Schematic sanity check required

[eSCHEn]

I know there's a few of the folks here who are good with circuits, would you mind sanity checking this schematic please? I'm normally adapting a circuit to my needs but this is from my mind so not sure at all! Appreciate it is very basic but then again, so am I

• Bits are all from the "what I have" pile.
• Supply is 12V 1.25A.
• Mist unit is ultrasonic based, draws 500mA at 5V.
• Fan is standard 12V computer unit. Draws 210mA at 12V and 80mA at 5V.
• For 5V I'm using the Nano's regulated 5V line. Arduino specs say it'll do 800mA and its AMS1117 is rated 1A.
• I can under volt the fan but it's too slow sadly.
• R1 and R2 are to limit draw to 5mA or so.
• R3 I'm trying to limit at 500mA for the mister unit. I don't have any low value resistors so was considering not bothering... The unit doesn't have a high inrush and is pretty stable at 500mA when running.
• D1 is for back EMF protection. Not sure I have it the correct way around

Whole purpose is to make a low-voltage fresh-air and humidity generator. I'll run it on a timed schedule for each provider, something like "every n mins run the whatever for y mins". No RTC needed that way. I need to get this up today if possible as the fruits are aborting on these mycelium

Any sensible help much appreciated! No AI though please, I'm rather tired of it and would like a human's input

 

[Seadog]

Reverse the diode. It's forward biased as drawn.

 

[ajlelectronics]

Don't even think about trying to pull half an Amp from the Arduino. Use a separate 5V supply and feed both from that.

 

[Misterg]

R1 and R2 are to limit draw to 5mA or so.

Hfe (DC gain) of BC337 (datasheet) can be as low as 60 at 300mA Ic, so you will need to drive them hard if you're looking for 500mA Ic

Vbe at 300mA Ic can be up to 1.2V

The data sheet doesn't have curves, so assume Hfe of 50 (say) and Vbe of 1.5V(say) at 500mA Ic.

Ib required = Ic/Hfe = 500mA/50 = 10mA

Max voltage drop across R2 = logic output of arduino - Vbe = 5V - 1.5V = 3.5V

R2 therefore must be less than 3.5V/10mA = 350 ohm. I'd use probably use 270R , but 330R *should* be OK. May as well use the same for R1.

R3 I'm trying to limit at 500mA for the mister unit.

If the mister will run from 5V directly, then you don't need R3.

You will lose about a volt across the transistor - is the mister happy on 4V? If not, do you have any power mosfets? (Or a relay?).

(Reverse the diode, as said above: 'arrow pointing up' if you have +ve supply at the top of the schematic.)

 

[daedalusminos]

There's quite a few things wrong:

Diode back to front as mentioned by @Seadog and you need to consider supplies, you can feed 5V in or 7-12V on Vin but not both.

I would have a 5V supply for the Nano and your mister, the 12V just feeds the fan.

I would replace the NPN's with N-MOSFETs suitable for logic level driving - add a series gate resistor (e.g. 100R) and a 10K 'pull-off' from gate to ground. If you do stick with the BC337's you need to consider transistor gain (160/250/400?), drive current and therefore series R.

You will need pull up resistors on the I2C lines if not already on the OLED.

What's the 10R series R on the mister for?

 

[daedalusminos]

For 5V I'm using the Nano's regulated 5V line. Arduino specs say it'll do 800mA and its AMS1117 is rated 1A.

It's not just the max current you need to consider, it's a linear regulator so you need to ensure power dissipation is not exceeded....

At 500mA and 12Vinput the AMS1117 will be dissipating 3.5W

From the datasheet, assuming it's in the SOT223 package then the thermal impedance (junction to ambient) is 46°C/W (best case - it depends on board design). So at 20°C room temp the die junction temp = 20+3.5*46 = 181°C. The max operating junction temp is specified at 125°C so you'll end up cooking the regulator.

 

[eSCHEn]

Thank you for all the replies, most useful; particularly how the calculations are done

I shall have a root about as I'm sure there's some MOSFETs somewhere here...

 

[eSCHEn]

A more thorough sift through the components yielded five of FDP7030BL (60 A, 30 V, 12 mΩ @ VGS = 4.5 V), some sub-1K resistors and a 120mm fan (old one was 80mm). I've taken on board what was said above and have produced this. I'm hoping there's less wrong this time

• OLED display has pull ups in place already.
• I've put two 100Ω in series as 5/100 gives 50mA and the specs maximum for a pin is 40mA. The next value I have is 1K so this will do.
• New fan still draws < 100mA but should have enough airflow.

