# Tag Info

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I don't find your circuit very helpfull in understanding / explaining what a Zener diode does. I would suggest you try this circuit instead and see if that helps you to understand: simulate this circuit – Schematic created using CircuitLab And forget about equations for a moment, and before you simulate this circuit just think about: what ...

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simulate this circuit – Schematic created using CircuitLab Physically, here's what going on. In a boost converter, energy is transferred from $V_{in}$ to the inductor when the switch is on and the diode is off. During this time, no energy is transferred from the inductor to the output capacitor. Now, when you increase the duty cycle, the ...

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It appears you are referring to the effect of a right-half s-plane zero on the transient response of a transfer function. If such a zero is on the real axis it acts as a differentiator and subtracts from the 'natural response' (i.e. the response without the zero). Depending on the numerator TF coefficients, this zero may cause an initial negative-going ...

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Hint: An approach to simplifying / solving this circuit is to redraw it into a more standard way (as other have said in their comments). You can start with labeling all of the nodes, I have colored them in the picture below. Any resister that has both ends connected to the same node is shorted out. Update: How about now? simulate this circuit ...

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Your new circuit (2nd diagram) should work just fine. You need to calculate how much power the SSRs will be dissipating. Assume that there is a 1 Vac drop across each SSR. Because you didn't mention in your question how much current the oven is supposed to draw, we can't estimate how much heat you need to get rid of. You may also want to put a SSR or ...

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First a diode(LEDs are diods) above a certain voltage is like an open circuit. The problem is that like every wire, the diode has a critical point after which it will "burn", basically some irreversible transformations occur. So you can say the diode can sustain a certain power. Now, power is related to voltage like this: P = I*V where I is the current, ...

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Just as Andy says, ceramic capacitors can exhibit microphonics, which is less than ideal in audio circuits. Large value, physically small surface mount MLCC ceramics also exhibit quite profound decrease in capacitance with applied voltage - take a look at this article from Maxim. This is a very good reason not to use them in filter circuits, where cutoff ...

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I'm not saying that they are any different to any other capacitor in general but, certainly, some ceramic capacitors can exhibit "certain features" that may make them problematic in some audio applications. Microphonics is a recognized problem. The capacitor behaves like a piezo electric transducer and, in a sensitive amplifier producing a big audio ...

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Yes, you can combine these. However you change that circuit in some way. If you did a circuit analysis before using e.g. matrices to describe the circuit, those aren't valid any longer. In real world you probably have reference potential points (i.e. earth points, housing, etc.) which are omitted in this simplified circuit but are often in a relation (i.e. ...

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You can only combine two single elements into one for sake of analysis if they are in series with nothing else connected to the node between them, or if they are in parallel with nothing else connected in series between them. So, do Xl and Z meet either of these criteria? If not, why not?

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The "canonical" form of a schematic diagram is a netlist. However, in order to compare two netlists, you need to make sure that the component names and the node names of the two schematics are identical, which may require setting up a mapping of some sort between the two. Then, you go through the netlist node by node, and make sure each node connects to the ...

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Yes. You print out the two schematics and then check each component and connection manually. As you have checked each one highlight it with a yellow marker. When you get all done the whole schematic should be all yellow. Along the way keep a pink highlighter handy. Whenever you find a mistake or mismatch use that color to highlight and this will be your ...

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If you would indicate what sort of receiving setup you plan to use it would guide the answers. A (continuously variable) slope delta (CVSD) modulator IC (special part) would have been one method in the past but they are rarely used these days. The data is free running or synchronous usually. An 8 pin microcontroller and bit of code (special part) could ...

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You could try the ADXL362 3 Axis Accelerometer. Sparkfun makes a breakout board and there are arduino libraries to make it work with not to much effort. You could continuously read and parse the data and record if you hit your threshold acceleration value. Then, your button input would trigger the arduino to look at the value and then turn the led on or off. ...

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Option 1: This is a perfect application of a microcontroller. Using a micro would allow you be a lot more flexible about the specific operation of your project. You could easily add features such as logging acceleration over the course of a trip, or specialized filtering of bumps. Option 2: If you want a solution based on discretes (either for learning or ...

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A typical digital microphone circuit is produced of a few parts. Power, Microphone + Analog Pre-amp, a Analog to Digital Converter, and the Microcontroller/Receiver. Often the ADC and Controller are one piece. The Preamp can be as well. For example, the OVC3860 Bluetooth Audio IC is an all in one IC that provides Mic power, pre-amp and ADC, as well as ...

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Get off you idea of "WITHOUT any special parts". An electret microphone will output a low level analog signal when some bias current is applied to the part via a resistor from a supply voltage. That low level signal then needs to be amplified to some level that is compatible with the analog to digital converter that will be used to produce the 1's and 0's ...

