Paul Newton

Made a terrible mistake going over "Getting Started with the Internet of Things"

Oops.

A nice puff of smoke (from the meter, not the board) and a horrible smell..no mosquito's tonight.

I assume you meant the potentiometer?
So the wiper must have been set to short circuit the 5V to the Gnd.
Time to do some fault finding....

If we assume you were powering the board from USB, then you may have damaged something between the 5V pin and the USB 5V input. Or, you may have damaged the USB circuitry in your PC.

PC
First I would suggest trying another USB peripheral (e.g. a mouse) on the same USB port you had the Netduino plugged into. Does the mouse light up and run normally?

• If so, the PC is OK, Netduino must be damaged.
• If not, the protection circuit in the PC may have blown. Try a different USB port.

In the past I have seen multiple PCs destroyed by a poorly designed USB JTAG unit, after the third PC died, we worked out why they were dying and subsequently always used a powered USB hub between the (new) PC and the USB device. We did not blow up any more PCs after that. Of course the PC manufacturer (<INSERT FAMOUS BRAND NAME HERE>) would never admit there was anything wrong with the design of their USB output protection.

Netduino
Looking at the circuit diagram, there is a fuse and a transistor between the USB 5V input and the 5V output pin. The transistor is used to allow the USB to power the board when there is not a power input on the Vin / barrel inputs. There is a also a PCB track linking everything up, this could also act as a fuse - lets hope that it didn't.

I think it would be worth you trying to power the board using one of the following methods:

• apply 7 to 12V DC on the barrel connector
• apply 7 to 12V DC on the Vin pin
• apply a regulated 5V DC on the 5V pin (this pin can be used as an input)

Then try connecting to the board and running a program using USB as normal.
If this works, I think we can assume your board has blown the fuse or the transistor. But, is still usable. We could probably hack the board to route power from the USB input to the 5V line if you really wanted to.
If it does not work, a track may have burnt out, or something else I missed has happened.

Hope this helps identify what is actually wrong.
Let us know what you find - Paul

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Hi Frankie, I'm not sure why, but I seem to have missed several updates on this post so I've got a bit out of step with all the updates and suggestions! That last diagram looks right. Before I saw your diagram, I drew the attached this evening to show you what I meant. 1/ The circuit from the example you started with. The transistor is fully ON, so it has a voltage drop of about 0.3V and most of the 5V supply is available to the load. The base of the transistor is about 0.7V above the emitter. In this circuit the base voltage does not really matter. 2/ Using this to drive the DIS. Note that the negative on the 12V battery is NOT the same as the Gnd of the Netduino. The transistor bridges these signals when it turns ON. 3/ I think this is what you had initially. In this circuit the base emitter voltage matters. The emitter must be about 0.7V less than the base, and the base can only be 3.3V at the very most. Hence the emitter can only be 2.6V. (3.3V - 0.7V = 2.6V) So the load only gets 2.6V not 5V! The rest of the volts are lost across the transistor which acts as a resistor because it is not fully ON. 4/ Using a second (pnp) transistor. Now one npn transistor turns on a pnp transistor. In this case the 0.7V base voltage drops are both next to the rails, hence both transistors can turn on fully and the load gets the full voltage. 5/ How it would be applied to the DIS. Now all the circuit uses the same ground. I hope this sheds light on the issues - Paul PS, should I ask why you are making big sparks?

Hi Mickey, Welcome to the forums! You will need to use a shield if you want to control motors, the Netduinos cannot drive motors directly. You might consider getting some cheap RC servos instead which can be driven directly from the Netduino PWMs. This may sound more expensive, but two servos may be cheaper and easier to use than two motors, a shield and various bits of hardware (gears etc.). The Netduino Plus will allow more easy connection to the internet through its ethernet port. However, I don't think you can stream video through the Netduino, and I don't think you can connect a USB device such as a webcam to the Netduino. Hope this helps - Paul EDIT: Just saw Dave's post below. I missed the "two motor track" statement in the original post - I was thinking about using RC servos to pan and tilt the camera. Not for driving a tank. Paul

Hi Galibore,

Reading the table on page 2 of the data sheet, the forward current of all the LEDs is 350mA. At that current the Red LED has a voltage drop of 2.4V, Green 3.5V and Blue 3.4V.

