New answers tagged voltage
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I was going to post a comment(especially since it's an old question), but it became too long.
That Application Note National AN-1852 (Which it located at TI now), describes in large detail the reasons for the inclusion of the op-amp to begin with.
It provides two totally different services to the circuit.
First it provides a low impedance 512mV bias to ...
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As others have noted, there's no problem as such with having both sides of an LED at +9V. It'll just be off, just as if both sides were at 0V. I do see a couple of other potential issues with your circuit, however:
Your voltage notation is confusing. Assuming that the triangle symbols in your circuit represent ground, which is at 0V by definition, all ...
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Assuming a buck converter that's 100% efficient, yes, you will be able to draw 1A from the output.
In practice, buck converters can be very efficient, often over 90%, so you'll still be able to get around 900mA output, which is rather better than the 500mA you'd get with a linear regulator.
An important part of the buck converter system is its input filter ...
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It's been a while and no one has decided to close your question, so I'l write something short.
I think you are talking about this schematic:
simulate this circuit – Schematic created using CircuitLab
With KVL, starting at the lower left corner (at the ground symbol) and working clockwise, I get:
$$0\:\text{V} + 5\:\text{V}-I\cdot R_1-V_{Z_1}=0\:\...
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No, you don't need a relay. There is never actually any reverse voltage across the "Ready to Arm" LED. It goes off because there is +9V on both ends of it, for a net voltage difference of 0V.
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The classic 4-resistor difference amplifier will do what you want:
simulate this circuit – Schematic created using CircuitLab
This circuit keeps the difference between Vout and Vref equal to the difference between V+ and V-, effectively adding Vref to V+ if V- is grounded.
The opamp tries to keep its two input terminals at the same potential using ...
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The total resistance for 200 mA at 20V is 100 Ohm. The wires should be quite long or thin.
But as answer: I guess you want to make some kind of heating. If you plan PWM control, high enough switching frequency should quarantee the temperature of the wire stays in certain range, no matter the current is alternating
being part of the time 0 mA and the rest ...
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Based strictly on the graph you provided:
The curve intersects 1.2V at about 16 hours, by my reading, and hits zero at about 25.5 hours. That's about 63% of the lifetime, or 63% of the coulombs. Without trying to calculate area under the curve exactly, we can observe that it's monotone decreasing, and therefore the joules available before 1.2V are more than ...
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It is hard to answer your exact question, since you want information on a scenario that has no commercial relevance, so companies are not going to test almost drained batteries just to know how much energy can be harvested from them.
Those discharge curves they publish are used to design battery-powered products: the curves allow to estimate how long a ...
3
Generally using a nearly-flat AA/AAA battery with a boost converter is not going to give you much, especially if the thing you're powering requires substantial power.
When the battery goes down, not only its voltage decreases, it's internal resistance increases as well, sometimes dramatically. So the battery may show, say, 1.1V when measured with a ...
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AWG 8 cable has 2.06Ω/km, so using 5m of it (or 0.005km) would be equivalent to 0.01Ω.
With a 12A current this equates to 0.12V of drop per cable, and 0.24V with both sides of the cable.
However, connectors can also add a significant amount of resistance, make sure that you are using the right connectors and they are terminated properly.
0.6V at 12A ...
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Remember that the "discharge" pin (pin 7) is essentially a copy of the output pin, except that it is open-collector — it can only sink current, not source it. But you could use it to control your LED without affecting the rest of your circuit:
simulate this circuit – Schematic created using CircuitLab
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For a quick and simple solution, (if you don't mind an inversion of the LED function), you could just add a pull up resistor to +5V then connect the LED between that resistor and the 555's output pin. For a more modest LED current use a resistor of 270 ohms or more.
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From the NE555 datasheet:
we can see that with a 5 V supply we can expect around 3.3 V at the output when loaded with 100 mA. Your LED of course consumes less current but that will not increase the voltage by much. So the behavior you see is to be expected.
I expect that even without the LED the voltage at the output will not reach 5 V! That's a ...
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Calculate the resistance of each based on their rated power and voltage. P = V^2/R. e.g. the first one is rated 60W, 220V
Once you know the resistance of each then you can work the resulting series circuit when the 380V is applied across the string.
Then, P = I^2R for each.
simulate this circuit – Schematic created using CircuitLab
simulate this ...
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Remember that for an NMOS transistor to conduct you must bring the gate voltage above the source voltage, and we usually expect the difference to be at least equal to the transistor's threshold voltage if want significant current.
The sources of your two NMOS transistors are connected to the NAND gate output, so the NAND gate output (the sources) must be at ...
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Most explanations disregard reactive power as "useless" power or power that we don't want to pay for.
Such explanations are wrong. Sources of such information should probably be avoided.
"Reactive power" is a common term for reactive volt-amperes (VARs). Reactive volt-amperes have one very important use, they provide the magnetic fields for every ...
