

You should always have a load connected (piezo, or resistor). Without a load, the inductor energy will have no where to go, and the voltage could rise until it blows out the MOSFET.

Mattman944
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  • Thank you Mattman! So how do I know what type of inductor I needed? How do you know that the cylindrical inductors are for filtering and the toroidal inductors are for higher power? They could be rated the same in inductance, correct? Also, thank you for talking about adding a load! I ended up burning one MOSFET by just hooking it up to power. – Alex Thomas Oct 11 '22 at 01:31
  • So, from my current understanding, I have the output of the N555 timer go to the gate of the MOSFET. This makes sure the MOSFET is only active for around 4.35μS at a time. the source of the MOSFET connects to ground, and the drain is connected to a capacitor and an inductor which is connected directly to the power source. initially, when the gate is off on the MOSFET, the power from VCC goes through the inductor and charges the capacitor. When the gate is turned on for the MOSFET, the power goes through the inductor to the drain of the MOSFET and to ground. This charges the inductor for power – Alex Thomas Oct 11 '22 at 01:41
  • If an inductor has direct current applied to it, or the circuit is unbalanced, then you need to check for saturation current. – Mattman944 Oct 11 '22 at 08:56
  • Do you have a scope? If you do, it is still tricky to analyze this circuit since inductors work on current. If you add a 0.1 ohm resistor between the MOSFET source pin and ground, then you can "see" the current with a scope. This is one of my interests, so I ordered the piezo that you referenced. I will build the circuit. – Mattman944 Oct 11 '22 at 09:05
  • Thank you so much @Mattman944! Could you please show me what the circuit is supposed to look like? I feel that I may also be misreading this circuit diagram somehow. Though I understand more now why I need the MOSFET and the inductor, that portion of the circuit still feels somewhat mystical to me. – Alex Thomas Oct 13 '22 at 02:15
  • This circuit is almost the same: https://www.electroschematics.com/ultrasonic-mist-maker/ The bad news is that it uses 20V! I have never used this type of piezo, so I can't confidently draw a circuit that will work. I need to build the circuit and experiment. – Mattman944 Oct 13 '22 at 08:03
  • I GOT THE STUPID CIRCUIT TO WORK!!! But now I have SO MANY QUESTIONS. I ended up using 8 AA batteries as a power source instead which was 12 volts. I don't know what the amperage is though. But I have a HUGE question as to how the AC signal is created. From my knowledge the N555 timer produces an oscillating DC square wave, but somehow when the output of the N555 goes into the MOSFET gate and the discharge from the MOSFET and output from the inductor go into the non-polarized 100 nano farad capacitor it produces AC signal! WHY????!!! – Alex Thomas Oct 15 '22 at 07:15