# Tag Info

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Best solution - replace the power source to get the desired voltage. If you can't replace the power source for some reason, add another transformer in between. There is another theoretical "hard way" - remove some wire turns from the secondary winding of the existing transformer. However, this is not a good idea for many reasons: may be ...

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simulate this circuit – Schematic created using CircuitLab Figure 1. Equivalent circuits. R1 allows selection of a bias voltage between 0 and -12 V. R2 limits the current from the bias to the next stage of the circuit. Note that your coarse pot wiper resistor is 100 kΩ and the fine is 1 MΩ. If they both feed into the same circuit the fine pot ...

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The picture shows two potentiometers, each of value 100k. The wiper of each potentiometer is connected to a fixed resistor, one of 100k and one of 1M. Each fixed resistor is connected to what looks like a thermistor (the circled things marked 10k.)

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Still the question is not answered. The current entering a resister is actually opposed by the ions in the resistor. The electrons loses the KE and travels further.The positive side of the battery of the higher potential side will attract all the electrons and forces the electrons to travel at the speed of light further. This loss in KE is known as the ...

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For what it's worth, I've just recently been disecting a string of 50 LED lights, advertised as "Constant On", meaning that if you remove one of the LED's in the string, the rest remain lit. Here's what I've found so far: The 50-light string consists of 2 25-light strings in parallel. Inside BOTH the plug and the socket is a small pc board ...

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$R1$ and $R2$ form a voltage divider, that 'looks' like a single resistance to a voltage at the sum point. This is called a Thevenin equivalent, from Thevenin's theorem that tells us that any set of voltages feeding resistors to a summed point can be reduced to a single source voltage connected via a single resistance. More here: https://www.electronics-...

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why are R1 and R2 in parallel? R1 and R2 are not in parallel. However, the impedance provided by R1 and R2 to the input is found by the parallel resistance formula. This is because the power supply has a much lower impedance than either R1 or R2. So, in calculating the input impedance, we treat the (positive) power rail as if it were ground. Then, in our ...

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The average voltage for a half-sine wave is as follows: $V_P=\sqrt{2}V_{RMS}, V_{Avg ~\lambda/2}= 0.45 V_{RMS} = \dfrac{V_P}{\pi}= \dfrac{\sqrt{2} V_{RMS}}{\pi}$ $V_{trig}=90 V = 37.5 \text{% of } 240 V_{rms}, \text{while } T=RC = 64 \text{% of input }(V_{avg})$ thus$V_{trig} \text{cycle time }= \dfrac{37.5}{64} RC$ But due to the 50 Hz ripple it ...

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Steps: Determine the breakdown voltage of the neon tube Calculate when that voltage will be reached at C1 using R1/C1 to calculate the time constant While it looks like R2/R3 form a divider that sets the time constant, before NE1 fires, it is an open circuit making the current through that leg 0, so there is no voltage drop across either resistor until NE1 ...

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The datasheet (with just 100 pages) is indeed not enough to obtain the details you need. You saw that sections 4.6 and 4.7 list general information regarding all operating modes and you also find short notes across other sections (e.g. "should be off during transition"). Search for "S32K1xx MCU Family - Reference Manual" (sorry, could not ...

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STM32F411 has OTG_FS hardware, which has internal pull-ups controlled by the software. Board designs using this uC don't need to provide external ones. I once came across a damaged STM32F407, which somehow got its internal USB pull-up damaged. I had to provide/solder an external pull-up resistor on the board. But this is an extreme situation.

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By definition, they are in parallel if and only if their respective terminals are tied together and nothing else is in the way. There is a more formal definition using topology but it's not very useful. R1 is in parallel with R2: pin A of R1 is directly tied to pin A of R2, pin B of R1 is directly tied to pin B of R2; R1 is not in parallel with R3: between ...

2

There are two parallel circuits in series with each other. One parallel circuit is made up of R1 and R2, the other parallel circuit is R3,R4 and R5. So, the circuit equation for total resistance would be: (1/(1/R1+1/R2)) + (1/(1/R3+1/R4+1/R5))

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I did this exact thing in a design for heat dissipation. But, my design didn't have a fixed voltage as input like yours, rather it had a wide input range.

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The resistors are split for better common-mode transient immunity (CMTI), i.e. for better separation of the two supply domains. The two sides are galvanically isolated, otherwise there would be no reason to use optodiodes (=isolators). If there is any noise (more like disturbances caused by motors, or high current fast circuitry) on one of the supplies, then ...

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The optoisolators serve several purposes: protecting the MCU from ESD: You'd need to design appropriate ESD protection. freeing the potential of the switches, so that if they get shorted to some other potential, no damage is done: Recall that such devices are installed by technicians who are often under a lot of pressure to get it done quickly, and ...

