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12

A relay is almost certainly NOT the right tool for the job. The first step in wiretapping an interface in order reverse-engineer it is to determine the nature of the signals you want to look at. For voltage signals, you need a high-impedance buffer amplifier (like the input of an oscilloscope) that will cause minimal disruption to the existing circuit. For ...


8

The answer to your header question is NO! Input 19.5x3.33 = 64.935 Watts. Output 12x5.5 = 66 Watts. So your new output power is higher then your input power which is against the law of conservation of energy. But according to your question details your equipment requires only 48.5 Watts. That is do-able but only if your buck step down converter (or ...


8

The advert says it needs 5 Amperes. That’s hundreds of times more than a Raspberry Pi GPIO can supply. You would need a driver, preferably with isolation, and a separate power supply capable of at least 5A. One solution would be a CPC709J, with a suitable series resistor for the LED. You may well cause disruption or damage even with that, but the chances ...


6

Yes, it is possible power of laptop charger is = 19.5 * 3.33 = 65 Watts your loads power: raspberry pi = 5*2.5 = 12.5 watts Display = 12 * 3 = 36 watts total power = 48.5 watts Total power < power of charger, you can use it. it will work


6

It is far cheaper and faster to design around an already made fully integrated component than to dedicate all the engineering resources to put together elements that are rather standard. Many systems require a full-blown CPU, lots of memory, hard drives, and other peripherals along with a tried and tested operating system. Having communication interfaces ...


6

BJT's as switches aren't complicated. One input is the value of your power supply voltage -- call it \$V_\text{CC}\$. Another input is the desired operating current of the LED -- call it \$I_\text{LED}\$. Another input is the worst-case voltage drop of that LED when running at that desired operating current -- call it \$V_{\text{LED}_\text{MAX}}\$. Another ...


5

I'll say, for two specific boards: I am puzzled by devices like "nvidia jetson tegra" or "Snapdragon 835 Mobile Hardware Development Kit" and many many others. you could actually look this up yourself, but, here we go: The Snapdragon platform is Qualcomm's processor platform for the smart phone market. It's, like the processor on the Pi, a System-on-...


5

Apart from not having enough energy to drive that circuit: Generating sparks near a piece of electronics is never a good idea. Although the circuit has protection on some of the I/O ports, especially the HDMI and USB interfaces, there is non on the GPIO pins. Those have the standard ESD protection which is designed for, well... ESD. It is NOT designed to ...


5

The micro-lightning arc HV noise generator needs 1 or Li-Ion cells to power it. If should never be operated near any computer. If it operates near an R-Pi, it will cause functional failure and possible damage to signal ports on cables acting as an antenna.


5

It's possible, with a mosfet, but there is a better way. It's typically not good to use a mosfet to burn up power with. A more typical way is to use a hardware timer to generate a Pulse Width Modulation or PWM. A PWM is a pulse type signal where over the course of a timing cycle the power is varied by varying the size of the pulse. I use this type of ...


4

From your description, you have the transistor connected as an emitter follower, like so: simulate this circuit – Schematic created using CircuitLab With an emitter follower the emitter vol;tage is always about 0.7 V below the base voltage, so you can't get more than 2.6 volts across the LED. You should instead put the LED between the collector and ...


4

If your load is located 20 meters from 2N7000 switch (as the comment says), the problem is in wire inductance. A 20-m wire has self-inductance of about 50 mH. When the N-FET turns OFF, there will be spikes of voltage with kV-level amplitude, which will kill the FET. You need to use a clamping diode, as it is normally used when controlling coils of ...


4

The R'Pi is a 3.3V MCU so has insufficient voltage output on the GPIO to use the IRF730 you have. To make any progress you need to be able to understand the components you are using. Your MCU. R'Pi output is only 3.3V. You should be able to measure this using a multimeter. Set the GPIO pin high and measure the output voltage, it should be 3.3V. Set the ...


4

I2C is bidirectional, you don't need bidirectional level shifters for the WS2812B which uses a single-wire asynchronous protocol. For example, 74LVCH1T45 is a suitable unidirectional level shifter that will work with a wide range of input and output voltages. Edit: Bidirectional level shifters with discrete MOSFETs that use 10K pullups will likely cause ...


4

Don't use a 9V battery to power your servos. 9V batteries have low capacity (won't last long.) Don't use a 9V battery to power your servos. 9V batteries cannot really deliver the current needed to drive one servo, let alone 3. Don't use a 9V battery to power your servos. You say your servos are rated for 4.6 to 6 V. 9V is too much. Don't use a 9V battery ...


