# Tag Info

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In accordance to what gsills said I will just go ahead and make the diagram for you, Think about this circuit. I will not specify the values of the Resistors and the capacitors. simulate this circuit – Schematic created using CircuitLab Think about how the transfer function of each functional block is related.

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The steep response you want will require a multistage filter. The most commonly used active filter configuration is Sallen–Key topology, which provides two stages per amp using just 2 resistors and 2 capacitors. Various filter responses can be created by varying the component values. Here is an LTSpice simulation of a 4 stage Butterworth high-pass filter ...

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Here's one way: simulate this circuit – Schematic created using CircuitLab This avoids having to wire to the mains (avoiding potential safety issues) and provides a 5V source for charging the battery when the 220V is present (you can use a Li battery charging chip such as a Skyworks AAT3681A, which requires no external components).

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Ok, your text is fairly unclear! Luckily I've got my crystal ball back from repair this morning and I have a rough idea what you want. For several reason I would suggest to use a car battery. You wanne have bright light and loud sound that lasts longer then some minutes. It's easy to find complete sirens and lamps for 12V and outdoor usage. In the winter ...

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The simplest circuit would be a 220V relay and a 3-4.5V buzzer. Connect the battery and buzzer to the N.C. (normally closed) and COM (common) connections of the relay, and connect the relay coil to 220V. The 220V will hold the relay open, and when the electricity goes off the NC and COM connections will be connected internally, powering the buzzer.

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In your two circuits, the dependent voltage source branch is transformed to a dependent current source paralleled with a resistor. It's "source transformation" and is an application of "Norton's Theorem". Source transformation also applies to dependent sources. So, the two circuits are equivalent.

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I want to series the solar panel up and have voltage regulator that can regulate it. You're not going to be using the LM317 as a voltage regulator in this case. It will solely act as a current regulator, also known as a constant current source. (CCS). How does the lm317t react when my solar panel hit shade and current goes under 200mah. I would ...

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Of course, for infinite YL (short circuit) the gain is zero. Hence, you are right and the authors of the book book made an error (typo?). Without any external load resistance (ZL infinite), which applies, for example, to a very (nonrealistic) inductance and a corresponding high frequency, the gain would approach a value that is determined by the internal ...

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If you want 90 MHz, you're going to need a much smaller capacitor, I would think. The capacitor and the inductor are critical to determining what frequency this produces. The page you linked seems to have a 22 nF cap in that location, which is 0.022 uF. 10 uF seems much too large.

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The idea is that you connect the unit to be controlled to the relay's switch contacts. For a reactive load (like an electric motor), some protection of the relay contacts should be used, e.g. a flyback diode for DC, otherwise arcing will reduce their lifespan and possibly make them stick together.

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The relays are shown unconnected because it's up to the user as to how he/she wants to use them. The schematics shown are timer circuits, so commonly you might hook the relay contacts to a light or to turn something off after a preset time (coffee maker?). It's not shown since the circuit is meant for general purpose applications.

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Have you considered a flame diode or triode? These are vacuum tube equivalents that operate inside the flame of an alcohol lamp instead of in a vacuum. This guy has built both, and the site has pictures of his devices and descriptions of the materials and techniqued he used. You will need a source of high voltage DC (he uses 200VDC) and you will have to ...

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The divider has an output impedance of 2.5M ohm, so it should not be connected directly to the ADC or you'll get considerable error. Why not use 10K + 10K 1%? Edit: According to the manual you linked in your comments, the maximum source impedance allowable for the board you have in mind (it appears to use a PIC) is 10K ohms. So 10K + 10K is fine, and ...

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You always need at least 2 electrodes (anode and cathode). If you only have 1 electrode, you can't build a circuit. In electrochemistry, the electrodes generally have some role in the reaction above and beyond just supplying electrons. This determines what materials they must be made out of. In the case of an LCD screen, the liquid crystal is ...

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Spice originally started with ascii type files (or text type files) describing connections between components - these are still the bottom line file used by all spice simulators I believe. Here's an example: - And is described by the following text: - Example_1 EXMPL01.CIR Vs 1 0 DC 20.0V ; note the node placements Ra 1 2 5.0k Rb 2 0 ...

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The problem I see with your schematic is that you have two ground references. Your AC power supply is grounded, as well as D2 cathode. And the bridge rectifier DC output rail's ground is in fact the anode of D2 (even though not shown), the cathode of D3 being the positive (Vcc). Also note that your load R1 is connected between these two points. Now, am I ...

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There use to be all these electronics "cookbooks" you can order them cheap on line. Opamp cookbook Cmos Cookbook TTL cookbook IC timer cookbook (I'm trying to think of others..)

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I would get a copy of The Art of Electronics. It's considered by many as the "bible" of electronics design. It's an old book but to this day it's a really good reference.

