New answers tagged

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I assume you want to build a proof-of-concept, rather than an optimised turbine that produces a significant amount of power. From what I understand, a generator behaves more like a current source than a voltage source. Because of this, you can take advantage of the LED's nonlinear voltage-current relationship and dispense with a regulator entirely, as long ...


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Here is a schematic for the case of sensor can't be floating. If input protection needed, see How to protect microcontroler analogue 0-5V input pins from high/negative voltage The voltage V5 (Vreference, offset) and resistor R4 can be changed if needed. Gain set with R2 = 1500 (Vinput V3 from -10V -> 10V, output full span (0 -> 3V).


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I'd just go with the 74HC4050. Simple, no calculations needed, plenty fast for an optical encoder. As with any CMOS, remember to ground any unused inputs.


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Datasheet page 31, part 10.1.1. It says, The fully-differential voltage input of the ADS111x is ideal for connection to differential sources with moderately low source impedance, such as thermocouples and thermistors. Although the ADS111x can read bipolar differential signals, these devices cannot accept negative voltages on either input. And also as you can ...


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ADS1115, 16bit ADC chip accepts input voltage range of 0 - Vcc. It does not accept negative input voltage. To solve this issue, make the input source floated with Vcc/2 as a reference point. But this solution is very basic and the accuracy is very low, depending on the Vcc quality and resistors tolerance. For better performance, get floating reference ...


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The ADC is supplied with positive supply and ground, so it cannot be used to measure negative voltages below ADC ground. The ADC inputs have protection diodes to power supply pins, and as the current is limited by the resistors, the protection diode to ground starts to conduct and that is the reason you have only -0.7 V below ADC ground at the input pin.


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The reason for this is that you can't apply negative voltages to this ADC. Its input voltage must always be 0V or larger. If you apply a negative voltage, the ADC's input protection diodes will start to conduct and clamp the input voltage to around -0.7V. You have to add an offset to the input voltage if you want to measure negative voltages, i.e. by adding ...


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Your biggest concern when using a small motor as a generator won't be getting too much voltage or current. Your problem will be getting enough voltage and current. My son has a small steam engine. Not that model, but one like it. It produces more power than your home made steam turbine is likely to. We built a generator from a small motor and used it to ...


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For just one LED, you can use a current limiter circuit built from 2 transistors. Here's an example (simulate it here): More here: Controlling High Current LEDs with an ATmega328


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To expand on an engineered solution with specs as I hinted in comments. As most know by now show me a good product and it will have great details in the datasheet. Likewise for DIY projects. The better you measure what you want, and tolerate, the better the outcome. Otherwise, it's trial and error (repeat). Light levels Here are some examples of the ...


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Your MCU has built-in ESD protection diodes on the pins, so you could just use those to clamp the voltage. Since they are designed to only protect the device during handling for assembly, you need to put high value resistors in series with the line to limit the current through those diodes and to not overload the driver. Sizing of resistor should be low ...


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Shift encoder output to MCU input voltage range? Your encoder datasheet shows that the output of the encoder is CMOS compatible, so it will be active pullup and pulldown. The easiest way to level shift is to use an N-chan FET, though you must remember that this will invert your signal. Set your Launchpad input pullup resistor on. For the voltage translation ...


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Perhaps calibrating LDR light-versus-resistance is unnecessary... Think of maintaining a constant light level, be it from daylight or LED source. Thus, the LDR will be maintained at a fixed value of resistance. LED will provide all the light in darkness, and no light during daylight. LED provides some light during day/night transitions. This might be ...


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If your optical encoder has a pull-up resistor on the phototransistor collector, adding only one resistor can solve this issue. Here, we can calculate the R as, $$ \frac{R_c}{R + R_c} = \frac{3.3}{5} $$ $$ R = 0.52 \times R_c $$ If your device requires very low power consumption, the external resistor will draw some constant currents. In this case, ...


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Most N-Mos Mosfets need the gate voltage to be 10V higher than the source voltage to be turned on. If you want the source voltage to be 9.6V then the gate voltage must be 19.6V. But your Mosfet has no part number for us to see its Vgs specs.


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I’m not sure what equipment you have available, but a sync-check relay (like this one) would only allow the generator breaker to be closed when the voltage magnitudes and phase positions on both sides of the open breaker are within narrow windows. Get somebody that knows what they’re doing to set it. By adding more complexity you can use a synchronizer. It ...


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N-channel mosfets don't work for high-side switching - you need a high voltage (gate threshold + load voltage) on the gate to open it reliably. Otherwise it is going to have unpredictable results which depend on the specific mosfet you use. You should put your mosfet so that it switches ground to the load, and then your calculations would make sense.