So if you want to use a 12V supply, you need to drop quite a lot of volts!

First lets work out the minimum resistance to limit the maximum current through each LED:
Red, R = V / I = (12 - 2.4) / 0.35 = 9.6V / 0.35A = 27.42 ohms
Blue, R = V / I = (12 - 3.4) / 0.35 = 8.6V / 0.35A = 24.57 ohms
Green, R = V / I = (12 - 3.5) / 0.35 = 8.5V / 0.35A = 24.28 ohms

The resistors are going to have to be quite powerful. Typical small wire resistors are only good for 1/4 Watt. The above numbers give the power in each resistor as:
Red, P = V x I = 9.6 x 0.35 = 3.36 Watts
Blue, P = V x I = 8.6 x 0.35 = 3.01 Watts
Green, P = V x I = 8.5 x 0.35 = 2.975 Watts

That's a lot of power to turn into heat! You will need to find suitable high power resistors to cope.

Next we need to add a variable resistor (potentiometer) in series with each of the fixed resistors.
Each variable resistor should go from zero to ? ohms to reduce the current below 350mA.

Looking at the forward current versus forward voltage curves on page 4 of the data sheet, the forward voltage curves stop at about 25mA. So lets say the minimum current should be 25mA.
For the Red, the curve shows about 2V at 25mA. This gives R_total = V / I = (12 - 2) / 0.025 = 400 ohms. R_variable = 400 - 27.4 = 372 ohms.
For the Blue, the curve shows about 2.375V at 25mA. This gives R_total = V / I = (12 - 2.375) / 0.025 = 385 ohms. R_variable = 385 - 24.57 = 360 ohms.
For the Green, the curve shows about 2.325V at 25mA. This gives R_total = V / I = (12 - 2.325) / 0.025 = 387 ohms. R_variable = 387 - 24.28 = 363 ohms.

At the lowest current the power in the variable resistors will be P = I² x R
Red P = 0.025 x 0.025 x 372 = 0.23 Watts
Blue P = 0.025 x 0.025 x 360 = 0.225 Watts
Green P = 0.025 x 0.025 x 387 = 0.24 Watts

When the resistance is reduced more current will flow and the power in the variable resistor will increase. Its a curve relationship, so I can't do the calculation in my head. But you get the idea, its more power than a cheap variable resistor will want to handle.

First thing I would do, is use a lower power supply voltage. With 5V the fixed resistors are going to need to be about 1 Watt each rather than 3 Watts - this is more manageable.
Then look for some high power variable resistors.

Hope this helps - this is why we use PWM to drive high current loads.

Paul

Hi,

Lots of members use Fritzing, it has three modes: schematic, pcb and layout. The layout view is popular for posting diagrams of hardware on breadboards because it looks real. It is also linked to a PCB manufacturing option where you can pay to have a pcb of your project built for you.

More recently I was introduced to circuit lab, it runs in a web browser. It allows a schematic to be drawn, and then the circuit to be simulated. You store the circuit on their website and can post links to it.

Both of these are free.

Hope this helps - Paul

ps what is "RAD"?

You know you're old when you tell a youngster they look sick and then have to explain that it is not a complement.

You can lead a chimp to software, but you can't make it write a driver.

