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Hot answers tagged resistors

30

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 ...

11

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 ...

9

NP = Not Populated. It's an optional component, presumably for an ADC reference voltage in this circuit. R27 and 28 form a voltage divider. C5 provides some de-coupling / filtering to eliminate VS noise. The three components will be added or not as a group.

8

If you are just looking for the mathematics, then it's just a sensitivity analysis. Suppose: \begin{align*} R_{parallel} &= \frac{R_1\,R_2}{R_1+R_2} & R_{series} &= R_1+R_2 \end{align*} Then, for parallel: \begin{align*} \% R_{parallel} &= \frac{\text{d}\,R_{parallel}}{R_{parallel}}\\\\ &= \frac1{R_{parallel}}\left[\frac{R_2^{\:2}}{\...

7

No, commercial off the shelf (COTS) can be used on cubesats, but YMMV. I have designs that have flown on cubesats with COTS parts on them in space in LEO orbits (most of the components were commercial parts from companies like TI and linear devices). However there are issues with using COTS parts in space: They could outgas/release volitiles (which is a ...

6

Your assumption that placing resistors in parallel will always reduce their tolerance is incorrect, in general. This would only be true if you could ensure that both resistors came from a population where the mean value was exactly equal to the specified value. You can not assume that this is the case unless you have measured the resistors yourself. If you ...

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 ...

5

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$\... 5 Initially you can ignore the diode, but once the voltage across the resistor and the capacitor reaches the forward voltage drop of the diode, Vf, it will be effectively clamped and all the current will be shared by the diode and the resistor. No further charging of the capacitor will occur. So run the simulation without the diode and draw a line where the ... 4 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 ... 4 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.) 4 How do I fix this? Add more LEDs in series. The proper answer is to read the datasheet for the LEDs and design your circuit appropriately. You should calculate the required power dissipation in your resistor. The LTL-307EE you show in your circuit has a forward voltage of 2V, so your resistor has to drop 8V. Your current at 100 ohms is 80 mA (the LTL-307EE ... 4 Fix your 200mA scale blown fuse. ( who hasn't blown their DMM fuse yet? hasn't done enough research ) I’m assuming you know which banana plug receptacles to use...for the 200mA scale... To measure current accurately, verify R ohms then in the circuit, measure its voltage drop and apply Ohm's Law. 3 NP means Not Populated. The part is not fitted to the board. The following abbreviations mean the same: NF for Not Fitted DNF for Do Not Fit NF and DNF are more commonplace, certainly in the UK, Europe and US. 3 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 ... 3 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 ... 3 Well, according to Ohm's law we know that: \text{V}_\text{R}\left(t\right)=\text{I}_\text{R}\left(t\right)\cdot\text{R}\tag1 And the power in a resistor is given by: \text{P}_\text{R}\left(t\right)=\text{V}_\text{R}\left(t\right)\cdot\text{I}_\text{R}\left(t\right)\tag2 Combining both equations we can see that: \text{P}_\text{R}\left(t\right)=\text{... 3 The energy stored in the cap =$\frac{1}{2}CV^2$= 100e-6/2*144= 7.2mJ with a half power time constant of$0.5P_{max}=\dfrac{\tau}{2\sqrt{2}}=~0.35\tau\\$ for RC = 10 x 1e-4= 1ms This means a start current of 1.2A, 14.4W and after 350us it is 7.2W in 1ms when the Vcap=64% Vin leaving 36% Voltage across the resistor and the square of that yields 13% ...

2

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 ...

2

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 ...

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.

2

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 ...

2

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.

2

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

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 ...

2

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 ...

2

Since the probability distribution will not be known, I'd like to suggest a different approach. Assuming resistors in a (terrible) uniform distribution between -1% and +1%, this is distribution of two 1 kOhm resistors in parallel: And this is for 1 kOhm in parallel with 10 kOhm: This clearly confirms the intuition (also detailed in jonk's math) that shows ...

2

R3 is not required for the functionality of the optocoupler. It is required by the circuitry connected to the OUTPUT. Since we know nothing bout that circuit, we can only guess. It could be there for isolation, short circuit protection, low pass filtering (with the input capacitance of the load), or it sets the DC current on an LED etc.

2

Three things come to mind. Use a larger resistor. Probably the 100 ohm resistor has 6 or 7 volts across it which is too much current for your LED's probably anyway. They probably want about 20 mA at most. You can try 220 ohms, or 470 or even 1000 ohms. This will make the light dimmer. Use a lower voltage instead of 12 volts. If that is possible in your case....

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