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Final review before ordering

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Peter -

As you are probably aware from my questions, my background is not in electrical design. All your comments are appreciated and most welcome. Having followed your posts ... it seems you are a very active member of the forum.

There are a lot of terms and new techniques for me to digest. Let me do my homework, and I will resubmit if you are willing to humor an old man learning new tricks.

I think you are right ... the contactor is not needed ... just look for power "off" from the native power ... and let the other power connector with the UPS handle the MKR 1000 and all the peripherals. I just need the UPS to work long enough to send a message indicating a problem has occurred, and maybe keep the "buzzer" active to draw some attention.

The "resistive divider" I assume is the same as a "voltage divider" ( a series of resisters to drop voltage)? Is the 100 ohm resister your suggesting from ground to D8 (the input pin) acting as a "pull down" resister. And the 68 ohm resister is the "resistive divider" to drop the 5V to approx 2.4V?

Have never worked with transistors ... so I am going to have to do some work before I can ask intelligent questions, but I think I will start with the "model of a simple switch" where the "base" receives a low current which allows a larger current to flow from the collector to emitter.

So my assumption at this time is connect the low voltage from an output pin (using the "set high" command) to the base of the transistor, and allow the greater current (5v) to flow from the collector through to the emitter. From there to the + buzzer, and back to ground.

I am attaching a part of the data sheet from Jameco for the 2n2222 which you mentioned. I just need to figure out what all the information means. If I am reading this correctly the information I need to understand is in the "On Characteristics" section.

I am using a Piezo buzzer running at 3 - 15 VDC ... drawing 10 mA at 12V. (Jameco part #2117430). 90 db is nice and loud, but the lower voltage will probably result in a lower sound volume.

I will try to deal with the "power supply" bypass when my head is a little clearer.

Thanks again -

Jim


Final review before ordering

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One final thought -

I looked at the spec sheets for the Arduino MKR1000 and the UNO.

The MKR1000 is showing a rated I/O pin current of 7 mA, and the UNO does show between 20 and 50 mA depending on the output voltage.

Will the 7mA be enough to switch the gates on the 2n2222 transistor?

The UNO data sheet

<img

The MKR1000 data sheet:

src="/uploads/default/original/2X/e/ee080378a058f506f8cd4e1e484a15e16807ce34.png" width="497" height="500">

Final review before ordering

Final review before ordering

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I'm retired these days so I have time to do what I want :slight_smile: and I have 40+ years of getting answers from folks to pay forward ...

Yes, in fact here is a quick schematic of both for discussion:

the voltage divider is 168 ohms over 5V so 29.7 and change ma flowing. Thus we see 2.9 V at the 100 ohm resistor and 2.01 V across the 68 ohm resistor with 5V (and of course 0V at both and no current when 5 V goes away). So the micro input sees between 0 and 3V and is happy. You are correct, the CPU in the MKR 1000 has lower output drive, I think 8ma minimum from the data sheet. That is more than enough for the 2n2222, the value of importance is hfe (or current gain) at 50ma collector current it is at least 70 which means it will sink 70 times the base current. So 5ma in to the base will mean the transistor could sink up to 350 ma and we only need 50ma so 5ma would be lots of drive and not strain the cpu pin. Now we need to calculate the base resistor value, the emitter base junction will take about .6v of the 3.3v input leaving us 2.7v across the resistor. So if we choose a 470 ohm resistor (rather than the 330 shown) we get around 5.7 ma of base current and everybody should be happy. The diode across the relay coil absorbs the inductive kick when the transistor switches. If you replace the relay with the buzzer (which isn't inductive) you shouldn't need the diode.

Peter

Ground plane erstellen, aber wie?

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"Copper fill" fills empty spaces that are not connected to ground.
"Ground fill" connects all grounds and fills empty spaces, but you have to manually "Ground Seed" the ground ends of parts that are going to be connected.

Google translate
"Kupferfüllung" füllt leere Räume, die nicht mit dem Boden verbunden sind.
"Ground Fill" verbindet alle Gründe und füllt leere Räume, aber man muss manuell "Ground Seed" die Bodenenden der Teile, die verbunden werden sollen.

Ground plane erstellen, aber wie?

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Thank-you Peter!
reason for writing in German was to quote the german dialogs. I'll try to translate them here.

