New answers tagged circuit
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DC analysis:
The voltage drop across the diode is 0.7 volts, and the DC current through the diode is only determined by the resistor:
$$IDQ=\frac{7V-0.7V}{4700\Omega}=1.34mA$$
For the AC analysis you can use the differential resistance
$$r_d=\frac{kT}{qIDQ}\approx\frac{0.025}{IDQ}\Omega=18.65\Omega\quad\text{(at room temperature)}$$
So for \$v_D(t)\$ we ...
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That is a protoboard/veroboard/perf board/etc.
Does a board like that require soldering? Is this at all similar to a "breadboard?"
Mostly, yes, you would solder it. You could also use wirewrap, but frankly, that kinda defeats the purpose.
It is similar to a breadboard, except that it has a different layout, and is used for (mostly) permanent or final ...
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These boards go by many names: PC board, prototype board, perf board, etc. Essentially they are a simple way for someone to quickly assemble a prototype circuit without waiting for a custom PCB or etching their own.
They come in numerous shapes and sizes with different types of pad connections. The particular one you link to is designed for use with ...
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The linked product is a generic pre-etched circuit board. It does require soldering to use, although the pads are connected together by traces as shown by the white areas on the top.
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Why does C1 need to be directly connected to the speaker?
Don't forget that C1 is is connected to the collector of Q2 too and this is actually critical to the circuit design. The fact the the speaker is connected to that node to is simply to convert the Q2 collector voltage to sound.
How does the capacitor discharge?
When Q2 is "off", the ...
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CircuitLab allows you to simulate this:
simulate this circuit – Schematic created using CircuitLab
Open the circuit, click on "simulate," and figure out how it works yourself!
Note that I had to simulate the inductance of the speaker (important in this circuit) by inserting a series inductor, as CircuitLab didn't include this in the speaker ...
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This circuit appears to be an oscillator. Likely it is designed to produce some form of siren.
I think C1 is connected in this way to ensure that the base of Q1 can be stimulated by the oscillations that will occur.
Correct.
As the voltage across the speaker increases, the capacitor will release more energy through Q1. This in turn causes the current ...
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Your professors analysis and lecture is very good. Your question above is out of context and as such you've really kneecapped anyone who is going to analyze it without them understanding that this is really just part of an analysis and rationale for how you do small signal analysis. He is taking you through replacing the non-linear elements and ...
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Some ideas for you
First things first - you need to detect how fast the servo is running. This can be done in several ways. Typical solutions are:-
Optical slot (disc with hole (or holes) fixed to shaft, LED/OPTO Transistor combo to pick up rotation - output a pulse every time light passes through hole
Reflective - same sort of idea but pick up from ...
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It's very likely that the unused states transition within one clock to a valid state; you should check this, but it looks like it to me.
In general, to make sure you get the behavior you want, you should include the unused states in the state diagram up front, and design the system so that they eventually lead to valid states under all circumstances.
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This does not make sense to me
Go back to the time domain and see that, for the real part of the load impedance, power is always absorbed. The average power is, thus, non-zero.
However, for the imaginary (reactive) part of the impedance, power is alternately absorbed and then delivered (the power alternates from positive to negative). For the ...
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Note that the LMC6772 chip you are using is not an opamp; it's a comparator that has an open-drain output. Since it can't actively drive its output high, the Schmitt trigger circuit won't work as designed.
The datasheet says, "Refer to the LMC6762 datasheet for a push-pull
output stage version of this device."
However, look at how you have the optocoupler ...
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The comparator you use has open drain output. This means the output of the comparator can only be actively pulled low. The calculations you use are for a push/pull output.
A possible solution is to increase the values of your resistors by a factor 100 and add a 1k pull up from comparator output to Vcc.
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In many applications, it is reasonable to assume that under any plausible fault conditions breaking one particular power supply connection will safely interrupt any fault currents. In such applications, a single-pole breaker is appropriate. In some other applications, however, there may exist plausible fault conditions in which either supply connection ...
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Using a micro controller seems like a wast for just a simple application.
My natural inclination is to agree. But there are other points of view. Here's an example of an equally simple application using a relatively simple microcontroller of the same family as that in your Arduino.
From robotroom.com dual fan control
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I'm assuming you are using brushed DC motors
Please correct me if I'm wrong but it sounds like you need to reverse each DC motor independently to give you the basic functionality you need so here is a basic DC motor reverse switch: -
Shown is a DPDT (double pole double throw) switch. It has two independent switches that mechanically activate together; ...
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It's embarrassing but I found the solution. Right click, add attribute... what I referred earlier was just the preview of attributes. Either way, I solved. Thank you for your kind help! – alkopop79 just now edit
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It sounds like you are not managing to select anything other than the default attribute name. Click the dropdown arrow for the "Name" field where it defaults to 'netname' and select 'value' from the list. Then enter your value in the large "value" text area. Then select the visible button and choose name and value or not as you need for this attribute (value ...
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The verbal description of the circuit sounds like this:
simulate this circuit – Schematic created using CircuitLab
Unfortunately, this circuit is unsolvable. VCVS1, configured with gain 1, is asked reproduce the voltage present across R1, while simultaneously being forced to 12V plus the voltage across R1. VCVS1 needs non-unity gain in order ...
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It's a big read and I hope I understand your basic question - maybe you haven't enabled the opto output on pin 7? Does this explain it: -
Look at the first two rows - A high in gives a low out and vice versa BUT only when enable is high. It does say there is an internal pull-up in the data sheet but try it and see.
Alternatively, the voltage you may be ...