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First, here's calculation of the power consumption on `R1. $$ P_2 = \frac{V_2 ^ 2}{R_2} \times 0.8 $$ To give same power consumption with steady DC current and voltage, $$ P_2 = \frac{V_{2eq}^2}{R_2} $$ $$ V_{2eq} = \sqrt{0.8} \times V_2 = 10.73V $$ If your simulation result still shows 11mV, I am not able to figure it out. I think it should be 11V. Thus, ...


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Most power supplies can be approximated as an ideal voltage generator (U0) in series with an internal resistor r. Let's consider a resistive load R. simulate this circuit – Schematic created using CircuitLab The current will be I=U0/(r+R) So the "usefull" power you get in your resistor is P=RI² = U0²R/(R+r)² Which is maximum for R=r So the ...


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Can we increase the power of the source by decreasing the time of the supply ? Unfortunately, the power of a source is just that but... can we trade off electric power for time? Absolutely we can. If the source has a limited energy then, we can choose to use that energy (within reason) how we like. It's average power that you cannot do much about. For ...


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Here's your circuit, reproduced in CircuitLab. The simulator gives a result of 2.299V at Vout. simulate this circuit – Schematic created using CircuitLab The error in your calculations starts where you "voltage devider formula for the lines PR1 and PR2" - because the "PR1" leg consisting of R1 & R3 is also affected by R5, and ...


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It is not robust to specify a maximum voltage across a zener since it is designed operate in breakdown, and has a very sharp curve. For instance, if you have a '10 V' zener, applying 9.9 V will result in negligible current, and the zener will be ineffective. Applying 10.1V would result in a very large current, possibly over the thermal rating (I*V) of the ...


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Voltage drop depends on where you measure it and your reference. Your question lacks two things which makes it ambiguous. The reference and the test points. Load Regulation % Error is listed in all datasheets for all regulated power supplies. It does not include distribution loss from Ohm's Law with associated cables and connectors. It is measured as ...


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The difference is context. Voltage drop is measured across loads, for example voltage drop across a length of wire. More current will cause more voltage to be dropped across that length of wire. Voltage regulation describes the behaviour of a voltage source, where an applied load causes the output voltage of the source to drop. For example in transformers ...


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Do you have an 'uninterruptible' battery-backed power supply (UPS) to cover generator starting during a real outage? If not, then your monthly testing should include this bad switchover behaviour, as it's what you need to work with during a genuine outage. Usually, a backup system that cannot tolerate an interruption sag will have batteries and an inverter ...


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Critical equipment needs to be on a UPS, otherwise there is no way to maintain stable power when switching to generator after a power outage.


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I have just come across this question and nearly 3 years late but wish to show the asked for equivalence. Labelling the circuit using the [passive sign convention](Passive sign convention). Applying Kirchhoff's voltage law for loop BACDB $$-5 +xR_1+yR_2+10 =0$$ Now let the subscript for currents produced by the 5 volt supply be labelled 5 and those by the ...


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Assuming a purely resistive load. Put your left hand on the the positive terminal and your right hand on the negative terminal of an 800-volt Porsche Taycan battery. Now switch hands. The current flow through your now fibrillating heart went in different directions but the power, either (volts)(amps) or (-volts)(-amps) was the same. We can conclude that in ...


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It's the tension between precise and concise. Given that the study of electricity and electronics is a major branch of theoretical and applied physics, it’s a big challenge to thread that needle with a satisfying description. Meanwhile, any explanation of electricity needs to be tailored to the audience. I’ll get into that more below. So on both counts it's ...


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One or two words is not really enough, if you want to state 'that there is a voltage between these two nodes'. That can be restated as 'if you connect a DMM between these two nodes, you'll get a reading', and 'if you take some charge from one node to the other, work will be done'. This second one is the most fundamental definition, if somewhat ...


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Voltage appears at both ends of wires with no load. But current demanded by a load will propagate like a wave at the speed of light and this time naturally depends on the length of the pair and the speed of light in the dielectric constant between the conductors. This only becomes relevant for very fast risetimes or very long distances.


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A Voltage (electromotive force) develops as electrons are separated. Electrons can be separated by magneto-electro-mechanical (generator) methods or electrochemical (battery) or by beta-particle decay of radioactive ☢️ materials, or tribo-electric methods triboelectric So, you can think of voltage building up as charge is separated. Although voltage is not ...


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You are misreading what it's about. Imagine you have a battery of 9 volts connected to two series resistors of 5 Ω and 4 Ω. Clearly there will be an amp flowing (a side issue) but, of relevance, we can say this: - $$\text{9 volts} = V(R1) + V(R2)$$ Now, if you rearrange that formula into this: - $$\text{9 volts} - V(R1) - V(R2) = 0$$ You have to respect the ...


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For cows voltage is good. The limit is 10k volts I believe. The girl in Texas died on a fence that was well under that voltage because volts aren't dangerous if they are pulsed no matter how high usually unless someone has a pacemaker or something. It is the amperage that kills. That man in Texas I believe is being charged with neglience of some type - not ...


