# Tag Info

2

How would I solve this circuit using mesh analysis? I assume the controlled source is a CCCS. The left most clockwise mesh current, $i_a$ is by inspection $i_a = 1A$. Clearly, the right most clockwise mesh current is just $i_b = i_x$. But $i_b$ cannot be found by KVL (as you've pointed out) so we need an auxiliary constraint to replace the ...

0

Are you waiting long enough for the output at Pin 3 to turn on? It would be 1/1024 the rate of the D1 signal, or 1024 x the duration.

0

For the 5V source you get: \begin{align} V_{O,5V} = 5V * \frac{2 \Omega \parallel 3\Omega}{4 \Omega + (2 \Omega \parallel 3\Omega)} \approx 1,15V \end{align} For the 10V source you get: \begin{align} V_{O,10V} = -10V * \frac{4 \Omega \parallel 2 \Omega}{3 \Omega + (4 \Omega \parallel 2 \Omega)} \approx - 3,07V \end{align} So in total you get: \begin{align} ...

0

In order to create a full palette of colors, the intensity of each color must be manipulated independently from the others - from zero through some maximum for each color - so I think your proposition is impossible.

1

simulate this circuit – Schematic created using CircuitLab this might just work if you sized your resistors and zeners properly. it would cause one led to turn on at a time, then stay on without changing too much brightness while the next one turned on if you got the transistors to be saturated just at the point when the pot reaches the zener ...

2

In general you'd need three knobs to set the three LED light levels. You could make a one turn knob potentiometer, read the voltage, and turn the voltage into some combo of LED's with a look up table. But it will most certainly involve programming and PWM.

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Neither of the solutions is correct, I think that the arrangement misled you. I've rearranged the capacitors. Does it help in calculating the total capacitance? The trick is, figure out how the current flows from A to B, where it branches and where it merges again. Case 1 simulate this circuit – Schematic created using CircuitLab Case 2 ...

2

Put it in the freezer overnight. Hot glue will both contract in the cold and lose it's adhesion. You can then chip it off while it is still cold. Cold spray doesn't work - it will quick-chill the surface but the core (and part sticking to the board) will still be warm.

1

I have seen technicians remove hot glue by swabbing the area with rubbing alcohol. Rubbing alcohol is also known as isopropyl alcohol. This doesn't really dissolve the hot glue, but causes it to come loose from the board and wire. In future, instead of covering the solder with hot glue, consider if you can apply the hot glue away from the solder area. The ...

4

Here's how the topology morphs:

2

With the switch open, you have the three rightmost capacitors in series, with the resulting capacitance in parallel with the middle cpacitor. With the switch closed, the two capacitors above and to the right of the switch are shorted, so can be ignored.

4

Hot but touchable is entirely within the normal operating conditions of an 7805. If you are concerned about the effect on your Christmas tree simply add a heat sink (which can be as simple as a 10 x 10 cm metal plate) to spread the heat over a larger area and thus reduce the temperature. (This assumes you are using a TO220 version. If you are using a TO92, ...

0

Yes, you can. However if both regulators are not putting out identical voltages then one will draw more current and get hotter than the other. The traditional way to fix this is to add a small resistor (eg. 0.1Ω) in series with each output. If you can touch the regulator without it burning you then it's probably not getting too hot. If your Christmas ...

5

No, but depending on what your power source is you may be able to do 2-stage regulation to spread the heat - for instance, if your source is 12V then lower it to say 9V first with a 7809 then drop it from 9V to 5V with the 7805. If you're even higher than that, then you could use a 7812 first to drop to 12, then the 9V, then the 5V. Either that or just use ...

2

Electrical circuits and micro-mechanical systems have the same differential equations if you write them out and it's much easier to picture mechanical systems. See here Basically: A resistor is like friction from a damper. A capacitor is like a spring. An inductor is like mass of an object. Zeta Zeta is the dampening ratio and a function of your ...

1

It's trying to, but it's being radiated away as heat by resistance in the circuit.

2

In the good old days of cables that consisted of a large number of copper wires, and switching stations with relays that effectively implemented a big crossbar, circuit switching meant that the provider put all the switches in the position that made a real (electrical) circuit from one client to another. The two literally had a cable for their own private ...

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Virtual circuit switching is a packet switching mechanism. In that a Virtual path is established between source and destination before the transmission of data. It is data traffic. Whereas, Circuit switching, happens in three steps, circuit establishment Data transfer Circuit Disconnect Circuit switching is voice traffic. Generally used in telephone ...

1

When examining a single bit in a byte (or register) using & it's important to remember that the result of the & will be a masked value, not the value of the bit you're interested in. For example, if you have the value 0b10110101 in a register and you want to examine bit 2, you could mask it out with 0b00000100: 0b10110101 0b00000100 & ...

1

Every circuit - even each amplifier - has frequency-dependent properties. Of course, in many cases (filters) this is a desired property. That means: Applying a voltage at any node (in most cases: input of the circuit) will cause a current distribution within the circuit that depends on the frequency of the input voltage. In particular, the signal (current or ...

0

Shown below is a circuit that simulates nicely, and the LTspice files needed to run it are here If there's a need for a circuit description, let me know and I'll edit my answer. Not right now though, my brain feels fried...

4

The circuit you show is a non-inverting amplifier with a gain of + (1 + 1.2K/9.1K) = 1.132, so you would expect it to output about 1064mV with 940mV in. If you want it to output 1000mV with 940mV in then you need to reduce the gain by changing the resistor ratio. For example, you could increase the 9.1K resistor to about 18.8K which would give about 1000mV ...

1

This is an absolute no-go. Not from the terminals on the outside of the battery, it is when you open up the battery-pack. Charging Li-ion batteries is mostly current controlled, with cell-temperature, voltage (change) and absolute charge as input parameters. A second (top-up) stage is voltage controlled. The battery should at al times being monitored ...

2

Here's a circuit which should accomplish what you want. It uses a dual 555, namely a 556 with two different rates. A CD4022 is used to count the button pushes. A pair of NAND gates debounces of the push button input, so the counter doesn't advance erratically. The first two outputs select one of the two rates using AND gates. The third output is steady ...

3

The easiest way to do this in small quantity is to use a microcontroller, such as a Microchip PIC10F or 12F series. There will be only a few parts in the circuit and a page or two of code. The micro should be under a dollar in single quantities. The best way to do it in production quantity is a COB CMOS die designed for just that function, a ...

1

For reference the TDA2822M datasheet can be found here. Those capacitors are connected to the power supply pin (pin #2). They are acting as bypass capacitors. Bypass capacitors are designed to remove AC noise from the DC supply voltage. Since capacitors are a short at high frequencies, high frequency noise is shorted to ground while the DC supply voltage is ...

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.

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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: ...

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