# Tag Info

0

To dim the running lights, they could be switched from being parallel across the 12V supply to being in series by using a relay with two sets of change-over contacts.

0

Just to add to my comment with the circuit diagram. The first circuit shows the DRL(day) connected to the common pole of relay 2. (The way I think you have wired your circuit) With relay 1 unenergized the ACC(+) has a path through relay 1 and relay 2 switches to DRL(day). However, when relay 2 is energized all that happens is DRL(day) and DRL(night) are ...

1

This DPST NO (2 form "A") relay solution will work:

0

You have a serious input and output voltage range problem : the input signal must stay within input voltage range (V- to V+ -2V) and the expected output voltage must stay within output voltage range (V- to V+ - 3.5V). Here is what I suggest : simulate this circuit – Schematic created using CircuitLab This way : The input voltage swings around ...

0

Could I interest you in building a bridge rectifier with Schottky diodes instead? Then polarity would be immaterial.

1

Thinking only the positive side of your circuit: assuming you use a normal diode with a voltage drop of 0.7V, GND and +12 reversed, the voltage across thq 1k resistor is U=RI => all remaining voltage will be over the resistor. Thus the protection doesnt work. One way of doing this (better or worse) would be remove he 1k resistor and using a diode with as ...

1

The diodes will have no effect to speak of on reverse polarity, and will provide no protection. If the device is a chip, it will act like a forward biased diode, and will take almost all of the current compared to your parallel diode with 1K in series. Edit: You can play with simulations below. With the left one I've simulated the 12V source (reversed) with ...

5

The most obvious problem in the circuit is that the + input of the amplifier is floating; it is not biased to any DC voltage level. You need to convey a reference voltage to this input. (This has to be done through a reasonably large resistor; if you just tie the input to a stiff reference voltage, the input will have no impedance to work against.) ...

0

You have no path for bias current. Add 1Mohm resistor from +3V to in+. Also the ring of audio input must be referenced to half of supply voltage, so add voltage divider consisting of two 10k resistors between +3V and 0V with the centre connected to ground (the grey line).

2

In general it is bad practice to apply input signals to a device before applying power. When you do this, you can cause semiconductors to 'latch up'. The semiconductor can act due to its physical build as a SCR (Silicon Controlled Rectifier), which can cause it to short out its pins. Subsequential application of power can cause excessive currents which ...

0

Depending upon the amplifier or other device... SOME devices can be destroyed by applying any significant input voltage without first appying power. OTHER devices are totally immune. I'm not sure whether this qualifies as an "answer", because it doesn't specifically address the ADA4004; I haven't seen that device's specs.

1

With no power supply, the transistors in the IC would not do anything, so the input would be "ignored"...and there would be no output. It requires a power supply to open the switches to let the signal in.

2

Just replace the MOC3083 with a visible LED (pin 1 is + pin 2 is -). Unless your drive circuit is very low voltage, the LED will indicate when the triac should be on.

1

You have the wrong charger. I have an 18 volt B&D drill with an HPB18 battery pack. The charger that came with it is 21.75 volts, 210 mA.

3

By any chance... is there another B&D wallwart in your possession? I don't think that's the right one for that drill, if the drill's battery is indeed an 18V battery. The wallwart shown looks like it might be intended for a cordless screwdriver instead.

0

The kVA rating is a measurement of power for the transformer (volts x amps) without taking the phase angle into consideration. The 2300/230 signifies the transformer ratio (primary/secondary). A 2300V (rms) input should give a 230V (rms) output. Assuming the transformer is 100% efficient then the volt*amp products of the primary or secondary should be the ...

5

you have not connected the body/bulk connection properly in your schematic. PMOS to VSS. But that circuit leaves the output node floating for the A=B=1 condition. Scan through the various solutions here, there is a robust TG version of XOR and XNOR, it uses 4 transistors though. If you look at all the possible states, you have 2 transistors each in 2 ...

2

That won't work. Vin Vout -2 6 0 4 2 2

0

If you scrape off the coating, you'll find divots that go around the resistor. l strip or divot is equal to 1 ohm, 2 is 10 ohms, 3 is 100 ohms, 4 is 1000 ohms and so on... it will get you in the right ball park, the rest is up to you.

1

Just a suggestion... use a pair of metal-film resistors as a voltage divider between +5 and ground. MFRs are a lot quieter than carbons. R1 should be exactly 150% of R2.

