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2

Somebody already gave you the answer but I can answer your question "what I am doing wrong here?" For Node 2, you forgot to include the current \$i_x\$ and you also forgot to include the current from the \$5 \Omega\$ branch. Additionally, you don't need to add the current from the \$100 \Omega\$ branch because that current is going into Node 1. For ...


1

Your equations are wrong, they should look like this, denoting currents as being positive when running into the nodes: Node 3: \$ v_3 = 20 \$ Node 2: \$ \frac{v_3 - v_2}{10} + \frac{v_1 - v_2}{110} - \frac{v_2}{5} + 2 \cdot \frac{v_3-v_2}{10} = 0 \$ Node 1: \$-2 \cdot \frac{v_3-v_2}{10} - \frac{v_1}{100} + \frac{v_2 - v_1}{110} = 0 \$ Calculating it ...


1

The CD4050 is a digital logic buffer that can be used to shift signal levels from high to low voltage. It is not a voltage regulator - however you don't need one anyway because the EK-LM4F120XL already has an on-board 3.3V regulator. But perhaps what you are really worried about is the signal levels between your 3.3V MCU and 5V peripherals. The ...


0

It seems I may have found a solution ( not fully tested as yet but looks promising). I used a Mosfet based low battery cut off controller and "resistor-ed" it to suit a 6v SLA battery. I researched a Jaycar Electronics Kit ; KC-5523 (called Battery Saver) and combined with a bi color LED indicator module kit (also Jaycar) I have a fully functional unit ...


0

A buck-boost switching regulator would work best. It would regulate down when above the output voltage, regulate up when below the output voltage, and have a pass through region when the input is close to the output voltage.


1

Here's a really cheap, simple way to do it in hardware; no code, no debounce, no input-to-output delay, it just works. TRIGA is one high-going input, TRIGB is the other, and OUT is, well, the output.


0

What you describe sounds a lot like a model on an analog computer. Start with the differential equations that describe the flow. Dig out a GOOD textbook on analog computing. They're still around, but they are NOT easy to find. Do your homework. You will wind up building a system of first-order differential equations that describe your model, scaling the ...


0

First see JRE's answer for the relevant equations. I'll be a voice in the wilderness and suggest that it's not crazy to make an analog model. I've done this a few times when trying to model heat flow in an instrument. (Temperature -> voltage, heat-> charge, heat capacity -> capacitance, thermal resistance-> R) The advantage for me is that I can use some ...


0

Put a diode across the coil, cathode to 12V_PV, and as everyone else has told you, get rid of R5.


2

I found some links to previous analog hydrology models - not the one I had in mind, and I was surprised how much stuff was done in that direction. These might help you see what is involved. They all work with ground water, but the principles should be the same. A look at some of the hardware should show you why everyone is suggesting you go digital. I ...


5

It makes no sense to try to model this with electronics. You'd need a whole lot of resistors and capacitors connected in a grid. That would be time consuming to build, expensive, and failure prone. Model your fluid flow on a computer. There are various finite element modeling and solving techniques that have been applied to fluid flow for many years. ...


2

A PIC 10F200 can do this job easily. Since this is working on a human time scale, the microcontroller will react instantaneously. Humans don't notice a delay up to a few ms or low 10s of ms, so the PIC 10F200 running at 1 MIPS is plenty fast enough to poll the two inputs, decide what to do, and light or unlight the LEDs accordingly. Even if it took 100 ...


1

You largely have it right. There are a variety of conversion, but voltage maps to pressure, capacitance maps to a tank or a compliant chamber, inductance maps to inertia of the fluid, and resistance maps to resistance. When you speak of a "porous medium", however, ganged elements may not be as enlightening to the questions you need to ask the model as a ...


1

Unfortunately current travels at nearly the speed of light, therefore you have to "emulate" the propagation by using delaying circuits. Without going too fancy, you can use capacitors to introduce time constants in your circuit, but it will look more like a tank. But if you think digital, you can use buffers to make the transition sharp. a. You can always ...


0

Why is R65 in there? Try shorting it out. The relay coil needs 530 mW, which would be 44 mA @ 12 V, and implies that the coil resistance is about 270Ω. A 100Ω resistor limits the current to 32 mA, and the voltage to 8.75 V. The latter is less than the pick-up voltage of 9 V specified in the datasheet.


0

First stage should be a low pass filter with the cuttoff frequency of 440Hz.So this low pass filter allows frequencies in the range 0 to 440Hz. Second stage is high pass filter with the cuttoff frequency of 220Hz. So this high pass filter allows frequencies in the range 220Hz to 440Hz.(Since the maximum frequency allowed by the first stage is 440Hz).


