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0

I think that a clarification is in order. When a switch, relay, transistor, etc. is NO or NC, it is in this state/condition, when the device is not activated. What this means, is that a NO switch (for example) will be closed when it is activated (a magnet is in proximity; door closed). With this kind of switch, you make a "closed loop," which will be ...


0

Do you have another resistor in the same range of known resistance? You could then use a wheatstone bridge to determine the resistance. Actually the best way would be to have a number of resistors of known value, each half the resistance of the prior one. Then you could do a binary search.


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If you were to use one of the 555-based measuring circuits, you could use an optical fiber line to get the output of the 555 into your microcontroller. I use a lot of stuff from Industrial Fiber Optics. They sell jacketed 1mm plastic fiber along with matching transmitters and receivers. I'd send power out to the soil sensor from its own wall wart power ...


0

You can certainly get op-amps (as @Whatroughbeast mentioned) that can do this easily. Analog Devices have a few that have fempto amp bias currents and you'll need to arrange the op-amp as a transimpedance amplifier (TIA) with a feedback resistor of about 1Gohm. You can get this stuff fairly cheaply (less than $10). I have used such a set up for monitoring ...


5

You can do what you want, but it won't be simple. Scratch that, it will be fairly simple, but it won't be easy. As von Clausewitz said, "In war, the important things are very simple, and the simple things are very difficult." It applies here as well. Actually, it's not as bad as it could be. Your nominal current is 250 pA. You can get op amps with bias ...


3

Existing equipment such as the Keithley Model 6517B Electrometer can measure resistances up to 10^16 ohms without using high voltages. This is possible because these instruments can measure extremely small currents, down to 10^-15 amperes or so.


0

It is called Fault Condition. Both conditions: Open door and Power Supply failure are covered.


2

Yes assuming V1 is a non ideal source with an internal resistance (r) in series with a load R. This can be modelled as an ideal source in series with r+R. The voltage delivered to the load will simply be \$V_r=\frac{V_1*R}{R+r}\$. In order to operate your circuit in the linear region it is recommended that you keep R>>r. The effective emf of the cell on ...


0

If your charger sits at 12.6 volts when it's outputting 2 amperes, then your resistor needs to suck up the difference in voltage between the charger and the battery while letting 2A through to the battery. So, from Ohm's law, $$ R = \frac{Vs-Vbat}{I} = \frac{12.6Vs - 10Vbat}{2A} = 1.3 \ \Omega\ , $$ and the resistor, run CW, will dissipate about: $$ P = ...


0

Hi, There are a few bugs, marked on the drawing in color. Think green :)


1

Normally Open magnetic switches can be put them in parallel rather than series when you want one wire pair to monitor several openings. Unfortunately if the wire is cut, it doesn't alarm, but this is taken care of by having each switch in parallel with a large resistor. If the wire is cut, the resistance changes upward, and you can tell that the wire was ...


20

N/C magnetic switches are unfortunately nearly useless for security. Since their "unalarmed" state is for them to be open, if the cable going to them is cut then the alarm will not trigger, and of course not be triggerable anymore. With a N/O magnetic switch the circuit is closed in the unalarmed state, and cutting the cable will trigger the alarm.


1

To invert your signal, use another transistor. simulate this circuit – Schematic created using CircuitLab Any of the BJTs will work. I'm personally partial to the 2N3904, but that's habit more than anything. Of the MOSFETs, I'd recommend first the IRF520, then the RFP70N06, and the IRF510 last, although any of them will work. If you use the ...


0

As L4 and L23 are in parallel their voltage is equal: thus VL234 = VL4. And similiarly VL234 = VL23. Now you can calculate IL23 with XL23 and VL23. As they are in series the current through L2 and L3 is equal, so: IL23 = IL2. Finally we need the voltages for VL2 and VL3: These can be easily computed using a voltage divider ...


2

It is called a common mode choke. It is a toroidal piece of ferrite with the wires of the cable passing thru the hole. The cable and the ferrite form a small-valued inductor, but only to the sum of the currents in the cable. This inductor increases the impedance at high frequencies. The purpose is usually to attenuate these frequencies so that RF noise ...


0

It's a (relatively large) ferrite bead, installed by the manufacturer on cables when the electronics at one or both ends of the end of the cable didn't pass EMC compliance testing (required for most commercial electronic products), and it was deemed cheaper to add the ferrite to the cable than to actually go back and do the job properly and do the filtering ...


0

The data sheet says: Modulation Input DC 0-5 V to Set Output Power Looks to me like the comment on 'matching' refers to the power supply rather than the control input. So you need a solid power supply capable of > 1.1A for the lamp and controller and an input signal suitable for a 3.5K input Z. A simple way to get that would be to use a 5V series ...


