# Tag Info

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You can assume the transistor is well saturated for a load less than about Ib/20. Since Ib ~= (5V - 0.7V)/21K ~= 0.2mA, ignoring the 100K, you get about 4mA. That's a fairly safe load, despite the Ic/Ib = 10 specification in the datasheet. So if you keep the load 1.25K or greater, you'll be fine. Probably 1K is okay too, at least if you don't need to worry ...

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In the same datasheet you have Ic=10 mA for vce=0.25v

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A simpler version, no transistors or extra resistors required. Assumes Standard value Forward Voltage Drops (Yellow 2.0Vf @ 20mA, White 3.3Vf @ 20mA). (9Vs - 2Vf - 3.3Vf) / 0.02A = 185Ω 200Ω is next higher value 5% resistor, giving ~18.5mA draw through any chain. simulate this circuit – Schematic created using CircuitLab

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In the simple series connection, the current will be limited to the current from the battery with the lowest current rating. Note: that may not be the smallest of the batteries, either in physical size or in total energy storage. Just for example, Nickel-Cadmium batteries generally have very low internal impedance, so even a fairly small NiCd battery usually ...

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Here's your detection circuit: simulate this circuit – Schematic created using CircuitLab The output will pulse 4.7V at the same frequency as the source when the bell is being triggered, which is appropriate for a 5V Arduino. Use a different zener diode if your Arduino is running at a different voltage. Note that R1 should be a 1/2W device or ...

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Consider a contrived circuit consisting of only a cell with constant voltage V and constant amperage A That's not possible. We can have: (1) A constant voltage source - the current is determined by the attached resistance. (2) A constant current source - the voltage is determined by the attached resistance. (3) Something in-between, e.g., a ...

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When short circuit resistance cause the power supply not to deliver enough current we can say it is a short circuit. Else short circuit resistance is a resistance that is connected parallel to the load.

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You cannot simply define a resistance for a short circuit. A short circuit is basically an unintended connection between two nodes, and that is it. Yes, typically they are low resistance paths, but they do not have to be. If you accidentally put a resistor across your VDD/GND rail, you are "shorting" your circuit because you are changing the expected node ...

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Microchip has devices qualified for fairly high temperatures (up to 150 C), which would seem to cover your requirement. If you need to go higher, there are suppliers for devices qualified for a brief nasty life at 225 C or higher, but they are 8031, not PICs. Naturally, the performance tends towards the dismal and the price astronomical. At some point ...

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I don't have borehole experience, but I do have a lot of generalized experience, including space environments. First, temperature. PICs won't do it. Go look up some data sheets. Their max is 85 C. You need MCUs rated for automotive or full mil-spec temperature (125 C). The same applies to all other components - full automotive (not industrial) or mil ...

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To control the speed of this AC motor you could use an AC motor controller module. Take a look at this one for example: http://electronics-diy.com/1000w-ac-motor-speed-controller.php

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The motors used in those kinds of appliances are typically "universal" motors, which are basically series-connected (armature and stator) via a set of brushes. The capacitor is there to help keep the electrical switching noise of the brushes from feeding back to the line cord and radiating into any nearby AM radios. It may or may not be effective at this. ...

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Why are you not driving the motors directly by L293D? Its a reliable working H-Bridge and also the it can make your circuit less troublesome.. There are other ICs avaliable for H-Bridge based on persons requirements..

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Here's a simple circuit that should do the trick. It's a kind of Wheatstone bridge with a comparator. The two transistors should be from the same lot- they'll typically be matched to within < 1°C. If you need to trim them so they're exactly balanced, you can trim R5 or R6. Take care as to the layout- the path from M1 back to the supply should not go ...