1

Try the LTC6992 from Linear Technology for size: -


0

Buy the cheapest microcontroller you can find, and implement the code you just presented. If you're super worried about delays then use interrupts to react to the button presses. I doubt the microcontroller would be too slow even with just polling since it won't be doing anything else.


1

So you want to be able to display the ALU result both before and after it is stored in memory, and be able to load that result into the ALU's a or b input? Connect all the data lines together to form a bus. Obviously this bus can only carry one 8 bit value at a time, so each device connected to it needs an input register (8 bit latch) to write (STORE) ...


4

That's simple — you just need to AC-couple the switch, like this: simulate this circuit – Schematic created using CircuitLab R4 and C4 are new components; their time constant is set to 10 ms, a small fraction of your timer period. Regardless of how long the switch is held, the TRIGGER input of the 555 will only see a short low-going ...


10

I believe the question you're asking is: "How can the 1.2k resistor have 8.3mA going through it if there is only 6mA going through the 1k resistor." The answer is that there isn't 8.3mA going through the 1.2k resistor. The zener diode will break down and conduct for any voltage above 10V. The misleading thing about this circuit is that the voltage at the ...


9

"These N-MOSFETs will turn on with just 5v from the Arduino." Are you sure? Do you expect them to conduct 60A with an Fds of 0.05 Ohm with 5V at the gate? (Just checking your understanding of the datasheet. They will switch on with 5V at the gate, but they will do less well than with the 'normal' gate voltage of 10V.) But your top mosfets have a problem. ...


6

To do what you want, you will need at least one diode in the circuit, between poles I and II, At the expense of having a voltage drop in one of the fans. When in position I, the diode will not let any current pass trough it, so only Fan I will spin. When the switch is in position II, Fan II will get power, and the diode will let current trough it (with ...


3

simulate this circuit – Schematic created using CircuitLab \$C_7\$, \$C_9\$, \$C_8\$ - are in series, thus: $$\frac{1}{C_7} + \frac{1}{C_9} + \frac{1}{C_8} = \frac{1}{C_a}$$ simulate this circuit \$C_6\$ is parallel to \$C_a\$: $$C_6 + C_a = C_b$$ simulate this circuit \$C_4\$, \$C_b\$, \$C_5\$ - are in series, thus: ...


1

I don't think there is a simple general rule that lets you break down any complex circuit into small modules to simplify understanding. The first steps I usually take is to understand what is the circuit's purpose and to find its "interfaces" (power supply, input and output signals). Then I try to examine the "path of the signal" through the circuit, so I ...


1

Here are the corrections you asked for, and some caveats: 1) a 555's trigger must be de-asserted before timeout or else the output will just follow the the button-press. 2) 1000µF is pretty ungainly and wastes a lot of power when - for the same time constant - it could just as easily be 1µF by making R1 820k ohms, a standard 5% value. 3) 9k isn't ...


1

If you just want to use 555 devices then you could wire the LED to a 555 in Astable configuration and use a 555 in Monostable configuration (as your diagram) to control the RESET (pin 4) of the Astable 555. So the monostable enables the astable for the required time for 2 flashes. See here for a 555 tutorial. Also: You need a resistor in series with the ...


0

If you need a bandwidth of B=440-220=220 Hz the center frequency will be app. at Fo=311 Hz. As a consequence, the required quality factor of your bandpass will be Q=311/220=1.4 Please note that it is NOT possible to realize a bandpass with such a selectivity based on the mentioned approach (lowpass-highpass series, or vice versa). Therefore, you either ...


0

A circuit that flashes more than once is "astable". What you want is your monostable to trigger an astable, let it run (for X blinks)... then disable it. look up an astable 555 circuit, and use it to drive the cathode (negative end) of the LED. When the anode (positive end) is held low, the LED has 0V - or 3.3V of reverse bias. When the monostable pushes the ...


0

I have some additional notes. When reading your questions I think you might well need some basic understanding of electronics. As the others already pointed out, the pins of an arduino are not fit to bear the current of a heavy load. But why is this a problem for you? This is because of basic laws of circuit theory. Have a look at Kirchhoff's Law for deeper ...


0

Both your batteries are shown backwards - the negative terminals of the batteries should be connected to the Arduino ground, with the positive terminals connected to the switches. If your voltage regulators are intended to be Zener diodes, they should be connected across the load, with a resistor in series between the [load and zener] and the positive ...


0

Battery pack 1 is backwards, the '5V regulater' coming out of D8 is disturbing. Please do not drive motors from digital microcontroller pins like that, and the battery there is also backwards. The 6V regulator shown is also weird. What is the SPST switch going there? Can you describe what the purpose of this whole circuit is? I may be able to draw a diagram ...