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There are several companies that do teardowns of Intel chips. It might be worthwhile to take a look at their reports if you can't find official Intel data. One such company: http://www.techinsights.com/

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Your theory is plausible. Brown-out voltage conditions can be a cause of these problems; but as Wouter pointed out, many computer power supplies can handle both under-voltage (brown-out) and over-voltage (spike, surge) conditions. Unfortunately, most situations are more complicated than the one you provided. Your statement, But as we all know the ...

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What you describe is a failure due to a too low power supply voltage. This is a well-known cause of crashes, which has prompted manufacturers to take action. One action is inherent in the power supply: it has capacitors that will provide the power to survive short power dip. The nature of a modern switch-mode power supply makes it very resilient againt a ...

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As a thought experiment, build a digital system (A/D -> microcontroller -> D/A) and try to program the ideal behavior. It's true that execution speed represents a delay so we could substitute an FPGA. Let's have the delay approach zero, since this is a thought experiment. My guess is there would be some code added to handle some desired behavior (like ...

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Short answer - no. There is no such thing as an amplifier with absolutely zero bias that can produce a perfect integrator. Longer answer. Yes, but you have to be very specific about what your requirements are. You determine what rate of drift is acceptable, you determine what your minimum cutoff frequency for a high pass filter, and you design from ...

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I suppose I know what you mean - and you are right. For an in ideal integrator we require a phase shift of 90 deg between input and output for ALL frequencies (including, for example 0.1 Hz). Because this phase shift requires a 20dB gain roll-off also for low frequencies (down to DC) the corner frequency would be in the vicinity of 0Hz (resp. infinite gain). ...

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This circuit: simulate this circuit – Schematic created using CircuitLab will do exactly what you want. Many times a designer will place a resistor in parallel with C1 to limit the gain at DC. If you don't want this behavior you can simply leave it off. Just be prepared to deal with your output hitting the rails in some circumstances.

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It would be best if you had a LED power supply with the proper output current. Given the low cost of these supplies this would be my preferred solution. However if you have a dimmer I think you could try driving it at 700 mA with 50% dimming, since the average heat dissipation will be the same and the datasheet shows that this LED (LXK8-PW30-0008) has been ...

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Using the definition: $$A = \frac{v_i}{v_0}\mid i_0=0$$ $$B = \frac{v_i}{i_0}\mid v_0=0$$ $$C = \frac{i_i}{v_0}\mid i_0=0$$ $$D = \frac{i_i}{i_0}\mid v_0=0$$ Output voltage is voltage over R $$A: v_0 = \frac{R}{R+jx}v_i \Rightarrow A = \frac{R+jx}{R} \\$$ Input voltage is voltage over jx + voltage over R  B: v_i = i_ijx + (50+1)*i_iR = ...

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No, only the current is moving in the opposite direction. The voltage across is still the same polarity, decreasing to 0.

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Well, if you can live with approximations and a big output impedance, it can be done with a simple resistive network: simulate this circuit – Schematic created using CircuitLab $R_{i1}$ and $R_{i2}$ are the internal resistances of the voltage sources, $R$ and $R_{0}$ are the resistors you must add. The thing work if $R$ is much bigger ...

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Sure there is. For a certain range you can use the same input stage as the op amp solution (voltage averager) and a voltage doubler such as the LM2765 charge pump. $2\cdot ({V1 \over 2} + {V2 \over 2}) = V1 + V2$

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Let us calculate the resistance of the LED at the operating voltage: R = V/I V= 21 v (typical), I = 350 mA, so R = 60 Ohm. The LED driver operates at maximum 24 volt with 700 mA, so it works with maximum 34 Ohm load. Any load with higher resistance will give an unpredictable current result that may damage the load. So you can not simply connect 35 ohm ...

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Start by analyzing the steady state. That means all capacitors are open circuits. The first opamp then just inverts Vinput with respect to ground. The second then inverts it again, but with a gain of 1/2. Therefore Voutput = Vinput / 2. Since both opamps are ideal and have feedback so that they operate in their linear region, the negative inputs of ...

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You don't specify the voltage across the shunt resistor or the accuracy required, but anyway the gain can be stabilized by use of negative feedback such as emitter degeneration. For this to work you need to design the circuit to have excessive gain and remove the excess with negative feedback. For example, if you need a gain of 100 you might design a circuit ...

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Your relay is rated 12V DC, so I am not sure if it will turn on if you use 9V DC. There is nothing stated in its datasheet about minimum voltage to turn on the relay. You may try, but nothing guarranteed. The picture that they give is wrong. The lamp (X) does not have power supply connected to it. It will never turn on. I found this picture and it may help ...