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Use 2 digital temperature sensors (e.g. DS18B20 based), connect these to an Arduino nano along with a relay module and write a simple code in C. This way you can easily re-configure it by changing the code (e.g. add hysteresis, maybe a humidity sensor etc.). Can't beat that with an analog circuit. It will cost you less than $10. 0 Usually to operate BJT in active region Base-Emitter Junction must Forward Bias and Base-Collector Junction must be Reverse Bias In above circuit if you assume it as PNP BJT. then Base is at lesser voltage level compared to Emitter and higher compared to collector 2 A very interesting question... Congratulations, @pyler! I am closely connected with the Miller theorem and, a few years ago, I spent a lot of efforts to reveal the idea on which it is based. In 2011, I created and fully wrote the Wikipedia page about the theorem. More precisely speaking, this page was dedicated not so much to the very theorem, but rather to ... 2 The crystal earpiece has such high impedence that it presents virtually no load to the CD4024. Normal earphones have impedence in the 16 to 32 ohm range, which would (at 9V) cause the CD4024 to provide far more current than it's capable of, potentially damaging or killing it. You need an amplifier with high input impedence which is capable of driving ... 0 I like this question, If it's just an RC's with no buffer/ opamp in between, then you need to think about the impedances. You'd like the low impedance in front. 1 You cannot use the standard (lumped element) circuit analysis to model the circuit at the time the switch is closed because dv/dt of the capacitors being infinite. So this has to be done in two steps. Apply conservation of charge at the instance when the switch is closed to figure out the state of the capacitors. That becomes the initial condition for the ... -1 The capacitors will explode or the wire will melt. The hydraulic analogy is a dam burst. You have emptied the capacitor into a dead short. 0 What you're trying to do is not well-defined, and is not allowed in circuit theory anyway. You need an initial state for the 1 F capacitor. And you can't just short together two capacitors at different voltages. That would imply infinite current and violate conservation of energy. http://www.users.on.net/~ithilien/tam/electronics/CapacitorParadox.html You ... 1 You have the right general idea, but you can't just consider the two capacitors as one 3F capacitor. Just before the switch is closed, the 2F capacitor will be fully charged and (I presume) the 1F capacitor is fully discharged. So when the switch is closed, the 2F capacitor will discharge and the 1F capacitor will charge. Remember that $Q=CV$ for a ... 4 Use the relaxation oscillator to make a ramp or triangle wave and then compare that wave's voltage to a DC voltage that you can control (for example with a potentiometer). 0 $$V_{th}=V_{\text{node3}} - V_{\text{node2}}$$ 0 Interestingly, an XNOR gate turns out to be much easier to implement in NMOS than CMOS; if the signals which are feeding to an XNOR gate aren't used for anything else, an XNOR gate may be implemented using two transistors and one passive pull-up. If the input signals are also used for other things, adding inverters on both of them would increase the total ... 0 I wouldn't bother with a buck converter to charge the capacitor. Just connect the output of the full wave bridges directly to the capacitor and use a shunt regulator to make sure the voltage doesn't go above whatever level you want. Just a zener diode might do. You don't really care what the load on the motor is when the capacitor is already full. ... 0 That second one is really clever! It recycles transistors from the XOR structure to invert A and B. I bet someone got a patent out of that. I'm not exactly a CMOS optimization master, so maybe it's more common than I think. It works like your first schematic, except that T6 and T9 are also used to invert A. You can almost do the same thing with T2 and T5, ... 0 Voltmeter reading will be less than 18V and more than 12V. Current flows from higher voltage to lower voltage. Connect center O Ammeter between +ve of the battery terminals. 18v battery will be charging the 12v battery till both battery voltages are same.Then the voltmeter reading will be between 18 and 12 volts. 0 The usual way to find a Thevenin equivalent for a circuit with dependent sources is to connect a 1 amp test source to the input. Find the voltage across the terminals, then divide by 1 amp to get the Thevenin impedance: $$R_{th} = \frac{v_{in}}{1\ A}$$ The Thevenin voltage is found in the usual way, by determining the open-circuit voltage $v_{in}$. ... 1 Very simple problem with an equally simple fix. Your entire circuit is floating. You need to use this for GND: Edit: Node 0 is the internal reference in SPICE for all voltages, and the ground symbol with the graphic '0' is connected to it. Every node in your circuit must have a DC path to node 0 (even if that path is a 100G ohm resistor). The ground ... 3 By Ohm's Law $$v_o = IZ$$ where $Z = Z_L + Z_C$ is the combined impedance of the inductor and capacitor. You have a non-zero voltage source so you have a non-zero current $I$ through this impedance $Z$. In order to have $v_o = 0$ you therefore need $$Z = Z_L + Z_C = j\omega L + \frac{1}{j\omega C} = 0$$ If you simplify this equation you will arrive ... 