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For my opinion, the simplest solution makes use of the classical feedback formula from H. Black: V2/V1=H(s)/(1-LG) with: H(s)=H1(s)*H2(s)=Forward transfer function for an open loop (in our case: H1=V3/V1 for R2>>infinite and H2=V2/V3.) Loop gain LG=Product of all three transfer functions within the loop (with V1=0 or R1>>infinite). Note that H(s) is ...

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The problem here is that the feedback network also feeds your first opamp. Specificly, the current I2 flows trough your feedback network of R3 and C3 creating a higher (less negative) voltage at U3. So this said, your first equation is not complete, because you also need to take I2 into account. I would start with the feedback network including the ...

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A transistor may be used to amplify current, voltage, or both. In the indicated circuit, it is being used to amplify both. When using the emitter as an "output", the output voltage will vary almost 1:1 with the input voltage. In some circuits, that 1:1 voltage behavior is very desirable, since the accuracy of the unity gain won't be significantly affected ...

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Anyway, all that you have to do is to calculate some of voltage and current equation with integration or derivate in the s domaine (Laplace). For example: $$U_3 = R_4(I_4+I_3)$$ $$\frac{d_{U_2}}{d_t} = - \frac{1}{C4}(I3+I4)$$ so $$I_4+I_3 = - \frac{d_{U_2}}{d_t}C_4$$ $$U_3 = -R_4\frac{d_{U_2}}{dt}C_4$$ in S domain $$U_3(s) = -R_4\times C_4\times S\times ... 0 Your signal already has a centre voltage of 2.5 volts and you are wanting to amplify the p-p value by 2 to take you from 2.5V +/- 1V to 2.5V +/- 2V. I really wouldn't bother - any circuit you add is going to amplify noise and signal together so there will be no perceivable SNR benefit and quite possibly there may be a small reduction due to the introduction ... 2 LED would indeed be brighter if you put it after because on the before transistor there will be hfe*Ib ammount of current, and it will have (hfe+1)Ib after the transistor. You will probably not notice this because hfe 100 or more in most cases and 1% extra current will not cause visibly more light. If you have a transistor with hfe of say 5, then you will ... 7 I was thinking that the current had to go though the transistor to be increased. This highlights a dangerous misconception. Current is the flow of charge. Charge, like energy, is never created nor destroyed. Thus, you will never find a device where the total current flowing into the device is not equal to the total current flowing out. In more formal ... 4 Replace the transistor with a simple switch, so you just have a battery, an LED, a switch, and a current-limiting resistor. Note that when you close the switch, current flows on both "sides" of the switch. "Why does the switch control current "before" it as well as "after" it?" Because it has to be: current in a simple closed loop has to be the same ... 7 Perhaps it's a little easier if you think of the circuit as being drawn something like this: simulate this circuit – Schematic created using CircuitLab Your finger acts as a resistor (a largely-unpredictable variable resistor), connecting the positive terminal of V2 to the base of the transistor. That allows a little current to flow through the ... 14 You are looking at a loop of current, so after and before don't have much meaning. Other than the tiny base current, the current in the red and blue paths are the same. Your concept of before and after in electronics isn't applicable. You need to understand some basics before you can make sense of this, but that would be too much to try to teach in a ... 20 The collector of the amplifier (the pin labeled "C") is actually AFTER the transistor (using the asker's reference frame, which is somewhat misleading.). The Base (B) is the input, and the active transistor creates a situation where the current on the collector is many times that of the base. So, the Base current is the input, and the Collector current is ... 0 3rd one doesn't work as XOR. Look at case A=0, B=0: Upper right MOSFET (p-channel) is turned ON. This yields 1 at the output which is wrong. Also look at case A=0, B=1: Upper left MOSFET (p-channel) is turned ON, which makes source of lower right MOSFET (n-channel) high. Gate of lower right MOSFET (n-channel) is high but its source is also high. N-channel ... 1 If you simply want to switch the current on and off at a certain rate, then you need an oscillator. A very common and cheap chip to do just that is the 555. Its full name is the LM555, but you will probably find it called just the 555 (or the 7555, a newer CMOS version). Many oscillator circuits (astable multivibrators) can be found on-line. You can vary the ... 1 High voltage clearance is a complex subject. Too many factors and standards to consider. In your case, I'd follow the IPC-2221A "Generic Standard on Printed Board Circuit". According the table 6-1. "Electrical Conductor Spacing" for a 80V difference between conductors we have: Internal layers --> 0.1mm (3.9 mils) External layers uncoated -->0.6mm (24 ... 1 It used to be possible, since the audio for the old-style analog TV was just an FM subcarrier. You could relatively easily modify the frequency coverage of an ordinary broadcast FM receiver to cover the VHF TV band. However, since all terrestrial broadcasting has now shifted to digital, you need an actual digital TV receiver chip, or the functional ... 0 Try applying partial fraction expansion to it, to get something in the form of$$ \frac{A}{s} + \frac{B}{s+0.5} $$Choose $A$ and $B$ so that the partial fraction expansion equals your original transfer function. Now the first term can be represented as an integrator circuit, and the second term as an RC circuit. You'll also need a summation circuit ... 5 The 33Ω resistor is causing a big problem. According to Ohm's Law, the voltage drop across a resistor is proportional to the current flowing through it:$$ V=RI  So when your motor is running and pulling 75mA the voltage drop across a 33Ω resistor would be (33 * 0.075) 2.475V. That leaves just 2.525V to run your 3V motor. However, that only holds true ...