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I don't think this will work without some gain in the system. Note that the circuit pointed out by @geniass uses multiple transistor stages before approaching an LED. Unless you have a very very sensitive ammeter, I don't think it will register the current generated by the raw RF signal.

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Please realize that "virtual ground" does NOT mean "ground (0 volts)". The term "virtual" is just an indication that the voltage at the inverting terminal is so small (Vout/Aol) that it can be neglected (set to 0 volts) during calculation. That means: This voltage is never 0 volts but we are setting it (during analyses) to zero.

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The concept of "virtual ground" implies that the Opamp must be capable of maintaining the "virtual ground". In the situation that you sketch: high frequency, high value of the capacitor, as long as the opamp is able to keep it's - input at ground potential, it will be a virtual ground. So assuming the opamp is ideal: the - input will always be virtually ...

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Just use a voltage divider. You may or may not need a buffer amplifier depending on what it's driving. Also, it may be possible to drive the divider from a digital output pin directly if the division ratio is large enough. 100k and 10 ohms would have an output impedance of around 10 ohms, which may be sufficient without a buffer. The input impedance ...

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Sure, you can just use a CMOS logic chip (gate, buffer whatever) and follow it up by a voltage divider. Say your circuit runs at 5V, you could use a 100K resistor and a 10 ohm resistor. That will give you a 500uV p-p square wave that goes from 0 to 500uV. If you need to offset it by -250uV that can be done in a few different ways. One way is to connect a ...

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The main loop is continually resetting the timer, so the timer will never expire and set the flag. In this case you should only restart the timer after the timer has expired and the timer flag has been set.

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These are the maximum collector current: 2n3906 => 200mA 2n2907 => 600mA 2N2222 => 800mA Try connecting your DC motor directly using power supply and measure the current using amperemeter. Use transistors with collector current rating higher than DC motor needs. For higher current, we need high power transistors and driver transistors. Look at design 18 ...

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I would say: Stage 1: monostable 555 circuit then take the output of that into Stage 2: invert the monostable output and "and" it with the 555 output (DeMorgan to simplify as needed). Then if you want to control the output pulse width you can use the other half of a 556 to make a "one shot" of the desired pulse width. There are some ill-defined details to ...

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Something that goes near to a formal definition can be found in this GoogleBook page. Excerpt: Another way to control the delay of an inverter is to add two extra transistors as shown in Fig.4.15a. This technique is also known as current starving. Lowering Vn and increasing Vp increases the effective drive resistance Req of the inverter and thus ...

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In this context "current-starved" means that the current through the circuit is indeed limited. A "normal" inverter is directly connected to the ground and supply rails. In theory it can draw as much current as it likes. In this design, there's a currentmirror output in series with these connections. When the current that the inverter draws is less that the ...

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MEK (methyl ethyl ketone) will dissolve at least some hot glues, but unsure how it would react with other components (asshat who owned my house previously used hot glue as caulk to seal all the storm windows shut, so did some experimenting)

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Yes No, practically speaking this will not happen because the 3V battery is not an ideal voltage source. It has a non-ideal source impedance, sometimes referred to as its internal resistance, which we can model with a resistor. For example, a CR2032 has an internal resistance of around 15 ohms and a forward voltage of around 3V. If you connect a 2V LED to ...

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The last of the "easy to find, easy to obtain datasheet" was the SMSC USB97C242. As memory sizes rapidly increased it was obsoleted a few years after its 2002 introduction. Closest thing to that you will find today is a dedicated USB to SD card controller. USB2240 comes to mind. Use with commonly found SD Cards. You could use a eMMC NAND memory device ...

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This can be solved using the data you have without any assumption, if (and this is not clear from what you said) the only thing you have to find is the active power provided by the voltage source, i.e. its average power. Let's use phasors and let's assume all quantities are expressed using RMS values ("effective values"). As you may know the complex power ...

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Z1 = R1 + jX1; Z2 = R2 - jX2; I=U/Z -> Z = Z1||Z2 You can't assume that phase displacement is 0, because it isn't. The only situation when phase is 0, is R1=R2 X1=X2.

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The thing that is horribly wrong that your transistors won't start conducting until the voltage at base reaches 1.3V. And that's a lot. You'll need to use either opamp with dual supply or a germanium transistor connected to a silicon one as a Sziklai pair. EDIT: You can stay with your original circuit, just add some bias to the capacitor: connect the upper ...

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In advance, I apologize for not having properly formatted equations. If someone would like to edit and make it look pretty, feel free! Otherwise I am going to make it look pretty in the morning. The simplest way to solve this would be with phasor analysis. Where: and . Let's call the voltage difference over the resistor/cap Vx and the voltage ...

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