Hi Randy, Welcome to the forums! The data sheet is a little confusing. It does indicate that you can use an RS232 interface, so yes you can communicate with it using a MAX 232 family chip. It also says in the first pages that the level is switchable between RS232 and LVCMOS. If true, this suggests you can connect the device directly to the Netduino without using a MAX chip. BUT, I can't see any further mention of how to do this. I was expecting to see different pins for the LVCMOS serial or an input that turns the RS232 on or off. There is a diagram showing separate RS232 and LVCMOS UARTS. The final confusion is that in tables of pin connections at the end of the data sheet it also shows I2C - but that is also not mentioned. I would start off by assuming you have to use RS232, and once you start using the device try and investigate whether the LVCMOS serial connection is possible. Its a shame when the data is not clear. Have Fun - Paul

Yup! Each digital IO is a digital IO. Most also have a secondary function - like PWM, UART, SPI. Its up to you to chose how to use them. Paul

Hi Mori, Welcome to the forums. Yes, you can use the Netduino to control relays. You will need to use a transistor plus diode and resistor per relay since the Netduino cannot drive a relay coil directly. You will have enough pins left to use for your four inputs too. Hope this helps - Paul

Thanks Mario I was just checking the wiring for the UART , Andre, you have connected the RX of the LCD to the RX of the Netduino. I think you need to swap them over so that the tx from the LCD goes into the rx of the Netduino and vice versa. Regarding your original question, "do you need the regulator" it all depends on how much current it needs. I think most of the small monochrome LCD displays are quite modest and would probably be ok driven directly from the Netduino's 5V output. So it comes down to: how much current does the LCD need to draw? You said 450mW , if that is the power disipated by the regulator, then that's about 110mA (I = P / V = .45 / 4 = 0.1125= 112.5mA). Which I think is fine. A lot more and then you are going to start to make the Netduino's regulator get warm. You might want to keep hold of the regulator in case you add lots more kit to the robot. Mine has now got three of the Sharp analogue range sensors and plans for three more digital ones. Also I have a block of 8 LEDs to tell me what the buggy is thinking. It starts to add up. If you keep it, you will need a couple of capacitors on it. One on the input and one on the output - they keep it stable. Hope this all helps - Paul And watch out for those pluses and minuses EDIT - I wrote the above before your last reply. They crossed over in space.

Too quick. Just followed your links I can see that the motor driver breakout board has what appear to be capacitors already fitted. Spotted that the Vcc on the motor driver is not conected. I think this should be conected to the Netduino's 5V. (Check that is the correct voltage - e.g not 3.3V) Almost 11pm here. Good night!

Hi Galibore, I spotted a few things about the diagram: 1/ I think the 6V battery is connected the wrong way round. I guess that's just the diagram, not the actual circuit. 2/ The batery terminal conected to the ground of the Netduino should realy be connected directly to the motor driver. As shown, the motor curent is pasing through the Netduino's PCB. This may cause various issues with electrical noise being aded to analogue and digital parts of the circuit. You do need to join the battery to the ground, but use a wire from the motor chip to the Netduino for that. 3/ The 7805 regulator needs an input voltage greater than 5V. If it is needed, connect it to the 9V battery. 4/ Check the data sheets for the regulator and motor chip, they should both have some capacitors fitted to their supply lines to reduce electrical noise. I think 2 may be the cause of your problems. Have fun - Paul

Hi, If you use the Netduino plus, you might also consider an ethernet to wifi adapter (instead of a wifi shield). I saw some around £30 on amazon. Note that I have not got around to trying one out myself. Paul

Thanks for pointing out the mistake CW2.
Why do they call it Master Reset if it only resets the shift registers?


Given all the good ideas above I think I would combine two of the ideas:
1/ Connect the notOE signal (pin 13) to a GPIO line - this will automatically be pulled high at reset or power on, and will tri-state the outputs of the 74HC595.
2/ Use resistors (preferably a SIL array of 8 resistors with a single common) to pull the outputs low when the device is tri-state.



When you are ready to use the shift register, write out a zero word first using SPI to the shift register and then energise the GPIO connected to the notOE pin. (The notMR pin could just be left high.)

Paul