# Tag Info

1

It's just because that's how this particular circuit is constructed. When you evaluate the circuit's response to the current source you "zero" the voltage source. A voltage source of 0 V is effectively equivalent to a short circuit, so the voltage source is replaced with a wire. After making that replacement, it turns out that the two resistors are ...

0

I totally agree with you Matthys Du Toit - perfect explanation on https://rayshobby.net/wordpress/understanding-24vac-sprinkler-valves/ .... just forget last paragraph in that link... We do not need complex circuit to solve the different current needs of that solenoid (~400mA to start electromagnetic valve and then ~200mA to keep it opened). Just had a 12V ...

0

The internal resistance and battery chemistry voltage sets the charging speed, when charging batteries usually you run current through the battery, then once it reaches a certain voltage you trickle charge it with the max voltage of the battery. The internal series resistance of the battery limits how much current you can put through what and how much self-...

0

The battery chemistry/technology is the main thing. As a general rule, charging a battery faster makes it age faster. That is to say, you'll get fewer charge/discharge cycles before you have to replace it. Sometimes people use the capital letter "C" to represent the amount of current that would (in theory) fully charge the battery in one hour. ...

4

There are already given several answers of RC integrators and RC differentiators and how they in certain conditions do something which resembles integration or differentiation. But a capacitor alone does both of them and it does them very well. That's because capacitors obey the following law: Charging current = Capacitance x The growth rate of the voltage. ...

5

What you're missing is the definition of RMS: $$V_{RMS} = \sqrt{\frac{1}{τ}\int_t^{t+τ}V(t)^2\ dt}$$ What you're calling the calculated area value is $$\frac{1}{τ}\int_t^{t+τ}{|V(t)|\ dt},$$ which is not the RMS of the signal. (in math terms, I believe you would say the square root and integral operators don't commute, so you can't just pull the square root ...

3

$\sqrt{\int_0^T s^2(t)\, \mathrm dt} \ne \int_0^T \lvert s(t)\rvert\,\mathrm dt$, simple as that.

5

Consider the simplest possible input to the integrator (a constant voltage) and compare the behavior of the two circuits. simulate this circuit – Schematic created using CircuitLab (I've added an inverter to the op-amp integrator to give an output that is the positive integral of the input voltage, for easy comparison). We know ideally the integrator ...

0

I guess you want to make your battery charging faster and you want to know what quantity should be bigger in your next battery charger that you are going to purchase or build. I guess you expect a clean single word answer - either "current", "voltage" or "power" with absolutely zero extra blahblah like "it's current, but ...

2

Integrators without op amps are only an approximation. Consider an ideal integrator with a time constant of 1 second. Put a unit input to it, and the output will rise (theoretically forever) at 1 per second in a perfectly straight line. Now use an op amp with a 1 Mohm input resistor and a 1 uF feedback capacitor. For a 1 volt input, the output will rise (...

2

You need to learn about solving partial fractions. That is what the K1, K2 and K3 formula is hinting at. If you haven't studied partial fractions then, it's going to be real tricky to split the equation that has the denominator s(s+8000)(s+20000) into three fractions that are added together as per the K1, K2 and K3 equation. Go and study partial fractions is ...

2

The "charging" speed. It will be the current, as the charging is current regulated and voltage is pretty much set by the chemistry (it does vary during charging). Most chemistry, besides NIMH will use the voltage to determine when to stop charging.

1

You may wonder why traces of some interfaces are not analyzed by the transmission line theory. Following is simple analysis of impact of trace length: Referring to input impedance of lossless transmission line, $$Z_{in}(l) = Z_0 \frac{Z_L + j Z_0 tan(2 \pi l / \lambda)} {Z_0 + j Z_L tan(2 \pi l / \lambda)},$$ if $l \ll \lambda$,  ...

1

The important thing is the rise time (not the pulse repetition rate) of the signal, compared to the length of the trace. If the signal can make several round trips of the line between the driver and receiver during the rise time of the signal, then we can ignore the transmission line effects. With a trace 200 mm long, which is about 1 ns electrical length ...

6

Any given core material, cross section and frequency has a maximum volts per turn. If you want to use the winding at a certain voltage, then you need enough turns to support that voltage. For instance, low frequency mains transformer steel will only run up to a peak field of 1.7 T or so before it saturates. If you had a core that was 10 mm x 20 mm, and ...

5

Yes - a simple RC circuit can be used (theoretically !) as an integrator - under the assumption that the time constant T=RC is very large. For example:, with R=100k and C=10µF the pole frequency is wp=1 rad/s and the range of intergation will approximately start for w>100 rad/s. What does the opamp? It not only allows to connect a load (without influence ...

1

The line is VERY loosely drawn at about when the signal trace length approximates a quarter of a wavelength of the signal it is carrying. However, much depends on shielding, characteristic impedance, how its driven and so forth. It also depends on the maximum frequency component of the rise and fall times of the signal. So if (say) you have a 100MHz digital ...