2

This will function okay, particularly if you use good resistors, and don't need more than some % DC accuracy. The 78L05 is not a very good regulator and its a worse reference so it's almost surely going to limit the overall DC system accuracy if not dealt with. You could measure the voltage with one channel of your ADC and deal with it digitally, but power ...

2

This idea will work providing you don't need to supply more than (say) 20 mA of current at 2V to other devices because it's likely that the op-amp wouldn't be able to do this - some op-amps will supply (say) 50mA such as the AD8605 (from memory) of course. DC offset is an issue if you want to use an op-amp so check the data sheet. Why not find an ...

1

This device does exist although it's not readily available in single unit quantities, its output amplifiers will get in the way and it's very non-linear. It is a Floating Gate MOSFET, used in Flash memory, EEPRom and the ilk. The programming charge can be variable though somewhat unpredictable as the FN tunnelling (Fowler Nordheim) will be variable across ...

0

Your mileage may vary depending on what your input is. I think your idea about using a potentiometer actually holds some water. Modding a servo motor so that you can read the voltage at the slide would probably be easy, but the hard part might be controlling the servo motor using purely analog means. I think that this is a similar idea to memristor logic. ...

6

This is not a practical answer unless you happen to work for a company with the resources of, say, Intersil, but the technology exists to make this work. Consider the ISL21080 type references which hold charge, hopefully for the life of the equipment they're installed into, based on a tiny capacitance isolated by quantum tunneling effects. Provided they ...

1

I left a comment and thought about it for a minute and will say with hopeful certainty that it does not exist - some drift away from the "sampled" voltage is not only likely but a certainty. Resolution is critical it seems, (as implied in your question) and this is why I say it doesn't exist. Noise is another factor that will reduce the fidelity of what you ...

3

The working principle of this oscillator is as follows: We have a classical tank circuit (L5||C1) which is connected in a closed loop with a two-stage non-inverting amplifier. This amplifier consists of an emitter follower (Q2) driving a common base amplifier stage (Q1). Hence, there is no phase inversion within the loop (condition for positive feedback). ...

4

The oscillation frequency is determined by C1 and L5: - $f_O = \dfrac{1}{2\pi\sqrt{LC}}$ And, for values of 3900pF and 100uH, it should oscillate at 254.8 kHz theoretically but there will be parasitic capacitance in the inductor and miller capacitance in the transistors that make the actual capacitance bigger hence lowering the resonance to 236kHz. I'd ...

1

As several people have remarked, I also believe this is about a job candidate in an interview, more than a 'correct' answer. That is off-topic, but I'll try to illustrate my view, because I think it might help others. My experience of interviewing literally hundreds of candidates is it is not about a correct technical answer. It is about how the candidate ...

1

This is not really an answer, but to help a bit your numbers are not correct unless I missed something. Assuming R1 is the "top" resistor of the divder (the potentiometer), and the bottom resistor is R2 then it appears as if you have done the calculations backwards. $$V_{OUT} = V_{IN}\frac{R_2}{R_1+R_2}$$ So, when R1 is minimum (5k), the output is going ...

4

A rather convoluted question, but I think what you need is an adjustable voltage source with a minimum output impedance of 5k. In that case, why don't you design your voltage divider to give you the correct output voltage and then buffer it like so:

-1

I dont have relvant to link. As i know there 2 ways to measure the DC current 1) hall sensor based 2) DC shunt based. for second method i didn't get any links. But for hall sensor based it is quickly straight forward.

2

While the above answers are correct, in the late 70's through the early 90's sometimes simple CPU's were constructed from TTL logic. While you may or may not want to get into such a project, it is interesting concept. Because the CPU is spread out and hand wired you can probe around and see it in its entirety rather than just as a small black box. Here's ...

8

For practical purposes of getting it working on your desk, you probably want an FPGA. In order to make sure it works, you should simulate it first with a program like Modelsim. This enables you to iron out the bugs before buying any hardware.

3

Texas Instruments, et al. have almost unbelievably extensive stables of digital logic functions available which can be assembled into pretty much anything you want. Go here and here , then check out HC (HCMOS) to get started.

-1

Easier way: X = -1/X multiply both sides by X: X^2 = X/-X = -1 X = sqr (-1)

1

An ideal diode analysis gives you an approximate answer: The voltage across the lamp is approximately 9.3-9.5 V. The diode carrying current I is reverse biased. For a real diode, there will be some reverse leakage current through the "I" diode. This means I could be anywhere from -1 pA to -1 uA, depending on the type of diode. If you choose something ...