3

I'm sorry to say, you need to go back to the drawing board. An Arduino Uno (i.e., ATMega328P) can only sink or source an absolute maximum of 40mA through any one of its IO pins, and Atmel only guarantee up to 20mA. That is not enough current to power (or sink power from) things like motors, electromagnets, etc. You need to switch the motors etc with ...


0

1) Solenoids are not inherently waterproof. 2) The resistivity of the water isn't given in ohms per meter, it's given in ohm-meters, where the resistance is measured between any two parallel faces of a one meter cube. 3)Re. Kirchoff; from here: "At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum ...


0

the question can be solved in another way... As, POS of F(A,B,C,D)=0,3,4,7,9,10,12,15. convert to SOP. SOP of F(A,B,C,D)=1,2,5,6,8,11,13,14. Now when u write down all the minterms you can group two terms together like keeping A'B' common A'B'CD'+A'B'C'D i.e.sop(2,1) to be equal to A'B'(C exor D) similarly others will also group up to form terms of xnor ...


1

No they are not generally waterproof. The current will corrode the copper away over time and your coil will fail, even though the current flow will be much less than you think ( probably < a few mA ). Even a small pinhole in the insulation of the copper wire will be enough to cause failure eventually. If you get a solenoid with a molded sealed coil and ...


0

If you have available the 34 bit serial data stream, data clock, and you must do it in hardware, a brute-force way would be to read the 34 bits into a 34 bit serial-in parallel-out shift register, then to broadside load the 32 desired bits into a 32 bit parallel-in serial-out shift register and shift them out as a single 32 bit word to the daisy-chained ...


1

In the last few pictures, it looks like you've configured the "sub flag" as an output, not an input. Since there's nothing driving that net, it's resulting in an error.


1

Should RxD be pulled up to 5 volt or 12 volt? To 5V, or else you're out of spec with with the Arduino. Are the values for R1 and R2 ok? Looks alright. RXD is an open drain pin. It will connect the resistor from floating +5V to GND when active. I don't understand what to do with the V_REN pin? My guess (from reading the datasheet) is that ...


4

For each 16 bit device (other than the final one) put a single D type flip flop on the output and use it's output as feed to the next 16 bit device. The d type will soak up each 17th bit or, looking in a different way, the d type makes the 16 bit device a 17 bit device.


1

The solution to this problem cannot be found by just trying all possible logic gate . It might be a bit complicated to understand. POS of F(A,B,C,D)=0,3,4,7,9,10,12,15. convert to SOP. SOP of F(A,B,C,D)=1,2,5,6,8,11,13,14. when A is 1 give us total 8 possibilities out of which the 4 which we are given have the property of even number of ...


2

You can find here a very well explained CPU made of relays for example. You can search on youtube for "computer transistor" for example to found computers made and explained at the transistor level.


2

Have you tried the home built CPUs webring of sites? This one covers a lot of what you are looking for: http://cpuville.com/


2

Rise time is determined by how much current flows into the capacitor. T = CV/I, so time increases when current is reduced. Anything that limits the charging current or bypasses it away from the capacitor will increase the rise time. To calculate the rise time you need to know the internal resistance and current limit of your power supply, and how much ...


0

There are plenty of circuits for controlling a motor via a MOSFET H bridge - just google "mosfet h bridge". Here's the first one in the list: - Here's the website I got this from - it'll give you loads of good ideas. Basically, the motor spins one direction when inputs B and C are taken to the appropriate logic level. To reverse the motor take inputs A ...


0

There is a nice white paper from SCHAFFNER explaining the 0.1/100 Ohm curves. Note that attenuation is measured as a voltage ratio (not a power ratio), so when working at different input/output impedances, you can actually have a voltage gain.


0

Passive circuits cannot produce a power gain but they can produce a voltage gain. A resonant circuit can produce a voltage gain equal to its Q. If the impedance levels are considered, there is no power gain.


0

There is no problem with your circuit. although I would suggest that you set pull-down resistors on the outputs. that's because the decoders usually set their outputs to high-impedance (high-Z) when they're not enabled. so the output may remain the same on the output node (because of node capacitance) and the wrong value may be read by the device that is ...


7

That is very likely a resistor network/array. And is used where you have a lot of parallel signals you need pulled up or down or even individual resistors. This picture is snipped from the Bournes catalogue. and comes in many internal resistor styles


0

You're wrongly assuming that such a machine would always loop through all the possible inputs; this isn't the case. You might enter a loop that does not span through the entire input space. For example, let's try to solve this cryptographical problem: Let H(n) be a hash function, defined thus: H(n) = n + 2. This function takes four bits in and returns ...



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