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I have finally answered my own question after a good bit of agony. I left out the part that I took a 100u inductor, unwound appx half of it, then rewound it appropriately to get the muctually coupled split inductor. It then acts like an autotransformer, doing exactly what it's supposed to, and transforms the voltage and current. The current flowing to the ...

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The best way to visualize the Thevenin equivalent is to assume you are physically looking at the circuit from the end you are calculating from. In this case you are looking in from the far right. (When I first studied this there would actually be a picture of an eye looking at the circuit from the one end.) When looking in from the far right end what ...

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We ignore the R10 because effectively there is an infinite resistor across the two output terminals. Which means you have the same voltage drop at the output as is dropped across R9 which is 16volts 22*(R9/(R9+R8)) (Just think of it as another voltage divider of an infinite resistor and R10. The infinite resistor gets all the voltage across it with nothing ...

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I think you are confusing Thevenin equivalent circuit with voltage and resistance. Thevenin equivalent circuit consists of an equivalent voltage source in series with thevenin equivalent resistance, and optionally load resistance / element which you assumed as load. In your case, the equivalent voltage is really the one appearing across R9 because R10 is ...

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You don't need to find Rth. Note that R10 is immaterial since the Thevenin voltage is found with the output open circuited so no current flows in R10. Without R10, the Thevenin voltage consists of a voltage divider, R8 and R9, and a source voltage of 22 volts. Just calculate the output voltage based on that configuration.

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Yes, but... You are removing the power switch and leaving yourself (and others) with no way to quickly disconnect power in the event of laser malfunction (fire) or immolation (laser setting something/someone in the room on fire). Lasers can be dangerous! Lasers can be dangerous, even small ones. Never point it someone's eyes. Treat them like a loaded gun ...

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Lee Valley Tools Secret Magnetic Door Lock is available on-line and sells for about Can \$9.00. However, the picture may give you ideas of where else it might be available. Lee Valley Tools does ship worldwide if you wish to purchase from them. FWIW - I think very highly of that company - they sell very nice stuff at premium prices but everything they sell ...

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Well, it's not electrical engineering just because it's magnets. I suppose an active version could be made, but there is this option: It doesn't appear to be available for purchase, but it doesn't look difficult to build.

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I have simulated the circuit with the following modifications: 1.) Remove C2 2.) Upper part of L1 with 30µH and lower part with 70µH. As expected, the circuit oscillates at app. 16 kHz. I suppose, the gain (with C2) is too large - driving the BJT into severe saturation. As a consequence, undesired effects (storage times etc.) have an influence and ...

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Power It is possible to get a very small amount of power from phone audio jack. You need to set up a bridge rectifier to turn the AC audio signal into a DC voltage, then use that voltage to charge a capacitor that will in turn power your circuit. Total amount of power you can distil from the audio output depends on a lot of factors like... What voltage ...

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You are calculating the one-bit parity of your n-bit input. The standard way to do this is with a tree of 2-input XOR gates, which minimises the gate count and depth.

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There is nothing wrong in your approach: sum = 0 for i=1:n loop sum = sum + a(i) end output = !sum But I don't see a need for full-adder here. You can implement this logic using half adder or XOR gate. XOR will be a better alternative: output = !(a(1) xor a(2) xor ... xor a(n))

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Simplest way I know is using light, put a phototransistor into the circuit as a low-side switch and shine a light at it, it will turn your LED on. This is good within line of sight. You can make this a fun laser tagging game my shielding the phototransistor so only a head on laser beam would hit it. If you have to use radio waves within a short range, you ...

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In your hypothetical example of trying to detect a 3VDC signal against a background of 250kVDC noise (say between the phase 1 and phase 2 power lines), imagine the signal is being measured by a well-insulated lineman who can put his DMM's positive lead on the "phase 1" 250.003kVDC wire, and the negative lead on the "phase 2" 250.000kVDC wire. Now this ...

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$(\overline Q P) + R$ $=\overline{(\overline {\overline Q P}) \overline R}$ = (NOT Q NAND P) NAND NOT R

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The best way for a beginner is to think about each term separately and how you would create that with a NAND gate. Let's start with the basic NAND gate truth table: A | B | Q --------- 0 | 0 | 1 1 | 0 | 1 0 | 1 | 1 1 | 1 | 0 Now let's look at each term. We have a NOT, and AND, and an OR in there. So, how can we make those with NAND gates? Start with ...

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If using just two 3-8 decoder chips: You would need to connect the first 3 data lines in parellel to the two decoder chips, then use the remaining high bit as an enable to the higher decoder chip and a disable the lower decoder chip. For example the 74HC238 decoder chip has three enable pins (two low and one high), your higher bit input would connect to ...