4

If I have understood everything correctly, you are safe with the 1200mAh battery. The nominal voltage is the same. The higher capacity is also an advantage. The C Rating is simply a measure of how fast the battery can be discharged safely and without harming the battery. (https://rogershobbycenter.com/lipoguide) This means with the 1200mAh battery, can ...


4

Do I need pull-downs for the level shifter inputs for when the GPIOs aren't being driven? Yes, I can almost guarantee this is the root cause for your problem. When you're not driving the GPIOs the inputs to the CD4504B are left floating, and it's always trouble to leave digital inputs floating. (By the way, why would you ever not drive the GPIOs here...?) ...


4

tl; dr: ZigBee isn’t a subset of WiFi (802.11). It is a different protocol family, just as Bluetooth is another different protocol family. ZigBee, BT and WiFi are not directly compatible with each other, but will co-exist in the same band. What Is ZigBee, And How Is It Different From 802.11, Really? ZigBee (802.15.4) is a family of protocols used for low- ...


4

Bit of background first: Generally, USB itself is not a bus over which a device can be really woken up if it stopped working completely, because it's a host-polling based bus, i.e. the host PC must inquire the keyboard in regular intervals. If that host is completely shut down, nothing one can do. However, being a host for a USB keyboard is relatively easy,...


4

It would be best if you provided current limiting for the GPIO port. It depends on the current gain of the NPN transistor, but at minimum a 200Ω series resistor from the GPIO to the NPN needs to be used to limit the current to less than 16mA. (and also use less than a total limit of 50mA on all ports)


4

If the train power supply is a relatively-constant 12 Vdc supply, simply feed the two 12V terminals into a bridge rectifier. The output of the bridge will be the same polarity no matter what the input polarity is. If you are concerned about the voltage drop in the bridge, create your own bridge rectifier using 4- Schottky diodes. Choose appropriate diodes ...


4

This is not possible as the two 5V pins are connected together and both referenced to the same ground potential. Either you should look for a 5V fan or get a boost dc-dc converter to get to the required voltage.


3

There are several methodologies to tune a PID, some can be fully automated and rely on some level of system identification. But most of them are just rules of thumb. For temperature control, you want to avoid the integral term as much as you can. It is necessary to remove steady state error, but any time spent far from the set point is time that the I term ...


3

The key word is development kit, or evaluation board. This is mostly covered by Marcus's answer, but I'd like to make clear what the process is. The target market is companies building what's known as an embedded system. Things like point of sale systems, set-top boxes, industrial control systems, scientific instruments, self-driving cars. In the case of ...


3

You are better off with a Raspberry Pi or a Beaglebone board, as either of those are full-blown computers that are directly running Linux variants. You would find that the standard implementation of tcpdump would directly work on them (if it's not already included in the base operating system). The Arduino Ethernet shield is basically a separate processor ...


3

If you use a load cell rated for 200kg full load, with an output of 2 mV/V, and a supply voltage of 5V, then your output voltage will be 10mV for a 200kg load or 5mV for a 100kg load. A change in the load of 2kg will cause a voltage change of 0.1mV. If you are digitizing this with a perfect ADC using a reference voltage of 5V you need at least 16 bits of ...


3

I made a small robot with 4 28ByJ-48 5V steppers, however I think the 5V is the minimum recommended voltage they will run on. By increasing the voltage to the motors from 8V to 11.5V - I noticed quite a substantial increase in torque, and I could run them at a higher RPM before they started mis-stepping. I was able to run continuously and over heating was ...


3

From your schematic, the current flowing from PB0 to T1's base-emitter junction is \$I_B = (5V - 0.6V) / 10\Omega = 440mA\$, which is extremely high for a MCU-pin's drive capability. Thus PB0 may limit the output (I'm not sure, but I hope). To saturate the transistor, a base current of \$I_{Bs} = \frac{I_C}{ (\beta_{min}/10)}\$ is sufficient (I cannot ...


3

Your thermometer doesn't seem to have any way to talk to the rPi. It is much simpler to get a DS18B20 probe for about $2 on the 'net, and google a tutorial to interface it with your Pi. It's a digital probe, so you don't need to amplify or condition an analog signal.


3

The two diodes (D2 & D3) work to OR the voltages, the idea being your 12V_IN will be slightly higher than 12V_BATTERY, so current will flow from 12V_IN to power the device when its present... Minus a diode drop. When 12V_IN is removed current will flow from 12V_BATTERY through D3 R1 is probably there to act as a current limiting resistor for charging ...


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