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1

What is the function of the capacitor in this circuit and why is it connected to the diode? My guess would be the diode is half wave rectifying the output of the transformer and charging the capacitor up. The capacitor provides the energy storage need for the racket to do the deed of dispatching the poor insect which has provide the connection to ...


2

There is a better solution for solderless board - "5eBoard": http://www.5eboard.com Eric


2

Use Y-Δ transform and find 7 Ohm. simulate this circuit – Schematic created using CircuitLab After applying the transform technique for R1, R3 and R4, the circuit is as follows: simulate this circuit


2

It looks like irreducible network - network which does not contain series or parallel connections that can be reduced. You can add some voltage source to X and Y terminals and do mesh analysis. It is explained in Wikipedia mesh analysis article, and also in this video by Darryl Morrel so I will not explain it here.


0

In the beginning, arguably the three most important things to remember about an ideal opamp operated without feedback is that it exhibits infinite input-to-output gain, that if the voltage on its non-inverting (+) input goes more positive than the voltage on its non-inverting (-) input, its output will be forced to as close to the positive rail as it can be, ...


4

Your opinion is correct. The LM1117 (like virtually all other linear voltage regulators) cannot instantly accommodate rapid/instantaneous changes in load current - inside there is a feedback system that attempts to keep the output voltage stable at a set point and this mechanism targets accuracy mainly and this means it has a high loop-gain so that slow ...


2

Of course, both resistors act as a voltage divider. However, you most consider the fact that you have TWO voltage sources at the same time (Vin and Vout). Hence, you must apply the superposition rule for calculating the voltage at the midpoint between both resistors: \$V_{n1}=V_{in}\dfrac{R_f}{R_i+R_f}\$ and \$V_{n2}=V_{out}\dfrac{R_i}{R_i+R_f}\$ with ...


0

If I understood your question well, in a voltage divider simulate this circuit – Schematic created using CircuitLab The output voltage Vout=Vin(R1/(R1+R2)). Thus the gain=1/(1+R2/R1)<1. But for your opamp, gain=-Rf/R1, thus gain>1, hence it is an amplifier unlike your voltage divider circuit, the reason why your Op-amp acts as an amplifier ...


1

You can think of this as an algebra problem in disguise. Remember making equations to describe lines on a graph? You've been given two points on a voltage vs. current graph for a linear circuit...


3

Think of a black box that you are told is a Thevenin source. You take two different pairs of current,voltage measurments. How would you solve that? Do you need to know if the inside is made of several sources with a mesh of resistors, or single voltage source with series resistance? How would you tell the difference between these two cases? Hint: much ...


1

I agree with Andy. The voltage in the inductor is always 0 V and therefore iL remains at 0 A. The solution of your instructor doesn't make any sense and it's probably for a different problem.


3

I would say that both answers are wrong. Imagine if the inductor were omitted from the circuit and any source of voltage or current were applied (via the 3 ohm series resistor) to what is basically a balanced bridge. What would be the voltage at the junction of the two 1 ohm resistors - it would be the same voltage as at the junction of the two 3 ohm ...


0

As you've figured out, these boards all do the same thing electrically. The relevant pins of the part are brought out, as well as the manufacturer-recommended grounding and bypass capacitors. (It would be wise to read the datasheet and double-check these yourself.) If you are going to use the ADXL345 on a multiplexed SPI bus, you need an additional logic ...


5

Yes they are part of the filter. The top section relies on R4 and R5 and C1 for shaping the frequency response of the op-amp circuit. The 100k pot has no part to play in the filtering other than guiding the current that reaches the inverting input of the op-amp from either the actual input or the op-amp's output. Maybe try researching Baxandall tone ...


3

A battery is not naturally connected to ground/earth on its negative terminal. If it were then you would make a circuit and the bulb would glow (should it compatible with 12V). AC on the other hand, tends to have its neutral wire grounded at some or several points in the power feed system and this means that a circuit is completed when you ground the bulb.


2

Actually this is a simple series parallel circuit. It is not a charge pump. Whatever AC voltage you put in at Vd will appear at Vl with some attenuation. A charge pump would have two diodes in there somewhere. Part of the confusion seems to come from the way the circuit is drawn. Perhaps it was drawn this way deliberately to obfuscate it and see if you ...


1

The internal resistance of a source can be determined by making two measurements. The voltage without any load and the short-circuit current. Dividing the two one gets the internal resistance. $$ R_{i} = \frac {U_{s}}{I_{s}} $$ In real life it's often not advisable the make a short circuit, instead one makes the measurement with two different resistors and ...


2

It is a simple series/parallel circuit, V1 does nothing if \$V_{D}\$ is ideal, \$C_{1}\$ can be ignored. \$C_{X}\$ parallel \$C_{2}=C_{n}=C_{X}+C_{2}\$ \$C_{N}\$ series \$C_{l} = (C_{N} *C_{l})/(C_{N} +C_{L})=C_{tot}\$ \$Q_{tot}=C*U=C_{tot} *V_{d}\$ \$V_{l}=C_{N}/C_{tot} *V_{d}\$ \$V_{N}=C_{L}/C_{tot} *V_{D}\$ Rest: \$Q=C*U\$


6

When we say an ideal current source has infinite resistance, we are not talking about series resistance, we are talking about shunting resistance. The model of a non-ideal current source looks like this: simulate this circuit – Schematic created using CircuitLab In this diagram I labelled the resistor as "G" instead of "R" to emphasize that ...


2

A perfect theoretical current source will drive it's current through an open circuit (if necessary) by self-adjusting the voltage at its output terminals to create thousands or millions of volts. Of course, in practise this is nonesense but who's talking about the real world? This is why a theoretical current source is an infinite voltage divided by an ...


2

My first assumption is that the intention is for the BJT to be operating in forward active mode (linear response), not saturated mode. More on that later. You cannot thevenize R1 & R2 because the lower end of R2 does not connect to an ac (small signal) ground; nor do you need to thevenize in order to calculate Vce. No, you do not need two loops for ...


0

Just some up the voltages and equate the sum to the supply voltage. Then solve the equation for \$I_e\$. $$ V_{cc} = R_3I_e + V_{be} + R_2I_b + R_1I_e = R_3I_e + V_{be} + R_2\dfrac{I_e}{\beta+1} + R_1I_e\\ V_{cc} - V_{be} = \left[ R_1 + \dfrac{R_2}{\beta+1} + R3 \right] I_e\\ I_e = \dfrac{V_{cc} - V_{be}}{R_1 + \dfrac{R_2}{\beta+1} + R3} $$


2

The answer is yes. My problem lied in the coil; not only was my initial coil diameter too small, but it needed to be a single tapped coil as opposed to two separate coils. Indeed, it works and I get a clear silence on 90Mhz. But the coil I made is extremely hard to control(it's very springy), so the only way I can get it to the right frequency is by ...


0

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.


4

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 ...


0

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).


0

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 ...


3

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.


2

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.


1

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 ...


1

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 ...


4

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.


1

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.


2

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.


1

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 ...


2

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 ...


2

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 ...


1

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 ...


0

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 ...


1

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..)


3

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.


2

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 ...


0

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 ...


1

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 ...


0

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 ...



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