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you have to design a 4x16 decoder using two 3x8 decoders. here is the schematic that may help you. simulate this circuit – Schematic created using CircuitLab the two squares are two 3x8 decoders with enable lines. the three selection lines of each decoders are connected together as common line(X,Y,Z) , the enable lines are ACTIVE LOW, they are ...

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It would be represented in 2's complement. For 4-bit numbers, the most significant bit represents -8 and the remaining 3 bits represent +4, +2, +1 respectively. Therefore the decimal equivalents of all 16 possible numbers are: 1000 = -8+0 = -8; 1001 = -8+1 = -7; 1010 = -8+2 = -6; 1011 = -8+3 = -5; 1100 = -8+4 = -4; 1101 = -8+5 = -3; 1110 = -8+6 = -2; 1111 = ...

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The blocks labeled "FA" are called full adders, and they are fundamental to how binary arithmetic is done at the gate level. The circuit you have presented here is an implementation of a 4-bit adder/subtractor. Combining addition and subtraction in the same operation requires the use of an alternative representation of binary numbers. This is called 2's ...

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I know you asked this a while ago, but in case you still need the answer, yes it is possible. According to http://www.cburch.com/logisim/docs/2.1.0/guide/menu/edit.html "Note: Logisim's clipboard is maintained separately from the clipboard for the overall system; as a result, cut/copy/paste will not work across different applications, even including other ...

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The upper TIP122's are wired as emitter-followers, so their output voltage can never be higher than their input (base) voltage. (In fact it will be ~ 2 x 0.6V lower.) When you fed the inputs with the full Vcc that was not much of a problem, but when you feed it with the 5V Arduino outputs you will get no more than 5V from the outputs. One solution is to ...

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It depends on if Vg represents an open circuit or a resistive load of some kind. If it's an open circuit, then you have the trivial case: there is no current and $I_{g}=0$. This case might be useful if you have voltage probes hooked up to nodes D and B. If there is a resistive load, then you must have prior knowledge of the load's resistance, ...

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The first step would be to look at the two resistors 22 Ohm and 75 Ohm. The total contribution of their resistance is their sum, and by linearity, you can either change their order (move one to the other side) or better, combine them into the total reistance of 97 Ohm. The paralell resistors needs some algebraic manipulation. Any equation of the form ...

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It often helps to first look for simplified answers among complex questions. If we first revise the circuit arrangement (steady state before t=0) some of the answers start to become obvious. simulate this circuit – Schematic created using CircuitLab In the more familiar arrangement above the voltage across the central resistor can easily be ...

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This doesn't satisfy everything that Dwayne speaks of, but for a simple tester it should serve your purposes. simulate this circuit – Schematic created using CircuitLab Q1 can be either an NPN or a PNP. This is really just testing the polarity of the base terminal by assuming that current will flow out of the base in a pnp and into the base for ...

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What you have drawn will work but it may not give you the actual result that you are looking for. Yes: one or other of the LEDs will light if the correct transistor that is working properly is installed into the correct socket. However, the LED will also light up if a transistor is inserted not correctly. For example, if you swap the B & C lead on ...

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From the data sheet the highest voltage on a GPIO is 3.8V. So If I have 5 Volts going through a.... lets say 1M resistor to the 3.3V input pin is this ok? The datasheet DOES explicitly cover the situation by stating an absolute maximum voltage. Above that voltage damage MAY occur. If the IC is operating mis-operation may occur. Or may not. ...

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You can't avoid that the battery voltage drops when you increase the current drawn from the battery, although using lithium or good NiMH cells reduces the voltage drop compared to alkaline cells or (the horror!) zinc-carbon cells. No way of cleverly connecting either the small LED or the high-power LED to the battery (using switching regulators, big caps, ...

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Instead of simulate this circuit – Schematic created using CircuitLab try simulate this circuit where the zener voltage is about 1/2 the normal battery voltage. This is about as simple as it gets. If you don't want the waste of battery power in the zener, you'll need do something more sophisticated, such as a voltage regulator or ...

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You can connect either DB9 of JP5. They are connected to the same lines in parallel.

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It looks like the two plugs are connected to the same places, however the JP5 is marked 'DNP' which implies it is not populated so you'd have to solder it on there (at best, at worst they've overlapped the footprints so it's not accessible without removing the 'DB9'). Looking at the Eagle file- in this case, it's the header behind the 'DB9' so you just ...