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There are many ways to achieve this in LTspice. What I'll do is step a voltage source from \$0\:\text{V}\$ to \$1\:\text{V}\$, to keep the computation simple. And I'll use the .OP card (not the .TRAN card here.) I can think of other ways, too. Perhaps someone will present a more prosaic approach. But this was easy for me. So here it is: I've arranged for ...


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The card is currently connected to V+ and V- only for taking measurements... I took a few measurements and some values are on the low side, while one of the pins gives me voltage where it shouldn't... 1 = slowly dropping from 7V (instead of 0V) This is expected. Pin 1 is connected to the Base of TR1 via a 10 μF capacitor. If power has been switched off for ...


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You start your question with a false assumption: that something like a 60 W light bulb has a constant power. Instead, it has a constant resistance, and is rated for 60 W wherever it is intended to be used. A resistive load (like an incandescent light bulb) rated for 60W is assigned that rating based on the intended supply voltage. Using Ohm's law we can ...


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Grounding -ve input to Vee=0V disables it. You can use a differential input to Vin+ and Vin- but ONLY if the DC gnds and supplies are floating with respect to each other. Otherwise do not! Also do not put a cap across the Vin- to the rail. That will destabilize the internal feedback and it may oscillate. You can't AC ground the Vin+ and use Vin- either. ...


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It's not exactly a violation of KVL. Consider what happens if the two supply terminals are connected by a resistor with zero resistance, which is what you are doing. Then the current through the resistor will be infinite, and you can use Ohms' Law to say that e = iR becomes 5 = infinity x 0. Well, that is certainly true, in the sense that infinity = 5/0, but ...


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I faced the same problem with an IoT project. I was using a 5 V regular USB power bank regulated "step-down converter" from a 12 V solar panel, but when the voltage of the solar panel reached 4.2 to 4.4 V the power bank froze, I do not know why. I did not want to use an active circuit to solve that problem or use a microcontroller. So I came up ...


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It does not lead to a contradiction in circuits where KVL is still applicable. KVL states that the voltage around a loop shall be zero, and a simple circuit where this model is useful looks like this: simulate this circuit – Schematic created using CircuitLab and indeed, the voltage around the loop is zero - there's a voltage drop of 5 V across the ...


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I agree with Rob. The data sheets for tubes like 6D6, 6A7, and 75 all list 250VDC plate voltage as "Max". Using a VARIAC set for 110VAC into a 1935 Zenith radio the plates all measure either at, or very close to, 250VDC. The radio works well at 110VAC. However, bumping the line voltage up to 120VAC does in-turn increase voltages beyond the tube's ...


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If the frequency is very high and the wires are long enough, we need to consider transmission line effect. When working with high frequency signals, transmission line effects should be analyzed for any length of conductor, not only "long" lines. The potential at different points on the same wire on each side of the resistor will have different ...


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In the AC case (low frequency), the electric potential of the points in the circuit is changing in time. For example, consider a sinusoidal source connected to a resistor. Given that the potential at any chosen point changes with time, do we just pick any arbitrary physical point, like we do in DC circuit and, and assign to it the 0 V even if we know that ...


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SOLVED: I contacted the author of the article mentioned in this similar question and confirmed that they failed to measure noise in 18650 with their instrumentation, so this noise will be at most the same internal noise as the instrument used. So the noise of the 18650 will certainly be less than a LDO with a noise of about 3 uV RMS. In my case, it is better ...


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Figure 1. Image source: Pololu files. The GND pins all have to be connected to supply common. The 5 V pins all have to be connected to supply +5 V. The DIN (data in) pins are all connected together and the data signal is referenced to GND. If I understand correctly, if I want to power them in series, I need 20 x 5 V = 100 V. No. That won't work for ...


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The device comes with simple battery holder which can take two 1.5V cells and connects them in series as there are only two wires out which gives 3V output. For some reason the device uses 1.5V motors so they can't be run at 3V. So by connecting a third wire in the middle of the batteries, you can have both +1.5V and -1.5V in respect to this third wire. It ...


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There doesn't appear to be any schematic for the build, which is unfortunate. The battery holder connects two cells in series, giving you 3 volts: A schematic of this: simulate this circuit – Schematic created using CircuitLab The "third connection" described in the tutorial is having you solder a wire to the conductor between the two ...


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They're picking off the wire that connects the batteries together in series to use as a ground for the motor so you can supply it with + and - 1.5V for a simple reversing circuit. simulate this circuit – Schematic created using CircuitLab


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If they are typical silicon diodes, then they drop about 0.6V when conducting. Other types of diode are available, but silicon ones are commonly used for rectifiers, so that's what I am assuming here. You have declared that the negative terminal to the load is at 0V. Nothing in the circuit diagram indicates that the bottom terminal of the AC supply is also ...


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