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BOM optimization is a likely candidate. The value 2k is probably used elsewhere and adding another value would mean one more line on the BOM and one more slot in the pick-and-place machine. Another BOM optimization could be availability. For example two 0603 resistors are easier to source than one 0805 (for power handling). You mention an eight resistor in ...

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If either side of the LED is shorted to either supply it will probably survive with two resistors. With only one, it probably won't. If the circuit has to be resistant to "knuckleheads" perhaps the 0.1 cent for the second resistor was deemed justified. Edit: The comment from mkeith is also very much valid. If you keep the number of different parts ...

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Is there any reason why this circuit would need two 2k resistors instead of one 4k resistor? No. I thought maybe it would be for power dissipation but unless I am making a stupid mistake, its only dissipating 1 milliwatt which even small SMD resistors could handle. Subtract the voltage drop of two LEDs from 24V. To be conservative let's say 20V. 20V/4k$\... 1 The cutoff frequency of an RC filter is determined by the total resistance and capacitance at the output. With R = 100 Ω and C = 3.62 μF the cutoff frequency is ~440 Hz. When you put 100 Ω across the capacitor the total (Thevenin equivalent) resistance is now 50 Ω, so the cutoff frequency rises to ~880 Hz. You see less dynamic error because the filter is ... 1 T=RC Case 1 RC = 100 * 3.62 uF f-3dB ~ 440 Hz A(50Hz) = -0.05 dB Case 2 RC = 50 * 3.62 uF . f-3dB ~ 880 Hz thus less Attenuation at 50 Hz. This is not really an anti-aliasing filter and no specs. no sampling rate was given nor BW ripple, band-reject attenuation and SNR from Nyquist noise. You should give a spec for any design: e.g. attenuation error ... 0 Besides picking sane resistance values w.r.t current consumption, you actually want different buttons in the ladder to produce even voltage steps on the ADC, which makes the conversion easier and increases noise immunity. Consider the following voltage ladder: simulate this circuit – Schematic created using CircuitLab Ideally, you would like buttons ... 2 simulate this circuit – Schematic created using CircuitLab Figure 1. (a) Your circuit. (b) The equivalent circuit which makes it clear that you've created a 2:1 potential divider. (c) The Thevenin equivalent has half the signal level feeding in and 50 Ω source resistance. By adding the load you have overloaded the filter to the point of altering its ... 2 On picture voltmeter connected in series with LED and internal resistance. Voltmeter input resistance is high, usually around 1 Mohm. So you restricted LED current and diode drop voltage smaller then normal. Voltmeter shows drop voltage on itself. So the rest voltage 12.6-11.24=1.34V is drop voltage on diode and internal resistor. 6 Your battery, LED and voltmeter are all in parallel. simulate this circuit – Schematic created using CircuitLab Figure 1. You're not measuring the voltage "after" (or "before") the LED. You are measuring the voltage across the LED-resistor combination. In this circuit it is the same as measuring the supply voltage. To measure the ... 1 You need 3 things. Get the datasheet for your thermistor so that you can see the resistance equivalent with their temperature. Choose Fan turn on temperature. Based on the turn on temperature and knowing your gate threshold voltage you can then select the value of R1 that will give you the correct gate voltage when the thermistor resistance fall on that ... 0 It will not light it up to maximum level (very low current due to high internal resistance of the battery), but using a Joule Thief circuit one may get some light out of this diode. It is simple circuit with 5 components, but needs a custom transformer. 4 The answer to this question is just to look in the datasheet. You'll see that this cell is hopelessly wrong for what you want to do. The Panasonic CR2032 states 225 mAh capacity and typical max. load current of 200 uA. So it can't remotely supply the load current. Even if it could, it would be flat in 40 mins. You must search for parts (battery) by their ... 2 If you want a tiny battery with a flat shape that can deliver that kind of power (but not for long) you'll have to use a pouch LiPo. 4 This won't work. You're asking a source of power (battery) that's designed to maybe supply 60 mW to supply 1 W. End of story – you need a different power source, no circuit is able to create power (physics wouldn't allow that). 0 LEDs are "semiconductors" which mean they don't act like resistors. An LED is a diode, and a diode has a complicated formula for predicting the current given some voltage. The complicated formula can be simplified using an exponential function. You can get even simpler to view the relationship as a "hockey stick" line, sort of like the ... 