# Tag Info

7

At low frequency, we can just use a piece of wire to connect electrical components, but why are special transmission lines and waveguides needed for high frequency circuits? Transmission lines are not needed in the sense that it's absolutely impossible without them, but it would be very very difficult to design and debug something that didn't use them, once ...

6

Your circuit doesn't work like you think it does. I simulated it, and got pretty much your results: simulate this circuit – Schematic created using CircuitLab At low duty cycle, the output looks like this: At high duty cycle, it looks like this: The difference between the two is just a few percent in the duty cycle. What you need would look more ...

5

Is there a way to make this system more energy efficient? If you are not bothered about determining the pressure immediately, you can turn off the resistive sensor for 99% (or more) of the time and, for 1% of the time take a measurement. This immediately reduces the current consumption to one-hundredth of what you currently have. This sort of trick is bread ...

4

The inductor and capacitor form an LC filter to filter out the switching noise either from the power supply itself or from the other components connected to the power supply. The resistor dampens the LC filter to prevent unwanted resonance. LC filters like this often use a high ESR tantalum capacitor instead of adding a separate resistor.

4

TI says that the TL074H and TL084H are identical. Older datasheets of the previous-generation TL074 and TL084 devices (without "H") have shown the same noise specification of 18 nV/√Hz for over thirty years, so it's likely that TI ceased to do separate testing for a high-noise version a long time ago. (When the manufacturing process is good enough, ...

4

Don't try to smooth power PWM signals with capacitors. Your 220 µF capacitor is an equivalent of 1.45 ohm at 500 Hz. It will waste a lot of energy when driven with AC. Filtering is a good choice only for low-power signals, and 220 µF is unreasonably large in this case. If your load can be driven by PWM, there's no need for smoothing. If your load requires ...

3

According to the datasheet, 10pF is too small. You need a minimum of 10k and a minimum of 50pF: Ct > 50 pF, up to any practical value and 10 kΩ < Rt < 1 MΩ.In order to avoid start-up problems, Rt ≥ 1 kΩ. I think there's a typo there. "Rt ≥ 1 kΩ" should probably be "R2 ≥ 1 kΩ." Given the minimum requirements, I get a maximum ...

3

I'll provide a hint. Construct a truth table for P. A B P ======= 0 0 ? 1 0 ? 0 1 ? 1 1 ? Fill in the blanks. Now what logic gate(s) can be used to generate that truth table? Post your effort into your question and then try to generate a similar table for Q and R. Tip: use the {} code formatting button (or four spaces on the start of each line) ...

3

First, ensure you don't need an isolated 10V output. Some inexpensive DIY setups use KB etc. DC motor speed controls which are designed to work with a pot, but the pot is connected to one side of the mains so you must add an isolation circuit. Assuming that's not required here, a better approach than a transistor would be to use an op-amp. You can simply low-...

3

Note: This answer was written before the question was changed to require a solution using TRIACs. simulate this circuit – Schematic created using CircuitLab Figure 1. Magic relay priority circuit. (Don't omit the fuses.) If L1 and L2 are on power is provided by L1. If L1 is off and L2 is on then power is provided by L2. If L1 and L2 are off then ...

2

simulate this circuit – Schematic created using CircuitLab Figure 1. Modified circuit. If your fan can be controlled that way then you only need two resistors. RLY3 slow. RLY2 medium. RLY1 fast. simulate this circuit Figure 2. Low-side switching of relays. The diodes aren't needed as there are diodes in the ULN2003. simulate this circuit Figure 3. ...

2

Since guaranteed offset error is relatively expensive to purchase, maybe you use a 16 bit DAC and another low Vos op-amp (with an ADC input on your MCU) and auto-zero the output.The ADC does not have to be particularly high resolution or accuracy since you can have a lot of gain in the op-amp and you’re only looking for something like a null. In practice you’...

2

Comment About Your Approach A serious problem with your design approach is the offset voltage of the opamp. Let's look at a precision, rail-to-rail I/O, opamp: LT1630/1631. It's offset voltage is $V_{_\text{OS}}=525\:\mu\text{V}$ (max). That's already more than half of your supposed reference voltage! There will be other sources of error that will pile up ...

2

Inrush current happens when capacitor is initially charged from 0V to the intended rail voltage. What should be the effective capacitance to be considered during this charging? -Since the voltage keeps on varying and effective capacitance for a ceramic cap varies with voltage The effective capacitance will be given by the total integrated charge required to ...

