27

You will need to check the datasheets for each part with unused pins. In many cases grounding them is a BAD idea, though in many other cases it may be CRUCIAL. And in some cases you SHOULD leave the pins floating. For example, do not connect outputs directly to ground, as this could cause a short. Do not leave inputs floating, unless they have internal pull-...


17

If you check the HW Design Manual then you'll see that it suggests the unused pins should be left floating. Please refer to Table 4 on p.15 and p.16.


13

One thing you CAN do when laying out the PCB is to ground all pins via a resistor on each pin. Then you can fit 0 ohms for a solid ground, 1K or 10K to stop the pin floating (without damage if something drives it, plus you can pull it high during test), or infinity (no resistor) if you must leave it floating. Final decision can be made for each pin during ...


9

A 555 is not really a precision device, and as you've found, if the resistor values or the capacitance value is too low you'll see relatively large deviations from the theoretical values, especially with the old bipolar type operated from 5V. Stray capacitance affects a 100pF capacitance value, and 150 ohms is low. Bypassing pin 5 has an effect on the ...


6

Unused output pins should just be left disconnected. For unused input or I/O pins you need to read the datasheet and/or make some discisions. As another answer points out the datasheet for your simcom device says you should leave them floating, so that is what I would do in the first instance. As a general rule though, a CMOS input should not be just left ...


6

Optical remote receivers such as IR Rx from Vishay/Sharp usually have AGC with a BPF and a Q=10 so you may want to be within 10% tolerance over temp to optimize gain. Consider a 38kHz XTAL chip $2 and CMOS inverter + passives https://www.digikey.ca/product-detail/en/abracon-llc/ABS25-38.000KHZ-T/535-10243-1-ND/2218056 The reasons for unpredictable k ...


5

Most (all that I used) of the impulse power converters have VOUT connected to the coil. This one has VIN connected to the coil. Apparently you are used to buck converterns, then. It might be worth a web search. A buck converter works by switching power on and off to the coil, so the coil input sees an average voltage less than the supply. Coils try to ...


4

1) The inductors are used to filter noise from supply voltage. As each supply node has its own inductor, noise caused by periodic gulping of current on one supply node cannot pass on to other supply nodes and cause problems taking the sensitive measurements. 2) That is a good question and there may not be an answer. It seems that the evaluation kit for the ...


3

The sync signal is not a clock or anything. It will just be occasionally toggled (once every minute maybe or only at bootup. Depends on what the system will need in the end). Your differential signal is passing through isolation transformers in your RJ45 magnetics. You cannot pass low frequency signals (or DC) through isolation transformer because ...


3

Here are the innards: This is a classic boost regulator. Figure 3. A classic boost regulator. Source.* The transistor shorts the inductor to ground causing it to "charge" up magnetically. when the transistor is turned off the inductor voltage rises to feed the output.


2

One simple way is to copy the device into your own library and modify its package there. If you continue using Eagle sooner or later you'd have to make your own devices. This seems to be a good time to begin. And while you at it I'd recommend adding pin names to the display package, it might be helpful later for connecting jumper wires.


2

Under normal operating conditions the diode D3 is reverse biased with only a small leakage current passing through it. If the input polarity is reversed the diode will conduct to minimize the voltage and protect the following circuitry. The forward voltage will depend upon the current available from the source - it could be about 1V at 30A. The following ...


2

1) What are these inductors used for in the design? Each inductor forms a LC low pass filter (L1 & C19, etc), to filter out noise originating from the 3V3 source to which they are connected to. The cutoff frequency is determined by $$ f_c = \frac{1}{2 \pi \sqrt{LC} } \approx 16 \text{ kHz} $$ 2) […] what specifications are more important when ...


2

The clue here is the signal names: x_P and x_N. These traces are balanced differential signals, which have a specified impedance. The red symbol on the schematic is indicating that these are required to be routed as a differential pair. This means calculating a geometry to establish the required impedance, and then the design software will try to maintain ...


2

As a point of reference an LTSPICE simulation of the 555 (at a transistor level) produced the following output frequency for your RC parameters. Seems that this simulation didn't manage to oscillate with that low-value reset resistor of 150 ohms as Spehro has suggested. It is always possible that this 555 model was developed to test other modes rather than ...


1

At least it is missing all bulk and bypass capacitors. The voltage regulator might be unstable. You also have chips gulping current at their respective frequencies, max being 1.2 MHz spikes, so any inductance and resistance in the supply wiring will cause voltage sags during each current spike, so chip supply voltages can have too much ripple for it to work ...


1

Might be an opengl issue. See https://forum.kicad.info/t/eeschema-not-responding-when-opening-schematicac/20621 (tldr you can switch kicad to software rendering within the config files. For details see linked forum post.)


1

Generally no input should be left floating. Check the data sheet to determine the logic level required for the application you have. If the input should be logic zero, ground it. If it has to be logic one, tie it to logic one. If the status does not affect your operation, check the name of it has a bar on it, meaning an active low input, connect it to ...


1

Looks to me like it has a switching power supply and can operate from either 5V or 12V. It's likely it might work okay on other (intermediate) voltages but it's not specified for operation other than at 5.0V+/-10% or 12.0V+/-10%. From the data: Also: That said, I would have a gander at the PCB near the input power connector and switching power supply ...


1

Such a simple passive summing circuit with too many inputs has another problem - its inputs are weighted and each of them depends on the number of the connected devices. If all devices are connected, the input gain of each of them will be only 1/6. If some of them are connected (other inputs are unconnected), it will vary up to 1/2 (when only 2 devices are ...


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