2 Your equation is incomplete. The actual equation is based on the sum of the impedances being zero, and to do this properly, you need to account for the complex nature of the impedances: $$Z_L + Z_C = 0$$ $$j \omega L + \frac{1}{j \omega C} = 0$$ If you multiply through by j, you get: $$-1 \omega L + \frac{1}{\omega C} = 0$$ Which you can then rewrite ... 2 $V_1$: Your KCL equation is right, but it looks like you either rounded too much or misunderstood something. $V_A$ is not -100V, it's -100.50251V. You can get this via nodal analysis: $$0.01*V_1 = \frac{V_A}{20k}$$ $$V_1 = V_A + 100$$ Thus: $$0.01(V_A + 100) = \frac{V_A}{20k}$$ $$0.01*V_A + 1 = 0.00005*V_A$$ $$0.00995*V_A = -1$$ $$V_A = -100.50251V$$ ... 1 Zero crossings of voltage happen where there are current maximums. That is because V = L dI/dT. The derivative is zero at maximums, whether positive or negative. Thus, when you add that wire and 0 The wire, being non-zero inductance and resistance. will do two things. It will (dramatically) change the resonant frequency AND it will dissipate a LOT of the ... 0 Just model your situation to a circuit's below, there are initial voltage on C1 and initial current on L1. And do transient analysis on it. If possible, you can transform the circuit from time domain to the s-domain. simulate this circuit – Schematic created using CircuitLab 2 You wrote your $I_3$ equation for the third mesh incorrectly; it should read $I_3=\color{red}-10\angle0^\circ\,\text{A}$. Solving this system of equations yields $I_1\approx-2.4805-5.4297j\approx5.9695\angle-114.553^\circ$ and so $I_0=-I_1\approx5.9695\angle65.447^\circ$. Note that $I_0$ is a phasor; for the time-varying instantaneous current we ... 2 The first sum is correct, the carry C is just the but that comes from the sum of the most significant bits, in this case 1. The overflow is straightforward to calculate once you know the values of C3 and C4 (C). The second sum is wrong, probably because you didn't sum the carry in (M). The least significant bit of the sum becomes 1+1+1=1, then you can work ... 0 Taking the Thevnin equivalent across the capacitor can reduce the circuit as shown in figure. Where the initial voltage across capacitor is Va/2. simulate this circuit – Schematic created using CircuitLab You can find $I_C$ from this. Now considering the circuit in the question, the current passing through the diagonal resistance $I_2 = ... 3 SparkFun has an Eagle library that has pads for single and dual row 0.1" headers from 1 to 40 pins. Here is one of several 1x03 headers (single row, three pins): The holes have a 40 mil diameter. The yellow square posts, which are for reference only, are drawn with 20 mil sides instead of the standard 25 mils, but expanding the sides out to 25 mils and ... 2 The pins in the common 0.1" post headers are .025" square. For .025" square posts, I usually specified .040" finished hole size (size after plating). For square posts, the hole must be slightly larger than the diagonal measurement of the post. 2 There are numerous problems with your schematic:- The ESP8266 is rated for 3.6V maximum. A 3.6V NiMH battery can charge up to 4.2V or higher and the Schottky diode may drop less than 0.4V, so the ESP8266 could get over 3.8V which will probably destroy it. You should insert a voltage regulator to keep the voltage down to a safe and constant value (eg. ... 3 This is not definitive, but it may be due to component values drifting with age. The network between TR203 and TR204 collectors is designed to drop approximately 4*Vbe plus twice the 14mv you expect, and is adjustable. Now if TR202 and TR203 are reasonably well balanced, both collectors will be around -1.3V (though TR2 may be a couple of volts different) ... 0 "EM Fields" said according to the data sheet of 1N4735A that uses Izt=41mA for testing, he found there was problem with the value of R1 (4.7kΩ) that limits the current to about 4mA. The value of Izt used in the compilation of data sheet is a choice of the company. It just tells reader that the data are gathered under such a condition. Some data sheets of ... 0 This is potentially a very tricky question, and you need to define the problem much more carefully. For instance, let's say that your trigger pulses last 0.5 seconds, and you get two of them, A and B. The rising edge of B occurs 2.75 seconds after the rising edge of A. A literal interpretation of your question would give a 0.5 second A followed by a 0.25 ... 0 It's not quite clear to me what your trigger pulse looks like, but you could probably do it with a 74HC123, perhaps using one half to hold the other in reset and generate a controlled length trigger output pulse from that half. There's a lot of possibilities (are runt trigger pulses an issue? what if trigger pulses arrive at 1MHz?) so you'll have to do ... 1 It looks to me like the voltage source at t=0 is just supplying some charge to the capacitor, and the differential equation would just be the decaying transient response of the RLRC circuit with an initial capacitor voltage of 8V (Vc(0) = 8V). So yes, at t -> ∞, i(t) = 0; there would be no voltage drop across any component, there would be no current flowing ... 1 Yes. Good job. (I wish all my students wrote this well.) 3 You have an equation for$v_o$of the form$v_o = kv_{gs}$but you need$v_{o}/v_i$. Use the hint (which is just KVL) to substitute for$v_{gs}$: v_o = kv_{gs} = k(v_i - v_o) and solve for$v_o/v_i$. For$R_o$you have forgotten the contribution of the dependent current source. To calculate$R_o$set$v_i = 0\$, apply a test voltage ...

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The top set of contacts will be closed when EB is active and open when it isn't.

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