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The diagrams mentioned by other answers on this page where a resistor is used can distort your signal significantly. The resistor will slow down the charging rate of the FET gates which will smooth out the leading edge of your inverted signal. Using an actual inverter or 555 inverter "squares up" the wave to keep nice sharp/hard edges.
A solution worth ...
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The burden voltage of a current meter isn't necessarily measured. A current meter essentially measures the voltage buildup due to a current through a known resistor. It is the voltage drop over the multimeter when measuring current.
The (shunt) resistor in the current meter is a know and static value, so if you know what current flows through it, you can ...
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If...
If the meter is 10Mohm then this implies 160kohm through wire??
If the complete situation is as you described, then that is exactly what it means.
However...
The input impedance of the multimeter is not constant with respect to input voltage and is not exactly 10.000000 Megaohms. Your resistance assessment is sensitive to both voltage ...
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For starters: move R3 from pin 8 to pin 7 and connect pin 8 directly to 12VDC.
With the current configuration pin 7 is tied to Vcc and it prevents the 555 from oscillating.
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Test the parts separately.
1) Pull the 555 pin 4 to +V with a 1 kilohm resistor. Does the noise start? If not, investigate the 555 in more detail. But assuming yes:
2) Pull the sensor transistor base to 0V (likewise). Does it sound? If not, check the transistor and its connections. If yes:
3) Measure Q1 base voltage at different temperatures. If it ...
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This might not be exactly right, but because the comparator has an open-drain output, you have to think of it as being something resembling the device surrounded by the dashed box:
simulate this circuit – Schematic created using CircuitLab
When M1 is off (high output), the voltage at OUT is determined by the R1/R2 voltage divider. An open drain ...
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It's the simulator's way of calculating, and we call this the transient effect. Remember, the simulator is based on linear approximations. If you can actually look at the source code of the simulator, its probably dividing some value by a very small number (i.e. 1/0.5). Think about your capacitor equation (1/jwC and C*(dV/dt) ) and try to think what ...
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Sounds like the back-EMF from the solenoid is disrupting the micro. There are a few potential issues / solutions here;
1st, read up on snubber diodes/protection used with relays/solenoids, it's been covered many times here on all the hundreds of variations on "how do I switch a relay from my *duino?" questions.
2nd, look at the smoothing and grounding of ...
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CircuitLab solves the circuit because it doesn't simulate effects like junction temperatures reaching beyond the limit, so that semiconductors melt.
A diode is not a fixed voltage drop. Current through a diode is related to voltage by an exponential equation. That exponential equation goes on forever: for any imaginable voltage, you can find a current. ...
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First step is filling a Karnaugh map. The first one for little bit help filling the map:
$$
\begin{array}{c|c|c|c|c|}
\text{I1}\cdot\text{I2}\cdot\text{I3} & \text{I1}\cdot\text{I2} & \overline{\text{I1}}\cdot\text{I2} & \overline{\text{I1}}\cdot\overline{\text{I2}} & \text{I1}\cdot\overline{\text{I2}} \\
\hline
\text{I3} & ...
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Same deal as Camil Stap's answer, but using a single 74xx00 series NAND chip (4 gates):
simulate this circuit – Schematic created using CircuitLab
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Your output follows this formula:
output = (I1 & I3) | (!I1 & I2)
A circuit would look like:
simulate this circuit – Schematic created using CircuitLab
For logic gates, you can use the 7400 series:
AND: 7408, e.g.
OR: 7432, e.g.
NOT: 74LS04, e.g.
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The circuit as shown is not viable - or you could analyze it in two phases, if you must:
Phase 1:
Each 1n4148 diode is rated for 200 mA continuous, 450 mA peak repetitive current.
When wired as indicated, each diode will drop approximately 1 to 1.5 Volts (Fig.3 in datasheet) before the current exceeds absolute maximum rating
As the supply voltage is 5 ...
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Calculate the peak current flowing through L1. If you are switching at a very low frequency that peak current tends to become V1 / R1. If you are using a reasonable switching frequency, the peak current in the inductor is mainly determined by the inductor.
When the inductor is open circuited that current has to flow somewhere and this is where the diode ...
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As of Version 6, the Eagle layout and routing program has been using XML for all of its files, including schematics and board layouts. You can download a freeware version here, generate a schematic and look at the corresponding .sch file, which will be in XML.
Prior to using XML, their files were all in a proprietary binary format. Now it is possible for ...
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Hold off for a while in case anyone knows different; but I have never seen such a format. The Electronics industry had its own common schema (EDIF) from the 1970s, which looked LISP-like for easy parsing, rather than HTML-like for ... I dunno, fashion?
But even so, netlists are frequently expressed in much simpler ad-hoc formats.
You could adopt a schema ...
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In order to design the circuit you consider two conditions. The first condition is the DC analysis. This is where you are mainly interested in setting the gate voltage so that the drain voltage is roughly at midpoint on the supply. In your example, midpoint is somewhere close to 6V because you have a 12V supply.
Why is it 6V? When you have 6V DC on the ...
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This is called small signal analysis and is predicated upon an assumption that your small signal doesn't disturb the operating point too much so that the assumptions still hold.
What is key is that that you use the bias points to determine the \$g_m\$ (which will depends upon \$I_{DS}\$ which depends upon \$V_{gs}\$ etc.).
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I recommend to have enough large and high-quality breadboard in the very beginning -- even though s/he is planning to keep things small. The thing is that you may need to create your own 5-12 converter that is the size of your small breadboard, not funny -- things I had to do when I started and then I had to wait long time for a better breadboard and/or ...
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