2

Impedance mismatch causes signal reflections, so for each edge in the signal, additional edges are generated as the echoes and echoes of echoes overlap it. This can be tolerated as long as the amplitude of the reflection is small, or the time delay of the reflection is short enough that this only leads to a bad shape of the transition edge, but doesn't ...

1

Impedance matching is always a concern. You always have to pay attention to it. However, with something like an I2C bus, there is a defined way to drive the bus (in this case open collector output with a bus pullup resistor of say 470R) which already takes account of this. Impedance matching is a universal electrical concept which applies any time you try to ...

10

You don't need the opamp, in theory. A simple, passive, RC circuit gives you am integrator or differentiator (otherwise known as first order low pass/high pass filter). One problem is output impedance and loading. The opamp allows you to create a similar circuit which has very low output impedance, and whose characteristic will not be significantly altered ...

0

Beyond inductor, it seems you are confused by voltage / current. An overall great analogy to electricity is water, it behaves quite similarly, as the current is the water current and voltage the pressure (height of the water). While most components (resistor -> small pipe, capacitor -> cup) can be easily represented in this way, inductor is a bit more ...

20

Number of turns on a transformer matters greatly. The turns ratio is one of many considerations in designing a transformer. The following are high-level considerations when designing a transformer. As Tobalt stated, magnetizing inductance is important. Too few turns will consume excess current, even without a load. You need a minimum of turns to prevent ...

2

Transformer design has to optimize a lot of different variables. And I am not an expert. But the transformer primary needs to have enough turns so that it acts like a big inductor and prevents excessive current from flowing. If you take a particular transformer core, you can either wind the primary with many turns of fine wire (high voltage primary) or wind ...

4

Obviously, B has some advantages indeed. Otherwise all transformers would use a tiny number of turns. More turns generate more magnetizing inductance $L_M \propto N^2$. Imagine a transformer with the secondary open, which is essentially a large inductor. If you connect an AC voltage with frequency $f$ to the primary, it will see an impedance of \\$2 \pi f ...

5

SPI can be faster as it is push-pulled lines, in the contrary to I2C that uses pullup resistors. SPI can be daisy-chained, but usually it requires the slave device to be of the same type. SPI will need a CS line for each device, while I2C works by addressing. The SPI software stack is usually simpler than I2C. I2C allows to have many devices on the same ...

1

SPI for bandwidth and full-duplex. Daisy chaining for SPI just makes it easier to grab data from a lot of SPI devices, but only for devices that support it and only for identical devices since their daisy chaining is only made to work with others of their kind. I2C for addressing (you can handle more devices without a chip select line for every device and ...

2

The output current from two isolated power sources in series will be limited to smaller of the two constituent supplies. A simple '4 diodes one capacitor' supply has a lot of ripple. The output ripple will be the sum of the two constituent supplies. Depending on your application, the current limitation may be temperature rise in the transformers or diodes, ...

0

The 3V will apply only to the inverter input. The "wires" themselves are "probably" similar whether the inverter is 3V battery input or 12V automotive input. Typical EL wire voltage/frequency (and thus inverter output) is 60-120V RMS at 0.4 to 2kHz.

-1

A FET and BJT alike are just single pole , single throw switches.(SPST or 1P1T) to a low V or ground (NPN or Nch) and high for PNP or Pch. They are both inverting logic switches in common emitter or source. It takes 2 to make a 2P1T or a 1P2T

2

It depends on what's connected to J5 and J12. Without knowing the details, I would say you'll likely have problems with J5 in this configuration when the FET is OFF. Think about voltages with the FET off: J5, pin 1 would be +5V and J5, pin 2 would be ~12V (through J12), and you'd have a -7V reverse voltage on the connected J5 device. If the J5 device can ...

0

Without any datasheets or schematics for your circuit, I can only make guesses. But first I have a couple questions. When you measured your power supply did you connect directly to the wires for a current measurement, with no other circuitry, i.e resistors? Did you check the power supply with it plugged into your circuit? Did you measure the current and ...

0

Replace the V, R combination with its Norton equivalent: a current source V/R in parallel with R. Combine the two current sources (add) then convert back to the Thevenin equivalent: voltage source (V/R-Icst)R in series with R. Then, instead of connecting the current source at t=0, solve for a capacitor charged to V discharging through R to (V/R-Icst)R ...

1

I will take a SWAG and assume it does not work because the 7805 is oscillating at a high frequency. I would suggest reading the data sheet for the 7805 you have and follow the recommendations it gives for the input and output. Not all 7805s are the same. You can also post a schematic of your circuit, pictures and frizzy things do not count. You probably will ...

