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8

Yes, they are buses. The address, and the data bus. They are basically a way of drawing lots of nets without having to draw each individual one. The label by the nets that connect to it define what connections are inside the bus.


0

Interestingly, an XNOR gate turns out to be much easier to implement in NMOS than CMOS; if the signals which are feeding to an XNOR gate aren't used for anything else, an XNOR gate may be implemented using two transistors and one passive pull-up. If the input signals are also used for other things, adding inverters on both of them would increase the total ...


0

That second one is really clever! It recycles transistors from the XOR structure to invert A and B. I bet someone got a patent out of that. I'm not exactly a CMOS optimization master, so maybe it's more common than I think. It works like your first schematic, except that T6 and T9 are also used to invert A. You can almost do the same thing with T2 and T5, ...


2

Great work! This is a very minimal circuit. They did a good job on keeping the cost down. Combining the circuit, and my knowledge of how these little lights work, I will provide this answer which is not totally complete. Basically, I think you are on the right track. At night, the solar cell is dead, so the current through R1, R2 is very low/zero. D1 is ...


1

Found my problem, i was just an idiot and put vcc to ground pin instead of ground haha.


1

The transmitter does indeed turn the DC into an oscillating signal, however it's worth pointing out that it's more of a sinusoidal signal than a typical square wave. That's the purpose of the capacitors- to resonant in a "tank" circuit with the inductors. That allows more energy to be transferred and there is less energy in higher harmonics (if the ...


1

Is that how this is working? Yes, you have a correct understanding of how it works. Also, are the resistors in the receiver circuit put there for the female USB? Why? Apple usb charging standard. Signals to the ipod what kind of charger and how much current it can pull. Also when I bought the diodes for the transmitter circuit, I could only ...


0

You can use a MC34063 to generate 12V at something like 100mA maximum. Follow the design equations in the datasheet for your each situation to calculate the component values. The circuit will be similar to this: If that's not enough current there are many other possibilities, just do a parametric search.


0

A (simple) amplifier cannot (generally) output voltages outside the supply rails but you could set (with resistors or similar) the output of a spare amp to -12V if you had one with a negative supply (below -12V) and use it as a low current reference voltage. A previous boss used a circuit with just one CMOS hex inverter IC (cannot be HC or HCT version type ...


0

Texas Instruments has a selection of inverting buck converters here which will get you a negative voltage from your 24V supply. Use a 7912L to regulate the negative voltage (if you need to) and that should do it.


1

If you do not have a negative voltage input, then you will need some sort of DC to DC converter to generate it. Depending on how much current you need, you might be able to get away with a simple charge pump, or you may need something a bit more complicated. You can generally buy complete DC to DC converter modules that you can solder on to your board if ...


6

The circle at a pin location of a digital logic chip indicates that the signal is active low rather than active high.


2

The LED probably doesn't illuminate because there is virtually no voltage across it (both ends at about 30V) but that's not the main problem. The transmitter schematic shows you are applying almost Vgs=30V to the gate of a FET with Vgs absolute maximum of +/- 20V. You almost certainly need to replace the FETs, as well as rethink the gate drives to the ...


1

Your bridge rectifier is not wired correctly, D3 and D4 should not be in series, and there's no connection from your bridge rectifier to the rest of the circuit's ground.


2

R3 is 94 ohms, and in parallel with the LED and R1. You would get at best 0.3V across the diode and R1 combined. Combined with the 1kΩ of R1, and you're not going to really see any significant current.


1

A typical LED has a forward voltage drop of about 2V for normal operation, which leaves around 28V (30-2) on the R1,R2 resistors, which gives 28/(1K+10K) = 2.5mA current flow, which is not enough for typical LED for normal operation. Check your LED parameters, and match the resistors accordingly.


3

its simple, the controller generate a square wave in S1, (probably its at VCC level at the beginning) your circuit U9 (Optocoupler) will work when the diode inside it is conductive at low level, so the transistor Q1 is operational (ON) and you can enable the heater. The LED D11 is conductive when the transistor Q1 is ON.


2

When SW1 connects the LED cathode to ground, the LED lights up (inside the optocoupler, in infra-red). The light causes the phototransistor to conduct. This raises the gate voltage of the MOSFET, and the MOSFET conducts. Current will flow through the MOSFET from the 24V supply, via the heater. So, to turn the heater ON, connect SW1 to GROUND. Note that this ...


1

For schematics point of view I suggest you trying ZUKEN E3.Series or Logical Cable (Mentor Graphics) For CAD point of view, there is CATIA (v5 or v6) of Dassault Systems and UG_NX of Siemens... These are the most common ones used in automotive industry...


0

As you've figured out, these boards all do the same thing electrically. The relevant pins of the part are brought out, as well as the manufacturer-recommended grounding and bypass capacitors. (It would be wise to read the datasheet and double-check these yourself.) If you are going to use the ADXL345 on a multiplexed SPI bus, you need an additional logic ...


5

Yes they are part of the filter. The top section relies on R4 and R5 and C1 for shaping the frequency response of the op-amp circuit. The 100k pot has no part to play in the filtering other than guiding the current that reaches the inverting input of the op-amp from either the actual input or the op-amp's output. Maybe try researching Baxandall tone ...


3

There is no difference. Many times the choice of where to draw the bubble is made as to where the signal level is active low. As such the symbols that you show can be implemented with the same circuit or chip. In similar manner experienced engineers will also choose to draw other logic gates on the schematic according to the actual usage as to where the ...


0

Use another multiplexer to read the 8 ID switches, selected by the same 3 bits as the others and with its output going to a single digital I/O pin. That should save 7 pins, leaving 3 spare for other things that you haven't thought of yet. You could use a digital multiplexer such as MC74HC151 or 74HC251, but it might be cheaper and easier to just use ...


0

There is a free and open-source CAD package called LibreCAD. It may do what you want, if AutoCAD is what you're expecting.


0

You do can multi-channel in Altium, using the sheet symbol. You create a schematic with ports representing the inputs and outputs to your repeating circuit, then place those on your main sheet, connecting up your ports. When you create your PCB, those circuits will be placed in 'rooms', which you can then setup a single room, then copy it to the other ...


0

For a very large number of copies use hierarchical design techniques to create rooms repeating circuit blocks a large number of times. After aranging rooms, then you can use Tools >> Rooms >> Copy Room Formats to make them all the same very rapidly. This copies everything about the previous template room including component orientations, tracks, vias, ...


5

I remember seeing a similar question here (but couldn't find it). The component seems to be a transmission gate, or analog switch. (http://en.wikipedia.org/wiki/Transmission_gate) Google-ing it helps you find the schematic: https://www.google.hu/search?q=transmission+gate+symbol



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