2

Basically, it has to do with the Maxwell equations and energy transport. The two important equations for this are: $$\text{curl}\,\vec{E} = -\frac{\partial\vec{B}}{\partial t} \qquad \text{and}\qquad \text{curl}\,\vec{H}=\vec{J}+\frac{\partial\vec{D}}{\partial t}$$ This means that an electric field creates a magnetic one and vice versa. If a harmonic current ...

2

About schematic A: It is more complete. It's easer to read. Component labelling is better, except for R101/R104. It contains a resistor (12kΩ) to hold the gate voltage low in the absence of an input connection. Resistor R101 (10kΩ) is way too large. It should probably be more like 100Ω. I would agree with 10kΩ if this were a biploar junction transistor. If ...

2

Two cascaded 555 timers (or a 7556) both in monostable mode with a differentiator on each Trigger input. The differentiators create short negative going trigger pulses from the falling edges of longer pulses. On falling edge of arriving pulse the first 555 outputs a pulse of adjustable length, adjustable with a pot. Falling edge of this pulse triggers the ...

2

Your LED has a maximum current rating: 700 mA. 3V just happens to (on average at room temperature) be the voltage at which this current flows. So, the moment you use a 1.5× higher voltage, you forced an exponential amount of additional current through your LED. That leads to a quadratic (of that exponentially increased current!!) increase in power converted ...

2

simulate this circuit – Schematic created using CircuitLab Looks like you already have a SPDT Relay, so now finish it off with a SPDT switch. This is how two-way lights are done in homes so you shouldn't have trouble finding a wall switch that is SPDT. Edit: find more about multiway switching here.

2

Most likely only the designers know that and the info might not be public, so we have to guess. Inductor and capacitor is used as a LC filter. The filter removes noise and ripple that might come from the power supply or the chip digital supply pins, so that the voltage that powers the sensitive analog power supply for the clock has less noise and ripple so ...

1

Your black negative image could not be seen so I converted it into a normal positive image with black lines on a white background. The inputs must be opposites high and low for the motor to run. If both inputs are high or low then the motor does not run. The emitter-followers reduce the output voltage.

1

Power the voltmeter from its own independent power supply, one that is not affected by what the user may do with the main output of the big power supply. This independent power supply shares the negative line with the negative output of the big power supply Feed the output of the big power supply to the measurement input of the voltmeter through a 1 kΩ ...

1

If you use two hinges... these should be made of metal and have a good enough contact to transmit the current for your LEDs. So you have your +5V and 0V, on on each hinge. No need for more complication. Since these will be exposed, short circuit protection is necessary.

1

Try something like this. Use magnetic contacts.

1

There are two things in the data sheet that suggest to me you have chosen the wrong device for your application. The schematic you have posted is directly from the data sheet and it specifies a minimum input voltage of 9 volts to achieve 3.3 volts regulated at the output. That's a whopping 5.7 volts of headroom that is potentially needed to make this chip ...

1

We pretty soon discover that it's a bad idea to mix this unrelated analog resistor ladder stuff with your power on/off functionality. So either you need a dedicated button for power on/off, or you need a DPST switch. This switch can then drive a MOSFET placed in series with your supply. You need some manner of latching function since the switch is momentary. ...

1

LEDs are driven with current, NOT with voltage. If its (LED's) threshold is 3V, then anything over 3V connected to it will act as short circuit with infinite (in practice, finite, but well over any spec) current. You always* therefore need a current limiting resistor. If your supply is 4.5V and LED drops 3V, then you will have 1.5V across the resistor. ...

1

in reality OA1 output (NODE1) voltage doesn't go all the way down to negative rail (-10 V) and outputs ~3 V When the voltage reference is bypassed to 0 volts via M2, the op-amp is then trying to regulate the output voltage to 0 volts. That's the first thing to remember. So, if the output of MOSFET M1 is purposefully controlled to be 0 volts, the gate ...

1

What you probably need is for the transistor to turn on so that it drops very little voltage between collector and emitter. This in turn means that your LED circuit receives the full 3.3 volts across it. Of course, this assumes that the LEDs are voltage driven and not current driven. However, your LCD is "defined" by an Aliexpress page and, they ...

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