0

After you get to this point: you have to solve a first order differential equation: https://www.mathsisfun.com/calculus/differential-equations-first-order-linear.html

1

You seem be be between a rock and a hard place: RPI GPIO swing is only between 0 and +3.3V IRF540 wants +5V to fully turn on. You might ask, "What is an acceptable voltage (above 0V) to keep the IRF540 off?". The answer is perhaps a few volts. So a kludge circuit would use a LED to translate the RPI GPIO ranging from (0 to +3.3V) up to (+1.8 to +...

2

In response to your comment to @winny's answer: Is impedance of 120 m-ohm equals to R = 0.120 ohm, and so U = 1.5/0.120 = 12.5 A and so the current of this battery is 12.5 A? You can draw 12.5 A into a short-circuit but a short circuit will have zero voltage and since P = VI you'll get P = 0 x 12.5 = 0 W. The Maximum Power Transfer Theorem says that you ...

3

The datasheet for a Duracell AA alkaline cell shows that its voltage drops very quickly with a 1A load. Beware: I bought some Duracell chargers that came with 4 AA Ni-MH cells and 4 AAA Ni-MH cells at Costco. The cost was lower than possible because the Duracell batteries were made in China with half the capacity of American or Japanese ones. I do not know ...

5

Short-circuit current of a new alkaline AA battery is in the low amperes. About 3A for a fresh Kirkland AA cell. 2.4A for a Panasonic Platinum power. Source: actual measurements

3

In the EE world, almost every question of this type will be answered by the manufacturers datasheet of the part in question. Here is the one for the Duracell AA you mentioned: From the impedance of the battery, you only need Ohm's law to calculate the peak current and power the battery can supply. I'll leave the calculations for you and your understanding. ...

0

Inverters work by changing electricity while maintaining power. Some power is "lost" by converting to heat (the efficiency of the circuit). Electricity isn't just the voltage involved but also the current. So as long as you meet both needs, you have not created power, just changed it. This keeps it in line with the laws of thermodynamics. By taking ...

1

Figure 1. A simplified inverter schematic. Image source: ElecCircuit. How it works: Q1 is turned on. Current flows from +12V through CT (centre-tap) to the top of the transformer. Q1 is turned off. Q2 is turned on. Current flows from +12V through CT (centre-tap) to the bottom of the transformer. Q2 is turned off. The result is pulses of current running in ...

2

That is an inverter. There are multiple ways how it can convert 12V DC to 220V AC. It can first step up the voltage and then output it as AC with an H-bridge, or it can first use the 12V to generate 12V AC with H-bridge and then step up the voltage with a transformer.

1

VO-, VO+ and 5V are the pins for output voltage. It looks like the 5V output is made from VO+ with a linear regulator but it looks like the components to make 5V from VO+ are not present, so unless there is something going on under the PCB, this power supply does not have a 5V output on the 5V pin. However, it might have 5V output on the VO+ pin. Perhaps if ...

0

Based on its label of SM-009-5V-02 I suspect it does have a 5V output on "VO+" on that version. You can (carefully!) measure that to confirm. However as commented, on other versions of the supply (just guessing, let's say with a 12 V main output), then a linear regulator (probably in SOT-223 or similar SMD package) could be fitted in position U2, ...

2

I assume you mean RC servos, with a three-wire connection consisting of power, ground, and a PWM signal. Can two 6V servos be connected in series to a 12V source, Not if you want them to work. First, the "top" servo in the string will have its signal line referenced to the "bottom" servo's power line; unless you make a special ...

0

Alex Forencich is right on. Coiling the heating wire is the typical solution to this kind of problem. It really simplifies this for you. You can use a higher voltage; more easily construct a constant current sources; your wire will have a longer life, since the power dissipated per unit length is lower... ENJOY

2

Do it all the time, most instruments (all the good brands) have VISA controls. With this on most instruments you can script most user actions (like pressing buttons) and also do other things like data acquisition. The interface is independent of comm (USB, RS232, GPIB or Ethernet). First you need to establish communications and once you start sending ...

1

One compelling reason that comes to mind is that many opamps don't like their input near to the voltage rails. There's also a few 24bit adcs that don't like it as well. The ADUC0841 comes to mind. Being differential in, the wheatstone bridge configuration just works.

1

Well, micro was smaller than mini! Why did Arduino call one of their boards the 'Nano'? and RaspberryPi the 'Pico'? Why are those companies called 'Arduino' and 'RaspberryPi'? So many questions! Here's a publication for your edification: https://link.springer.com/chapter/10.1007%2F978-1-349-07674-1_17

3

Back in the day there were mainframe computers, typically occupying a large room with several 19” racks. Then there were minicomputers which would often fit into a single rack, the likes if PDP-11s and VAXen. Then came microcomputers such as the Sun-2, which used microprocessors, being more or less single-chip CPUs. The physically larger processors fell ...

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