And again, thanks to all for the replies and assistance this is not my bag so lots of learning for me here.

 

[daedalusminos]

Blimey, chunky FETs!

The 10k pull offs should be placed between gate and ground.... it's to ensure the FET does not turn on whilst the Arduino is initialising.

 

[Misterg]

• I've put two 100Ω in series as 5/100 gives 50mA and the specs maximum for a pin is 40mA. The next value I have is 1K so this will do.

Just for info: There is no DC current flow into the gate pin of a MOSFET (there is a gate capacitance that needs to be taken into account in fast switching circuits, but not here). The only reason for including a gate resistor in a circuit like this is to stabilise it against high frequency oscillations. A single 100R resistor would be fine, but it should be mounted as close to the gate pin (pin 1 in your schematic) as is sensibly possible*.

As above for the 10k resistors - the source pins (pin 3 in your schematic) should be connected directly to ground.

(* For example if the mosfet is connected to the board by flying wires, put the resistor at the mosfet end of the wire, not the board end.)

ETA: Is the new fan happy on 5V?

 

[eSCHEn]

Those were the least chonky MOSFETs I have - the others were 100A, 260A and 343A . I got them all years ago for an idea that I never implemented (story of my life lol).

As far as I understand MOSFETs (please correct me if wrong!):

• They have a capacitance which means when you energise the gate it is initially a dead short as the 'capacitor' is filling.
• You want to fill this up as quickly as possible so you're not drawing current whilst in the linear region. Ie you want to get into the saturated zone ASAP.
• Once the MOSFET is energised, the trench is full and so it will continue to be open and allow flow. The 10K is drain the charge out and close the gate.

I think I have the 10Ks in the correct place now. I've highlighted a couple of points (a and b) to show where I think I would be incorrect in placing them - that would make a voltage divider instead wouldn't it?

The Nano can do 40mA on a pin and 100Ω would give 50mA. I would adhere to specs but as the regulator chat above shows, those are not always the be and end all. If it was software I could simply check the source for the truth but hardware is not as strict . Is 50mA on the pin "okay" because you know it from experience? That's not a jab at all BTW, I just want to know how you got there . It would make me very happy to use less components, if only for less soldering.

Schematic so far, I haven't removed combined R2 or R5 as I would first like to understand the 40mA/50mA power question above

And again, thank you for your time and responses. It's hard work doing this solo!

 

[Misterg]

As far as I understand MOSFETs (please correct me if wrong!):

• They have a capacitance which means when you energise the gate it is initially a dead short as the 'capacitor' is filling.
• You want to fill this up as quickly as possible so you're not drawing current whilst in the linear region. Ie you want to get into the saturated zone ASAP.
• Once the MOSFET is energised, the trench is full and so it will continue to be open and allow flow. The 10K is drain the charge out and close the gate.

Correct, but the capacitance is small (<2nF) and can be ignored as this is essentially a DC circuit.

I think I have the 10Ks in the correct place now. I've highlighted a couple of points (a and b) to show where I think I would be incorrect in placing them - that would make a voltage divider instead wouldn't it?

Yes, they're correct. They actually form a voltage divider where they are now, but the ratio is 10,000 / 10,200, so it'll only drop the voltage by 2% (1% if you had a single 100R resistor) and can be ignored.

The Nano can do 40mA on a pin and 100Ω would give 50mA.

No. It's driving a mosfet. There is no DC current. The transient effect of charging the gate capacitance is over in about half a microsecond*. You can ignore it. The 100R value comes more from tradition than detailed calculation. It just needs *something* there to kill any oscillations. The value only needs to be chosen by design when you start getting into the 10s of kHz range. One 100R resistor per mosfet will be fine.

*time constant = R x C = 100R x 2nF (say) = 0.2 µS.
Capacitor will be 90% charged in ~2 x time constant ~0.5 µS

 

[eSCHEn]

Got a spare moment to update the schematic, I think this should be everything now. I took the opportunity to move the 10Ks to the other side of the 100R. I think this is where I should have done it above?

 

[Misterg]

That should work.

It doesn't really matter which side of the 100R resistors the 10K resistors are. Use whichever makes it easiest to keep the 100R resistors close to the mosfets.

 

[eSCHEn]

And just to tie this off, a picture of the fruits in question now growing nicely