  • First I click on 'routing' and choose 'ground plan' (Massefüllung)
  • subdialog window opens offering:
    Copperplan (top/bottom)
    Groundplan (top/bottom)
    Seed for Groundplan
    ... etc

  • clicking Seed for Groundplan a new window opens telling how to choose a seed
    (already done before by highlighting an element with a ground connection or
    alternativ by highlighting a ground track)

  • clicking 'Routing' 'Groundplan' same dialog as for clicking 'Seed for Groundplan' opens
    on the bottom one can choose 'OK and Copperfiing' 'OK' 'break' => no 'OK and Groundplan' to choose here?!

  • clicking 'OK and Copperfilling' fritzing does excact this: producing a copperplan whith no ground connections :slight_smile:

What do I missing or doing wrong?

Ground plane erstellen, aber wie?

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Hi Old_Grey
thank-you for the hint. However how do I this
"you have to manually "Ground Seed" the ground ends of parts that are going to be connected."

Sorry this might by a stupid question but didn't find out :frowning:
Thanks!

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Ground plane erstellen, aber wie?

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Click on the part in PCB view, right-click the pad you want to ground, and select "Set Ground Fill Seed"

I'm no ground seed expert so i don't know how the dialogue box works.

Final review before ordering

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Thank you Peter -

All good stuff ... let me absorb these concepts, and I will reply after doing some breadboard mockups to get my understanding a little clearer.

One quick question - If I look at the voltage divider diagram on the left, I would think that 5V input going to the digital input pin would only see 68 ohms of resistance (before the branch) and not a total of 168. The 100 ohm resistor is "after the branch" ... what am I not understanding. I can see that anything after the total of the two would see a voltage drop based on the combination of the two resistors.

I will do some mockups and think about it some more.

I=V/R
I=5/68
I=.073

7 mA correct?

Jim

Ground plane erstellen, aber wie?

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Thank-you Old_Grey!
I've tried exactly this several times before with no success. Maybe someone will try this procedere on his/her machine and will report here.
Btw. after selecting a part in PCB view and right.clicking a ground pad the popup box dosen't show "set ground fill seed".Iit just shows "order", airwire, delete.
Maybe something is wrong in my schematic. All ground connections are tied together and connected to one ground symbol "Ground1" from inspector.
Thank-you for patience

Ground plane erstellen, aber wie?

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Upload the .fzz here using the 7th button.

Final review before ordering

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Yes, but :slight_smile: I (and I expect most experienced folks) tend to forget that newcomers don't have the other laws that apply hard wired in to their brains. In this case kirkoff's law comes in to effect because this is a resistor network. We need a bit more complete (but still not anywhere near reality such as a spice model) drawing of this equivalent circuit. We still have the external voltage divider R1 / R2 but the cpu input pin is a bit more detailed, specifically its input resistance (which is probably really higher than a meg) and the substrate diode which is the real issue in over voltage on the input pin. This drawing contains a more accurate representation of both the voltage divider and the 20 feet of 26 wire to the sensors (which we will use later).