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Current through an inductor causes a 90° phase-shifted voltage over it. The more current, the more voltage, the more reactive power.
In contrary to a resistor, neither the reactive power nor the induced voltage is lost to the circuit as heat, so you can do neat things with it. The simplest example are autotransformers. Look that up. A more complicated ...
1
We can design the heat-removal from the Zener. Shall we do so?
We'll assume the leads of the Zener are copper, and are 1mm square. Yes, they likely are
round, but I'll let you insert a square-to-round correction factor. Copper, in the default thickness of PCB foil, which at 1 ounce/squareFoot is 1.4 mils or 35 microns, has
thermal resistance of 70 degree ...
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Probably the simplest solution would be a zener diode in series with the load, chosen to drop just the amount of voltage you want dropped (not the voltage you want to get). No resistor is required. In your case, a zener of around 11V would do. The zener will need an appropriate wattage rating, as it will be dropping up to about 1.3W.
simulate this ...
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You're using a linear regulator which simply "burns off" the excess voltage.
The current does not change and remains the same so you can draw up to 1.25 A at the output of the regulator. So after the regulator you're limited to 5 V, 1.25 A so 6.25 W.
When you draw that 6.25 W there is 12 V - 5 V = 7 V at 1.25 A meaning 7 V * 1.25 A = 8.75 W dissipated in ...
3
You are able to use a fixed voltage drop of about 11 Vdc at about 120 mA. This is fairly easy.
simulate this circuit – Schematic created using CircuitLab
The transistor is a Darlington device in a TO-220 package and has a reasonable gain of greater than 1000. The Vbe drop is about 1.2V. Choose the appropriate Zener diode for the desired voltage ...
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In a linear voltage regulator as the LM117, all the voltage drop × current is turned into heat. That's about 9W in you case. You can draw 1.25A@5V from the 5V output.
If you wanted to draw more current on the 5V side than it is supplied on the 12V side and produce less heat, you had to use a switching regulator. There are some manufacturers which produce ...
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The LEDs at the end of the strip tend to not follow suite because of the 5V drop across a 5m Strip. It would appear they are not receiving the proper data. I connect both ends of the strip to the same PSU. Also If using an Arduino or R Pi you can limit the amount of current the LEDs will use. You'll find it in the library you are using or on the spec ...
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The ground connection MUST be common to all device involved*.
Without a continuous circuit there can be no current flow.
The ULN200X is effectively just a group of Darlington open collector transistors in a package. You can "safely enough" drive each section with a correct voltage input (varies with the X in 200X) and then drive outputs that the Darlington ...
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You can drive the ULN2003a with 3.3V outputs- they don’t put any voltage onto the driving circuit.
Of course if you mess up the circuit or the grounding your chances of ruining the Pi are much higher than if you use 3.3V only.
You need series resistors for the LEDs and you should make sure that the maximum total current can be supplied by the 5V supply. ...
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Although the negative resistance is veiled in mystery, in fact it is a quite simple concept. It can be easily explained by analyzing the voltage drops across resistances.
The positive resistor subtracts its voltage drop from the input voltage thus decreasing the current while the (S-shaped) negative resistor adds its voltage drop to the input voltage thus ...
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This is a variation on Dave Tweed's circuit. If you use a BJT you can get much lower voltage loss in the common-base (or -gate) amplifier -- like less than one volt.
You probably want to look around for a "super beta" transistor, or maybe a Darlington, and you want to be careful about supplying adequate base voltage -- the 0.7V that we all think about ...
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As @PeterJennings said, the power dissipated by the polyswitch when it is conducting is a function of current. The polyswitch has its own resistance, which works against the current to generate a voltage, and, hence, heat. The switch then trips when it gets hot.
The 240V part of the rating is the voltage the switch can withstand when it is off. At a ...
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The quick answer is that the power is not dissipated in the fuse but in the load. If, for sake of argument, the fuse has an internal resistance of 0.1 ohm then at it's full rated current it will be dissipating I^2R W = 1.6W and experience a voltage drop of IR = 0.4V. The rest of the voltage, be it 15.6V or 239.6V is across the load it is protecting.
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Well, first of all we are trying to analyze the following circuit:
simulate this circuit – Schematic created using CircuitLab
Finding first \$\text{R}_\text{th}\$ we need to look at:
simulate this circuit
Now, we see that:
$$\text{R}_\text{th}=\frac{\text{R}_4\cdot\left(\frac{\text{R}_1\text{R}_2}{\text{R}_1+\text{R}_2}+\text{R}_3\right)}{\text{R}...
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Your \$I_{sc}\$ is the current produced by the source \$V_S\$ when the output is shorted.
It isn't the current that goes through the short.