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Open-circuit voltage is the maximum output voltage with no load applied. short-circuit current is the maximum output current with 0 ohm load. Assuming the V-I curve (graph of voltage output vs current output) is linear, then the source impedance is 4.5V / 18mA = 250ohms. In other words, this solar cell behaves like a 4.5V battery with 250 ohms in series. ...

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It means if you were to measure the voltage across the leads coming from the solar cell with a multimeter with no load on them the voltage would be 4.5v. If you were to connect the two leads to your multimeter in a current mode, essentially shorting them, it would read 18 mA. So this is the theoretical maximum current the solar cell can output (in ...

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It means what it says: if you have the cell unconnected driving nothing, there will be up to 4.5V across the terminals. If you connect the terminals together through an ammeter, it will read 18ma. There will be some sort of standard assumption about the light on the cell during these measurements. This does not mean you can get 18ma at 4.5V; drawing current ...

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The logic for this can be easily designed into a 256 bit memory lookup table. Back in the day there were plenty of small programmable PROM type memory chips that made this easy to implement. These days it could be implemented into a portion of a larger memory element. If logic is going into an FPGA then this could go into an initialized block RAM element. ...

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If you have a 4-bit subtractor or four full subtractors, then connect A and B as inputs such as to produce the difference B-A . Then the borrow-out pin of the subtractor will give you the answer.

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OP says: You may use any MSI modules as well as any other gates Therefore you want to save yourself a lot of work and use a 4-bit MSI (medium scale integration) IC comparator such as the CD4585,or the CD4063. The pinout below is for the latter. The inputs connect to A0-A3 and B0-B3. Ignore the three lines (xxx)IN, they are used for cascading. The ...

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you can use below circuit :

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Option 1: Design a unit to compare two 1-bit numbers. Then add a cascade from a previous stage to the unit. Then cascade multiple units to compare two 4-bit numbers. Option 2: Use a Karnaugh map, boolean reduction, or similar methods to come up with an equation that performs the operation, then implement it.

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Always be very careful when designing circuits that are ment to put a current trough your body. It would be dangerous to let the current for the LEDs actually flow trough your body. What the people at the museum probably did is use some kind of amplifier. The resistance of a human body (including the skin) is somewhere between 10 kilo ohm and 100 kilo ...

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Vee should be connected to Ground, or to a negative supply. As this is a CMOS part, all digital inputs MUST be connected (possibly via a resistor) to either Vcc or Ground at all times - otherwise the inputs may float to a "maybe" state, with unknown consequences. Connect all S0 - S2 and /E to ground with 10K (or so) resistors. Also connect A0 to ground ...

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My best guess is that in the original unit, the LEDs were all connected in parallel. Although it is bad practice to run the LEDs from batteries without a current limit resistor, Asian manufacturers do it all the time. In the schematic that you show, you have connected the LEDs in series. Assuming that these are Red LEDs, you would need at least (3 * 1.7V) ...

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I'd say there are two possibilities. First, please check the original unit. In addition to the batteries and the LEDs, there ought to be a resistor. This will limit the LED current to safe levels. Without any resistor at all, it's entirely possible that you've simply destroyed one or more LEDs by hooking them directly to the battery. The second possibility ...

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Because you want the LED to react to audio from a MP3 player or similar, I suggest that you have a look at a recent thread on this forum: 155914. I showed a circuit that uses 2 transistors: one transistor is wired as a diode and biases the 2nd transistor just below its' turn ON point. It should work well for you.

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OK since you already have it as an electrical signal you might try something like this. simulate this circuit – Schematic created using CircuitLab

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This is a very general question but I will attempt an answer. Take your example of a mobile phone powered by a battery. We can model the loads seen by the battery as a number of parallel circuits: one for the display, one for the keyboard, one for the controller, one for the radio transmitter/receiver, etc. Each circuit will draw enough current from the ...

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They have fixed voltage and current... They don't! You have usually fixed voltage and a maximum limit for current. When a part of the phone needs more current, total consumption will increase. Since power is product of voltage and current, the power consumption will increase as well. When you go over your maximum current limit, your voltage is going to ...

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Most diacs have a breakdown in the 30V range. You could use SIDAC, which has a breakdown in the 7~9V range. If you don't need the bidirectional characteristic, a programmable UJT such as the 2N6028 could be used. But for most circuits requiring a threshold, comparators, op-amps used as comparators, discrete transistors and zener diodes, or even ...

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It may be that your battery has dropped below the combined forward voltages, or that one of the LED's has failed in an open circuit. To see how much current is actually flowing, measure the voltage over the resistor. Since the resistor is 82-Ohm, and I=V/R, every volt dropped across the resistor equals 12.2 mA.

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