0

You can't; it's not a problem with the circuit. Android 4.3 added support for the Bluetooth Low Energy (BTLE) protocols used by this device. Earlier versions don't support BTLE, and most of the phones running earlier versions won't have the appropriate Bluetooth chips to support it anyway.

3

Those look like they were actually drawn with either a schematic stencil or very carefully by hand. "By hand" is something people used to do before software. With practice, the art can be learned, but that's time consuming and lame. Just make a schematic in Visio and then apply a sketch filter with Photoshop.

1

What value of $x \text{ equals} \dfrac{1}{-x}$ ?? Rearrange to $x + \dfrac{1}{x}=0$ Rearrange more to $x^2 + 1=0$ Solve for x like a quadratic: - $x = \dfrac{-b+/-\sqrt{b^2 - 4ac}}{2a}$ $x = \dfrac{-0+/-\sqrt{0^2 - 4}}{2}$ $x = {+/-\sqrt{-1}}$ This implies that the only value for x is +/-j because the square root of -1 is -j

4

Both these equations are correct because $j = \sqrt{-1}$ We have $$Z_{cap} = \dfrac{1}{j \ \omega \ C} = \dfrac{1}{\sqrt{-1} \ \omega \ C} = \dfrac{1}{\sqrt{-1} \ \omega \ C} \cdot \dfrac{\sqrt{-1}}{\sqrt{-1}} = - \dfrac{\sqrt{-1}}{\omega \ C} = - \dfrac{j}{\omega \ C}$$

-1

Charging up a capacitor quickly and discharging it a bit more slowly is something that you probably do every day without noticing - charge your phone much? That battery acts a lot like a cap. It all has to do with the time constant, which is a product of resistance and capacitance, simply put. So, to charge fast, what kind of resistor would you have to use ...

1

-j=1/j, so you can use any of the expressions,, 1/j = (1/j)(1) = (1/j)(j/j) = j/(j*j) = j/(j^2) = j/(sqrt(-1))^2 = j/-1 = -j,, or 1/j = j^-1 = (exp(pi/2))^-1 = exp(-pi/2) = cos(-pi/2) + j*sin(-pi/2) = 0+j(-1) = -j the reason 1/j=-j and 1/2 is not equal to -2 is the euler's identity,,

2

Separate the variables and integrate: $$α = \frac{V}{RC} + \frac{dV}{dt}$$ $$\frac{dV}{dt} = α - \frac{V}{RC}$$ $$(\frac{1}{α-\frac{V}{RC}})dV = dt$$ $$RC\int \frac{dV}{αRC - V} = \int dt$$ $$RC\ln(αRC - V).(-1) = t + C_1$$ $$\ln(αRC - V) = \frac{-t}{RC} + C_2$$ $$e^{-t/RC + C_2} = αRC - V$$ $$e^{-t/RC}.C_3 = αRC - V$$  V = αRC - ...

0

Well, you know that to charge a capacitor instantly will require infinite current, right? And we know that all real-world power sources have some finite resistance associated with them, as do capacitors (ESR). However, as your intuition suggests, if you just pop your capacitor across a supply you are depending on the parasitic resistance to limit your ...

0

Did you consider that the LED of the optocoupler limits the voltage at the output (pin1) to its forward voltage. I.e. the voltage divider that provides the reference voltage (at pin3) is always loaded not only by the bottom resistor (500 Ohm) but also by the branch via 300 Ohm resistor and LED. Maybe this causes the threshold to be not where you think it ...

0

The fluid-analogy is flawed in some ways, e.g. due to allowing open circuits and the fact that the fluid molecules actively move through the tube while the electrons in the conductor actually merely nudge their neighbours, who in turn nudge their neighbours and so on, which propagates much faster than the actual electron movement speed (which is slow). So ...

1

One analogy I liked to use for understanding current is a train. Think of the electron as connected carts on a train. If I push one cart (even just a little) this first cart will move since they are connected. Then parallel tracks or larger carts may describe other properties.

3

Falstad physics applets: Circuits with visible electricity If we use the water analogy, we find many interesting things that are hard to understand ...about water! For example, water flows slowly through a hose, yet the hydraulic energy flows incredibly fast. Also the hydraulic energy is counter-intuitive: it can easily flow backwards relative to the ...

5

JYelton answer is very good in my opinion, but I will add alternative answer. Electricity is unique and there is no complete analogy between electricity and something else. Is not just about moving electrons. Electricity is complex and "contains" many diffrent phenomena magnetic fields electric fields ionic reactions chemical reactions There is no ...

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