0

It seems like they could mean two things; one being that the dimming supply must be a voltage source stiff enough to supply 0 to 5 volts into a 3500 ohm load, and the other being that the output impedance of the supply must be 3500 ohms. In order to find out which, you could put a 3600 ohm resistor in series with your 5 volt supply output and feed the ...


0

To answer the other half of your question, RC Phase-Shift Oscillators work through essentially a delay means. Here is a good explanation of phase shift. An OpAmp is used as the amplifier, because it has an "inverting input." All other parameters being equal, increasing the +input causes a + increase to the output, while decreasing the -input causes a + ...


1

This is one of those pesky cases where the standard procedure for finding the open circuit voltage or the short circuit current leads to "funny" results. Another way to compute the Thevenin equivalent resistance of a network N is the following: (1) Disable all independent voltage/current sources inside N, i.e. current sources replaced by open circuits, ...


0

Seems nice try. Instead of using logic gates you may use basic micro controller like 89c51. And also programming also not much difficult. The things you need 1. Microcontroller 2. Toggle switch (for input ) 3. Lights LED ( if u go for lamp need driver ) 4. Other miscellaneous like crystal, reset button, power supply In my view it looks like a simple and ...


1

Converting a design "sketch" (which is really what you have drawn above) into a schematic involves three steps: finding real-world parts that match what you are trying to do, making modifications to the circuit based on the first step, and then drawing the schematic with some sort of schematic capture program. For the last step, I am using the EAGLE ...


0

I found this thread becouse I was having the same exact issue. I added a 100nF tantalum cap across the power rail and like magic - it stopped!! Note: I am using this on a PCB and not a breadboard.


0

As a first impression, the impedance of that coil will be very low - so low you will probably have problems exciting it with a normal audio amplifier. Speaker coils use the iron from the speaker to have more inductance. If the coil can be 20mm, at 0.3mm trace width, you can have about 15 turns. Try that in this calculator. It gives about 5 uH way too low


0

A coil energized with current can create a force of attraction to an iron object. It can also repel a magnet if the polarity is correct. That force is: - Force = \$\dfrac{(amps\cdot turns)^2\cdot \mu_0\cdot A}{2\cdot g^2}\$ Where A is cross sectional area of the electromagnet g is gap to your magnet \$\mu_0\$ is 4\$\pi\$ x \$10^{-7}\$ Let's say the ...


0

He uses the thumb tacks to close a simple button circuit. The tracks are soldered to the individual button inputs. Then he uses a wire connected to the common terminal (likely common ground) to "press" the buttons by sliding the over the tacks. You can see the tack wiring on imgur about halfway down the page and the way he presses the keys at 6:40 in the ...


0

In both cases the potential difference between the middle and the top is zero volts, therefore, there is not current flow (assuming ideal circuit).


1

Some capacitors, especially ceramics will derate or have lower capacitance as the applied voltage goes higher. However, most electrolytics will have nearly the same capacitance independent of the applied voltage. The voltage rating is the maximum voltage that should be applied to the capacitor, it does NOT imply that you will only get the rated capacitance ...


1

This is because using nodal analysis and the superpostion princicple (for independent sources), you would normally analyze the circuit once source at the time, with other voltage sources shorted, and then add the results. Obviously, shorting one source also shorts the other, hence the error. The error makes sense - the circuit cannot be analyzed as currents ...


1

You can analyze this circuit by inspection (and without having to resort to superposition) if you consider KVL around this loop: The KVL equation would be -5 V + (-10 V) + \$V_{150}\$ = 0 where \$V_{150}\$ is the voltage across the 150-ohm resistor (with positive reference terminal at the top). Rearranging, \$V_{150} = 10 \mathrm{V} + 5 \mathrm{V}\$. ...


0

Here's a more-or-less easy way to look at it:


1

An alternative method to solve this would be using the superposition theorem. It isn't really required in this case, but in more complicated setups with multiple sources it is certainly useful. Here is how you do it. 10V source 5V source 1. simulate this circuit – Schematic created using CircuitLab Looking at the diagram above it is clear ...


1

The voltage sources are both in the same direction Therefore your calculation for K is wrong. It should be: K = 10V + 5V = 15V from U = R*I we derive I = U/R and end up with: I = 15 V / 150 Ohm = 0.1 A The special case of this circuit is that all resistors are connected directly to a voltage source and the voltage drop over each resistor is given ...


1

Your voltage equation is incorrect. Both voltage sources have the same polarity with respect to the resistor, therefore they add.


