11

This is an acceptable way to equally share the reverse voltage in a string so that -15V may be shared equally to -5x3 when AC is applied within this range of <= 15Vp Or multiples of this in larger voltages. How did they select this value? Red LEDs are often rated at 1uA leakage and White 10uA due to the dielectric with reverse voltage -5V max. Thus this ...


6

Those are called pull-down resistors. They make sure the signals are 0V when the switches are open. Without the pull-down resistors, the inputs are floating, meaning that you have no control over their voltage. They can be pretty much whatever, and in practice their voltage will vary with the circumstances. Some RF disturbance or you touching the circuit or ...


4

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.


3

Thanks to the comments on my post I found my answer. The transistor needed to be wired to the GND end of the relays rather than the 5v side.


3

Briefly, there are two reasons: there's more to resistors than current limiting, current limiting is still needed around integrated circuits. In the digital world, you'll often encounter pull-up and pull-down resistors. These serve to set a "default" voltage for a pin or bus, when the chip is not actively driving it. This is the case for the ...


3

Saying resistive, capacitive, or inductive just means which behaviour is dominant. It's not a precise term and one of those things students might try too hard to ascribe precise formal definition to when it is inappropriate. In my opinion it is rather unimportant. Things are what they are with varying shades of grey. So just combine Xc and XL since they ...


3

If this is from the aliexpress 500 resistor kit, you read them with a multimeter. First, all the colors are some shade of brown, and they don't have the proper offset so you don't know which side is which. But even if you try both values, and somehow can tell the colors, you might notice both options are still not what the multimeter says. That explains why ...


2

Five band resistors with a fourth band of gold or silver form an exception and are used on specialized and older resistors. The first two bands represent the significant digits, the 3rd is the multiplication factor, the 4th is the tolerance, and the 5th is the temperature coefficient (ppm/˚C). The third band is the multiplier so it is just grey = 8 , red = ...


2

If you ignore the green band, it all makes sense- 0.82\$\Omega\$ 5%. The final green band may indicate something like non-inductive, which would be sensible for a low value current sense resistor. It does not appear to be a standardized code. It could also be fusible. If the current sense resistor is blown up, there is likely some other serious issues such ...


2

I’d say it’s to ensure that the LEDs switch off cleanly; without the resistors the LEDs could continue to glow dimly for several seconds after power is switched off because of residual charge in the driver; this can be annoying in a dark room and can be observed even with very tiny currents. As an aside, the datasheets for many white LEDs warn against ...


2

The input would read high if not for the resistors. If you put the switches to GND rather than Vcc the circuit would work (with the switch logic reversed, obviously), however it's not ideal because the open inputs will be a bit sensitive to EMI. It would be fine on a breadboard demonstration with switches. The internal circuit of the 7402 (from the datasheet)...


2

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, ...


2

The resistors are the R part of the Ohm's law. You should really start with it, because your assumption that "they restrain the flow of current to protect the components" isn't a complete "picture". Now when you are asking to describe ADS1115 board, instead of submitting a picture of it, you can searched for a schematic instead. Because ...


2

There are too many reasons why your circuit fails. Lack of low impedance current loop. Lack of differential bias threshold for each Vbe = 0.7V (consider 0.6 your bridge reference +/-0.1 for gradual ON/OFF) This design I made solves both but is still weak due to the relatively low resistance of the bulb compared to the battery ESR for a Panasonic Alkaline ...


2

Your circuits are better described as logic indicators rather than gates, since their output is light from an LED rather than a logic level. In this spirit here is an 'XNOR' indicator:- simulate this circuit – Schematic created using CircuitLab R3 and R4 produce a fixed reference of 2.5 V. When the two inputs are different the junction of R1 and R2 is ...


1

There are two concepts to learn here. The LED internal junction voltage is sensitive to heat so a 3W heat source to a 3W heatsink might rise say 60'C on the internal junction if done right then the internal voltage lowers and it draws more current if the source is constant. (.e.g. -4 mV/'C rise x 40'C = -240 mV per LED) Then the internal LEDs also have bulk ...


1

First, I will present a method that uses Mathematica to solve this problem. I know that this approach is not 'smart' but this method will work all the time, even when the circuit is way complicated than this one. Well, we are trying to analyze the following circuit: simulate this circuit – Schematic created using CircuitLab When we use and apply KCL, ...


1

Your transistor is an emitter follower. A BJT is not controlled by the base terminal. It is controlled by the base terminal and emitter terminal relative to each other. BAT2 is referenced to GND and therefore applies voltage to the base relative to ground. However, the base relative to the emitter is what the transistor cares about. It does not know or care ...


1

Concerns: 3 pin vs 2 pin the 3rd pin is a dummy plastic pin. The supply is double insulated. 3V to 4.5V their specs are vague but the 0.05A current and power consumption will reduce with voltage but also increase lifespan somewhat. I read they are rated for only 200h continuously. (due to vibrator contact arc erosion, I expect) However if 4.5V* 0.05A ...


1

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.


1

The manufacturer generally specifies a footprint (or multiple) for their parts. In the case of your 0.5W resistor, it's the standard reflow soldering footprint for the 1812 package size. The resistor will be able to withstand the rated power indefinitely with the manufacturer recommended PCB footprint and an ambient temperature of 70°C (as specified in the ...


1

If you have a 5 V DC power supply connected to a 1Ω resistor and a coil with negligible resistance in series, then you will draw 5 A, and dissipate 25 W in the resistor. You will need a 1Ω resistor that's capable of dissipating 25 watts. Such a resistor will be physically much, much larger than the typical 1/4 watt resistors you use on circuit boards. In ...


1

Assuming it's just an LED with a series resistor for 5 to 12V, then the suggested resistor might work. But it will flicker very noticeably. But not all LEDs can handle the reverse voltage when the AC voltage goes negative. If it can't, then the LED may fail instantly. It would be much better to use a bridge rectifier as well as the resistor.


1

The construction of this rotary switch reminds me of the switches in a decade resistance box. The resistors are most likely wire-wound resistors. They may have been specially wound or even trimmed to the exact value required. Given that this instrument operates on AC signals in the audio frequency range, it's likely that these are non-inductively wound ...


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