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With the 1kohm base resistor, the transistor will be 'fully on' at 3.3v (base current is about (3.3-0.7)/1000 = 2.6mA), so that isn't the problem. The 33ohm resistor however will drop 2.5v at 75mA. As such, there's less than 1.8v left for your motor. You should remove it. You can limit the current by either lowering the supply voltage (use your 3.3v in ...

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Let's say $R_1=330k\Omega$ and $R_2=470k\Omega$ For charging, you can analyse the transfer function of the circuit with the switch closed. ${V_{out}\over V_{in}}={{R_2\over sC}\over{R_2+{1\over sC}}}\cdot{1 \over {R_1+{{R_2\over sC}\over{R_2+{1\over sC}}}}}={R_2 \over ... 1 For the closed switch (charging period) both resistors are active (in parallel). When the switch is open the 330k resistor is inactive (discharging period). Hence, the time constant for discharging is larger (470k*0.2µF). The situation will be different when using a voltage step from 0 to 5V and back to zero (duration 0.5..1 sec, for example). In this ... 0 Well look at the circuit and try to understand what is going on. When the switch is closed you will be charging the capacitor (along with letting current flow through the parallel resistor) through the small series resistor. The small resistor will result in a small RC time constant. When the switch is open and the capacitor is already charged then you ... 2 Breakdown voltage of FR4 is more than 300V/mil. Creepage (surface clearances) might be more of a concern, especially if the PCB may be in a bad environment (dust + humidity, for example, or mold). If possible, put grounded "guard" conductors between 80V traces and 3.3V traces if they have to be adjacent on a surface, and try to limit the current on the 80V ... 2 20 mils separation between the 80V and other low voltage signals or the GND is not enough clearance. I have just recently done some PCB design work that has an 84V power rail. I've had to ensure that the clearances between any 84V net and other signals is over 47mils and preferably even more. I can refer to some supporting information on this amount of ... 2 Things like that do exist, and they're called, generically, "latches". There are a huge number of hits here, which should get you started, and if you want to go the mechanical route, here are a bunch of latching relay circuits. 3 What you describe is simple latching behavior. There are lots of ways to get that. You can wire up a relay so that it keeps itself energized once it is energized. A SCR is a electonic device that inherently latches once triggered, and stays on as long as some minimum current is supplied. There are various ways of wiring up transistors to do this. ... 3 It's possible to buy magnetically latching relays, if you don't want to do this electronically. If you're OK with an electronic latch, an SCR (Silicon-Controlled Rectifier) circuit would do the job. Or maybe a set-reset latch (SR flip-flop) would do it, if you don't need to drive a relay. 0 When Vi>2.7, your diagram is correct. So ,Vo= (Voltage across the R2) + (Voltage drop across Diode = 0.7V) + (Voltage from the DC source = 2V) Use KVL to find the voltage across R2. Since it is effectively a voltage divider with a voltage , Vi - 0.7 - 2 =Vi-2.7 So the Voltage across R2 is , VR2 = ( Vi-2.7) (R2/(R1+R2) So, Vo= ( ... 2 There are many answers already. This is simply another approach which does not involve finding the equivalent resistance. First, one can see by inspection that the there is$2A$through each$5\mathrm \Omega$resistor since there is$10V$across each. Second, due to the symmetry of the$10\mathrm \Omega$resistor network, the voltage across each ... 1 I agree with Ignacio Vazquez-Abrams. And here I provide the graphical exaplanation. Blue section consistst of two 10R resistors in parallel --> 5R Green section consistes of two 10R resistors in parallel --> 5R Blue + green are combined together in series --> 5+5 = 10R Now we add the yellow section to it. 5R, 5R, 10R --> 2R I = V/R = 5 A 0 The circuit is redrawn in 1, below. Then considering in 2 that the wire connecting R2, R3 ,R4, and R5 is across the output of a balanced bridge, there's no voltage across it and no current through it, so in 3, below, it's gone. Then, in 3, R1, R2,and R4 are in parallel as are R3, R5, and R6, so in 4, below, they reduce down to a pair of 4 ohm resistors ... 6 The two sets of 10Ω are in parallel and series, so 10Ω || 10Ω + 10Ω || 10Ω = 5Ω + 5Ω = 10Ω. And now that is in parallel with the two 5Ω, so you get 2Ω. And that into 10V is 5A. 2 Use "something else"- symmetry. It's easy to see that the parallel resistance is 2$\Omega\\$ so the current follows to be 5A.

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