1

Think of it this way: Since the middle branch (containing the "top" 50 ohm resistor) is open, no current will flow through it. Therefore, you're dealing with a single series loop, and you're actually measuring the voltage across the "bottom" 50 ohm resistor. All you need to do is find the total current through the main loop, and use Ohm's Law to find how ...


2

Another approach is to take each one of the branches with batteries and substitute them with their Norton equivalent, i.e. a \$5\Omega\$ resistor in parallel with a \$ \dfrac{2V}{5\Omega}=0.4A\$ current source. simulate this circuit – Schematic created using CircuitLab Then you get three paralleled \$5\Omega\$ resistors driven by two paralleled ...


0

Ground is just a reference point. Just move the ground to point A and calculate the voltage at B. The relationship between A and B will always be the same, regardless of where you put ground.


1

You can solve this by stepwise reduction. 1) The left side of the two batteries is connected, hence the right sidse of the batteries have the same potentional. Connect those two point with a wire. 2) Now you have two same-voltage batteries in parallel. Remove one. 3) Now you have two 5 Ohm resistors in parallel. Replace them by a single suitable ...


0

Your outright guess that node A is 2V with respect to node B is not correct. There are resistors in series with the two batteries that must be taken into account.


0

I don't think anyone's actually read the question properly. "As expected, this causes the speaker to pulse each time the positive wire breaks and unbreaks. However, the frequency is very low and I was wondering if it would be possible to have a circuit that "adds some frequency" to the current frequency so that I would get a continuous beeping instead of ...


2

A mixer might be what you're looking for. A mixer multiplies two signals together. \$x_{out}(t) = x_1(t) \times x_2(t)\$ if \$x_1\$ and \$x_2\$ are two input signals. From the multiplicative property of the Fourier transform, this means that in frequency space, the output contains sum and difference frequencies of the input. Say \$x_1(t)\$ is the signa ...


1

Assume your signal is a square wave and that its frequency is \$f_0\$. This square wave is actually the sum of sines of frequencies \$f_0\$, \$3f_0\$, \$5f_0\$, and so on. You can design a band-pass filter around one of these higher-frequency harmonics to produce a continuous sine wave of a higher frequency. These sines are known as the signal's harmonics, ...


2

Could be leaky, could be pickup of AC hum from the mains or RF from AM radio etc. as the transistor will act as a detector. To tell the difference, put a 1uF ceramic (low leakage) capacitor from base to emitter and see if the issue disappears. In any case, you can reduce the sensitivity simply by connecting a relatively high-value resistor from emitter to ...


0

"Singular Matrix" issues are usually resolved by adding a 1GIG resistor to ground. It means "a current is too small to calculate." Try adding one from net d:u3:u2:1 to ground. (I can't read it either.)


0

You could use use buffers after the switch as your solution #1 with an R-2R network. A 74HC4050 or 74HC4049 depending on whether the switch is inverting or not. Or buy resistors that are in the 1:2:4:8 ratio- there are only four parts or so- there's no need for a network. Eg. 10K/20K/40K (two 20K in series)/80.6K. You don't (necessarily) need a constant ...


1

Just a single switch is not enough. Actually, two switches is not enough. You need a memory element, such as a relay, plus a switch at each end of your weight travel. The following will do what you need. It consists of a relay (double pole, double throw), two limit switches (one normally open, the other normally closed), and a diode. simulate this ...


0

Just because something is old doesn't mean it doesn't do the job. The advantage of a van de graaf generator is there's less risk of electrocution, since it only holds a small charge. You'd get a quick shock, but with any luck, won't kill you. And it can actually be quite accurate; a given machine will fairly reliably build up to a repeatable voltage. ...


0

I remember doing lots of analysis like this at university! Ah, they where the days. There are only two times a voltage is the same at two points is if 1)there is a straight line connecting the two points together and no components are in that line. 2) Where a resistor is in the line and no current is drawn. For example an unconnected battery which ...


0

Umm, about your SCR dimmer: It's not really a dimmer. It's more like a chopper. The logic goes like this: Wait for AC input to cross zero volts Turn off (this is hard-off, no current at all except leakage) Wait for desired off-time (decreases with higher power) Turn on (this is hard-on, like a wire) Repeat from #1 This makes it much more efficient than ...


0

When the wiper is at P the resistance is (0Ω || 10Ω) + 1000Ω = 1000Ω. At Q it is 1000Ω || 10Ω ~= 10Ω. Therefore the resistance decreases as the wiper moves from P to Q, and subsequently the current increases. Therefore all graphs are potentially correct. In order to determine which of the graphs is correct, write out the equation for the current in terms of ...



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