For why the voltage divider works the way it does, we need to add kirkoff's law to ohm's law. Essentially kirkoffs law says the sum of the currents flowing in to a node equals the sum of the currents flowing out of the node. For us that means resistors in series add their values (so the 100 ohm and 68 ohm resistor provide 168 ohms across 5V). The current flowing in the 1M gate resistance provides so little current that it can be ignored (if it was 100 ohms the the cpu input pin would see 50 ohms rather than 100 and the voltage would change in an undesirable way). Essentially the values (under normal conditions) of the components in the input pin for the cpu make have no effect on the voltage divider and the voltage at the input pin will be 3 volts. The problem with supplying more than 3.3 (actually more than 3.9 or so) volts to the input pin is the substrate diode that connects from the pin to the 3.3V supply in the chip. If we supply 5V to the input from a low resistance source (such as a 5V output driver that can source 40ma) the diode will conduct and attempt (and if it has enough current will succeed) to drag the 3.3V power supply and the chip to 5V. High current and damage to the 3.3V chip are likely to occur here. I know this from unfortunate experience because I tried to do some 3.3V level translation with a 74hc series chip (with the mistaken belief that it didn't have the substrate diode, which in fact is only true for some members of the 74ahc family). Luckily I had disconnected the 3.3V device that I was driving before trying this so all I blew was the 74hc device and the driver pin on my arduino CPU chip as it dragged the 3.3V regulator up to about 4.5 volts through the 74hc chip.
A 74ahc chip in the same circuit works fine (and the document that stated 74hc did too has been removedfrom the site where I found it :slight_smile: ). With that done, now on to your sensors 20 feet over 26 wire ... To the left in the drawing you see an again simplified representation of what that wire looks like electrically. It has resistance (about .8 ohm for 20 feet of 26 according to an online calculator) and inductance and capacitance. The resistance might affect the voltages seen at the sensors, it is also small enough it may not matter. There are two considerations here: first as noted the sensors want somewhere close to 5V (or 3.3V) for power and may malfunction if the power drops below about 4.5 or 3V respectively. Bypass capacitors at the sensor end will help with this (as will the fact that the sensors don't draw much current typically. If you can using two wires to supply 5V and ground to the sensor end of the cable and then with bypass capacitors on the sensor end of the 5V lines use a 3.3 V LDO regulator to provide the 3.3V to the sensors. The other related issue to be aware of is signal level noise immunity. A digital input typically accepts between 0v and .7V as being low and 2.4V of higher as a high (and voltages between .71 and 2.3 are undesirable for a variety of reasons, and should be transitioned through fast). The capacitance, inductance and resistance in the 20 feet of wire conspire to make this difficult. The effect of the three is to slow down the transition from a 1 to a 0 (and visa versa). If there is sufficient current flowing (which there likely isn't in this case), the resistance can produce a voltage drop in the wire where 0V at the sensor shows up as .8V at the cpu end (the .8v being dropped by the .8 ohm resistance of the wire if there was 1 amp of current flowing). The effect this can produce is similar to what bypass capacitors help with on the power supply leads: it works 99% of the time but %1 fails because a 0 was read as a 1 due to voltage drop in the cable. It would be a good bet to try the sensors on the cable preferably in the place where it will be used to make sure it works acceptably. If not you would need to consider line driver options to get the data reliably from the sensors to the cpu (although I suspect this may work fine as is).

As it stands you won't get plated through holes. Your pcb is set to single sided and it needs to be double sided to get the top copper hole to plate through to. If you have no objection to double sided boards, that would fix your jumper issue as well. If you change to a double sided board on both you will get the top side pads plated through holes and on the cpu module can route the jumper connections on the top layer getting rid of the jumpers (I have been assuming you wanted to keep the design single sided, which perhaps you do).

Peter

Disable Auto Update

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I would also like to know how to disable the automatic update feature. In an education environment, users don't have the permissions to update...

Creating parts with movable schematic elements


[Solved] Cannot load Parts

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I have solved my issue (which seems to have been a couple of problems).

Fritzing on macOS stores data in a number of locations:-
1. Within Fritzing.app /Applications/Fritzing.app/Contents/MacOS/fritzing-parts
2. User files, typically ~/Documents/Fritzing
3. Configuration files ~/Library/Preferences/org.fritzing.Fritzing.plist
4. ~/.config/Fritzing/backup

There seemed to be a number of issues:-
1. .config/Fritzing only seemed to contain very old files from an old version of Fritzing, so I deleted it.
2. ~/Documents/Fritzing seemed to have 2 parts lists, viz. my_parts.fzb My Parts.fzb
3. When I deleted and reinstalled Fritzing I had 2 MINE parts, and was unable to delete either, and still was unable to update.

I backed up, then deleted ~/Documents/Fritzing ~/Library/Preferences/org.fritzing.Fritzing.plist
I reinstalled Fritzing (yet again) and it worked, and updated OK. I had to manually set Preferences.
I gradually restored all my User files, manually combined parts into my_parts.fzb and it now appears to be operational.

[Solved] Cannot load Parts

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Thanks! The feedback may let us help the next person that runs in to a problem, and happy that you managed to get it resolved.

Peter

Disable Auto Update

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With no suggestions from anyone on how it can be disabled, the only thing I could suggest trying is blocking access to github in the firewall (either on the machine or in a perimeter firewall). A Mac user that was having update problems due to an old ssl version discovered if the network was disabled, Fritzing didn't try the update allowing him to use frtizing.

Peter

Electronic Designer Needed For A Short Project

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Hi Everyone, Jonathan here, I am looking for an Electronics Designer for a short project that I am willing to pay for for a few days, maybe weeks depending on how good their work is. Please E-Mail me for more information . Thanks Jonathan - E-Mail: info@asianreplicas.com

Creating parts with movable schematic elements

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