2 Why not look at the LED specs in its datasheet? The absolute max allowed current is 30mA but 20mA is used for its forward voltage that is typically 2.0V but can be 2.6V max. Simple arithmatic shows that at least (2.6V x 3=) 7.8V is needed but one series resistor must limit the current to a maximum of 30mA. 1 If driving a LED with a series resistor, you need a voltage that is above the forward voltage of the LED at your desired operating current by a sufficient margin to account for LED variability. Unfortunately component manufacturers often under-specify their products. In your case they give a graph for the typical forward voltage at different currents, but ... 4 Most of the answers forgot to state the most important thing: You don't talk about the correct voltage supply for LEDs, you talk about supply current. That's the reason for having the LEDs in series instead of parallel! For manufacturing reasons LEDs have a Vf (junction drop voltage) extremely variable. Also being them diodes have an exponential current ... 4 Typically, modern red 5mm LEDs drop about 1.7V-2.1V at 20mA, depending on the chemistry and size of the die. That means that more than 2 of them in series will mean it's difficult to predict the current. If you use individual resistors and connect them in parallel across the 5V, you'd want (for 15mA) about 220 ohms each and the current (using my Vf values) ... 21 Each red LED drops about 1.5V. So three of them takes 4.5V, leaving only 0.5V for the resistors. 0.5 / (47+47+47) = 0.0035A or 3.5mA. That's rather low. If you were to remove two of the 47 ohm resistors, you would get 0.5 / 47 = 0.011A or 11mA. That's brighter, but still less than most LEDs can handle. 9 LEDs are nonlinear devices -- their current vs. voltage curve is exponential with voltage. In practical terms, it means that essentially no current flows at low voltages, and then the current very rapidly rises. In the case of your LED, that "low voltage" is around 1.6V. If you look at the datasheet, figure 2, you'll see this. In fact, that LED ... 4 If you connect them in series, their forward voltages (Vf) add up. Red LEDs have a Vf of about 2.0V at rated current. Three of them in series would yield Vf of 6V - more than your supply. So use two in series at most. Also, you need only 1 dropping resistor. Knowing that 5mm (T-1-3/4) LEDs are usually rated for 20mA, we can set the current as follows: R = (... 5 5mm LED has max current 20 mA. Drop voltage usually 1.5V. You can connect 3 LEDs in series with resistor 47 ohm. 10 mA current is enough. Leave only one resistor. 0 You can only use one limiting resistor for several LEDs only if the circuit is "dynamic indication" type, ie. only one LED is lit at a time. Otherwise the current through the resistor will be divided by the other LEDs. 1 5.1K is what is recommended by the manufacturer (Aosong Guangzhou Electronics Co) under normal conditions. High cable capacitance might require a lower value. 0 In general the LED won't be as bright with 100 Ω as with 50 Ω. Assuming that the circuit has no additional purpose for consuming the current, it will still 'work'. In electronics you will find few exact answers to questions like this. This is because the requirements are not strict, and because the conditions vary. For example, the efficiency of the LED, the ... 0 LEDs are rated for their ‘typical’ current, which ranges between 10 and 20mA. You can however obtain useful brightness at much lower currents, as low as 1 or 2 mA, as the eye is more sensitive to lower intensities. High-efficiency LEDs will perform especially well at these reduced drive currents. Experience: set-top boxes. 1 Question How to design an alert system to do the following: Detect if any one or more of the LEDs below are open or short circuited, Report the bad LEDs, if any, and whwther they are open or shorted. Answer To make things as simple as possible, the OP's specific circuit is abstracted and generalized below. But the OP's specific LED is not going to be ... 3 9V in 10 ohms needs a current of 9V/10 ohms= 900mA. 9V in 100 ohms needs a current of 90mA. A little 9V Name Brand alkaline battery cannot produce 900ma but can produce 90mA if it is brand new but for only a few minutes. Your battery has no name brand so it probably is garbage. 3 Your battery is very, very dead. It has an internal resistance of around 500 ohms. A fresh Panasonic alkaline battery (consumer grade, not industrial) has an internal resistance of less than 4$\Omega\\$.

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Connect the black (COM) lead of your voltmeter to ground. Now, connect the red (+V) lead to each of the test points, TP7301 through TP7303, in turn. If the voltage you read is equal to BG_SUPPLY then the LED is open, if the voltage you read is zero then the LED is shorted. Since each test point corresponds to a single LED, it is clear which of the LEDs has ...

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simulate this circuit – Schematic created using CircuitLab Figure 1. Modified schematic. The "sinewave" is PWM unless it has been filtered in some way (which you don't want in this application). When the PWM goes high (5 V) R1 provides almost 10 mA to the base of Q1. This should be enough to drive it into saturation so that the collector (...

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You guys are awesome! I've decided to go KISS on this one and order the direct replacement https://www.aliexpress.com/item/32905012786.html?spm=a2g0s.9042311.0.0.7f7a4c4dO42Uqh but i did learn a few things with your replies.

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