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Assuming that the only problem is the voltage — the DRV_OUT can otherwise handle the solenoid current, you could use the big brother of the common logic level conversion circuit:
simulate this circuit – Schematic created using CircuitLab
Otherwise, you'll have to use your 2-transistor circuit — but be sure to add a pullup to Q101's gate, ...
3
There's no reason to drive the GPIO pins high at all. Just configure them as open-drain — drive the pin low when you want to activate the corresponding "button", and tristate it otherwise.
As long as the fob voltage is not higher than the MCU voltage, it will be fine.
And running the fob at 3.3 V should be fine, too.
1
Since DC motors are also generators, the reverse voltage raises proportional with the RPM. So the higher the RPM, the higher the reverse voltage -> less current. Also the ambient temperature is most probably not stable, which leads to fluctuating current measurement results.
5
The voltage is zero after the resistor because you have defined it to be zero -- it is the circuit common (or ground).
If you are talking about the RLC circuit on the front page, it is because the L and C are energy storage elements that are charged by the voltage source before the start of the simulation. The simulation begins at the opening of the switch. ...
0
The situation is very simple:
Because we know that the voltage at point A is 7 volts higher than the voltage at point C. And also That the voltage at point B is 1.8V higher than the voltage at point C.
Therefore we can easily find the V_AB voltage.
Can you do it?
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Short Answer: Yes, you can use the comparator with a voltage divider. Just make sure you choose an appropriate reference voltage and hysteresis level. However, it may be better to use an appropriate charge controller.
You should not use a GPIO input in place of a comparator. Not only is there no way to determine the exact threshold voltage, it will draw an ...
2
Using a battery that is not brand new so its loaded voltage is 8.0V and using a 3.2V blue LED, then the current is (8V - 3.2V)/220 ohms= 21.8mA. No problem when most 5mm blue LEDs have a maximum continuous current of 30mA.
2
Here is how this works -----
simulate this circuit – Schematic created using CircuitLab
I picked a 180 ohm resistor, because at 20 ohms/volt we are guaranteed a
50 milliamp intersection at the left axis. Thus we scale 20 ohms/volt by 9 volts, finding a 180 ohm resistor provides that current.
Note other LEDs will have lower forward voltage, moving ...
0
Firstly, I would recommend not using a voltage divider to get your different output voltages. You should look into using stacked zener diodes with a current limiting resistor on the top of them. This would only be good too if the different voltages are going into a high impedance load as well. If you you don't have a high impedance load with your current ...
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Now that we've gotten your setup explained in the comments, we can explain your problems.
Switching power supplies and breadboards don't mix. Breadboards have parasitic capacitance between pin rows, and parasitic inductance in the rows - not to mention in the wires. The contacts are also often weak and don't hold well - the resistance varies just by ...
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KVL is a law, not a suggestion. Trust Kirchhoff, trust your instincts. If you are getting a different answer using mesh analysis then we should take a look at your equations.
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The V1P8DIG and V1P8ANA provide 1.8V to digital and analog sections respectively. They are internal LDO's that drop the main VCC to around 1.8V, which the IC needs to function. If you already have 1.8V then the LDO's don't work, so you'll bypass the LDO's. My guess is the LDO's are provided because most people use 3.3V for Vcc, you would be less likely to ...
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In general, a 60 amp branch circuit should be constructed using components; conductors, connectors, and outlet devices that will be protected by a 60 amp circuit breaker of a type that is suitable for that service. All branch circuit components should be certified by an independent testing laboratory in that regard. The standards that apply to the ...
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It means that Vos is referenced to the ground pin which usually doesn't need to be defined. The Vdmk signal is not referenced to ground, it's referenced to the OS signal (which I think is also Vos). This could also mean that both are a sum of some kind. Either way the Vos/OS signal goes from 1.5 to 3.5V
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In general, set up your KVL and KCL equations with the assumptions that quantities have a "higher" voltage on the "+" end of the resistor or voltage source and that currents are positive in the direction of the arrow. By being consistent with this, your units should always make sense (positive voltages across resistors, etc) as long as you did your systems ...
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Because you get a negative \$V_O\$ value which means that the voltage at the right side of the top resistor is higher by 11.9 volts than the left side.
This also means that in "reality" you have this situation:
And you can find the voltage across the current source do the KVL.
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Your first question is a bit unclear. The marking for \$V_O\$ means that there is a voltage named "\$V_0\$" that is the voltage across the resistor, and it is assumed to have a higher voltage on the left side of the resistor.
For the second question, I assume that you mean the polarity of the voltage across the current source. Whenever you are looking for a ...
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The stall current will vary according to \$I_{stall}=\dfrac{V_{supply}}{R_{terminal}}\$, but the nominal current is unchanged, as it is the current that can continuously flow through the winding without creating excess heat.
When changing voltage supply, the affected parameters are:
no load speed
nominal speed
no load current
All others remain the same.
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