# Tag Info

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The power rating is a combination of higher tolerance , lower tempco and higher max operating temp. Thus these variables define the Pmax. I found that% tolerance is the biggest factor due to max temp R error. The exceptions increase cost dramatically for high temp, low tolerance error, low temp. coefficient. They are not all the same. So you may find 1/8W ...

0

It's simple, the through hole resistors have the leads connected to the board, the smd resistors have pads which are directly in contact with the body of the resistor and the body itself is pretty much in contact with the board. That allows them to dissipate heat by transmission way more efficiently, plus when they are black they also dissipate by radiation ...

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SMD 1206 defines the flat/projected size of the component / in other words, it defines the footprint. But they do have differences when their wattage is different. How much heat the component can handle not only depends on size, it also depends on other factors: component matérials used, board layout, ... . I could find SMD 1206 resistors that can handle ...

1

Can “this resistor can handle inrush current from a ceiling fan motor?” maybe NOT. When you choose a smaller speed, with a smaller capacitor, the motor appears much lower resistive since lower back EMF AC Voltage is lower, the motor stall/starting current can up to 10 x max rated current or 100x power drawn thru the same resistor. This power drops ...

1

You read the data sheet. The maximum steady-state current depends on the resistor rating and (at least above some minimum) the ambient temperature. Surge ratings should be in the data sheet. Fusible resistors are designed to fail “open” in a fairly predictable manner, without undue flame, smoke and other drama. I would think that a fusible wire wound ...

3

Soldering surface-mount components is tricky, and requires the proper tools and experience. Indiscriminate use of a heat-gun is more likely to destroy the mother-board than to fix the issue. Lumps of solder or burned traces may also make repair difficult or impossible. The best solution would be to find an experienced technician with the proper equipment, ...

3

A MOSFET gate looks like a capacitor. The resistor will slow down the MOSFET turn-on and turn-off, and it would protect a delicate driver pin (unlikely, but possible), but in the end the thing will still switch. In that circuit the resistor is to slow down the MOSFET switching, to reduce the speed that the drain voltage changes, which in turn reduces EMI ...

0

Sorry for the answer to an old question, but it could use some additions. While Andy is technically correct, it is in fact recommended that current is measured as OP shows in 2nd picture, ie from one sensor, NOT between them. The reason is twofold: the current only flows through the resistors, so if you measure between them you do not minimise the ...

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simulate this circuit – Schematic created using CircuitLab This should operate about 350 mA for a 1W LED (initially) and then decay towards 10% of this after 24 hrs. However AAA Alkaline batteries are expected to decay in < 2hrs at this rate and the product page indicates 24 hrs of service ( including dim service) LED's of this type are rated ...

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This diagram is incomplete as it is missing bypass capacitors. Typically they would be 100 nF to 10 uF. The 1K resistor limits charge current to just 12 mA max, much less if battery is close to full charge. This would be an unregulated 'trickle' charger. The diodes are used for reverse polarity protection.

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The two diodes (D2 & D3) work to OR the voltages, the idea being your 12V_IN will be slightly higher than 12V_BATTERY, so current will flow from 12V_IN to power the device when its present... Minus a diode drop. When 12V_IN is removed current will flow from 12V_BATTERY through D3 R1 is probably there to act as a current limiting resistor for charging ...

2

You are trying to create an adjustable voltage power supply? I'm sorry to say that this design will not work. As you read, the output impedance (also called output resistance) will be very high. There is a certain amount of current that goes through the voltage divider resistors in your circuit (ignoring the output), which you can calculate using Ohm's Law....

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In your initial equation (which appears to be KCL applied at the "center" node), there are a couple of errors. A couple of important distinctions: Node voltages are different than element voltages. $V_A$ is an element voltage (its + and - are placed across an element). The center node has a potential (voltage) with respect to ground, I'll call it $V_{... 1 One other way to integrate signals worth mentioning is to set the window time to the period that you want to integrate and then ctrl+click on the signal (can even be resistance) and it will give you and average and RMS value for the window time. 0 What does ±200 ppm/K mean? These resistors ideally (are designed to) have a temperature coefficient of 0 (zero) meaning their value does not change at all over temperature. Of course this is impossible to produce reliably and in large quantities. So there will still be some temperature dependency. That's what the ±200 ppm/K describes. The manufacturer ... 7 That is likely a wire-wound power resistor with half the windings in the opposite direction so it is non-inductive. That is what that final black band means. Making it a 23Ω Non-inductive resistor. 1 In general, you should never leave digital inputs floating, unless instructed otherwise by the chip manufacturer or designer. Depending on the logic family, a floating input could be interpreted as a high, a low, or undefined value and could even lead to damage to the device by passing too much current through having push-pull transistor pairs both turning ... 0 The correct way to wire things is this: Pin-button-Gnd. No 5V involvement. In the code: byte button= 2; byte buttonState; void setup(){ pinMode (button, INPUT_PULLUP); } void loop(){ if (digitalRead(button) == LOW){ // button is pressed, do something } // or buttonState = digitalRead (button); if (buttonState == 0){ // button was pressed, do ... 0 The lm393 output pulls low, but does not have any transistor inside of it to pull high. Instead, to use it, you have to provide some external way to make the output go up. The resistors you show pull the bases of the output transistors high when the lm393 output is not in a low state. This insures the base-emitter junction of the transistors is reversed ... 4 The LM393 output can only be either pull the output to ground, or open circuit. If the output would be open circuit, then the base of Q1 and Q2 would be floating. R3 and R4 pull the bases of Q1 and Q2 to the supply rail when the LM393 float their output so that the bases are not floating. 3 Looks like NMOS transistor being used in depletion mode as an active resistive pull-up. See Wikipedia: Depletion-load NMOS logic. 2 Any 1k resistor will have to dissipate the 118 milliwatts. The resistor rated wattage just tells if the resistor can handle it or not. If all else remains the same, they all will end up at the same temperature. As resistors that can handle more power are larger and contain more mass, they just take longer to reach that temperature. To reduce the temperature, ... 0 Point 1: The Watt rating of a resistor refers to its maximum power dissipation. The heating you are referring to is joule heating given by P = I^2 R As you can see the heating is not related to the rating of the resistor. The watt rating is an absolute maximum To determine the amount of heat the resistor will generate / can take you will need to look at ... 0 Does a parallel RC circuit introduce a signal delay? Short answer Yes. but this is NOT the major cause of the delay you are seeing in your circuit. For an envelope detector there is always a slope on the trailing edge due to the RC discharge curve. Typically this can be made short enough to not be a concern. On the leading edge of an envelope detector, ... 1 There must be a typographical error somewhere. You are correct to doubt 10 Ohms. Since Zeners become low resistance like a voltage source, you might simulate it as charging a battery with series resistance. Of course, it would not generate power. This is useful to remember because all real Zeners, batteries and voltage sources have a series resistance. (... 0 I believe this is a black first resistor. Reading the top (horizontal) resistor right-to-left. Not as obvious in this photo, but to the naked eye, the left (tolerance) band is definitely gold. The first band looks 100% black. It's reading a consistent 148 Ω, but the left lead heated up and separated from the board. As I read it, it's black, ... 0 Your circuit is indeed a poor implementation of a peak detector and is not an implementation of an envelope detector at all. As already pointed out you CANNOT drive the input pin or your opamp negative, and if you do you may actually cause inversion in the IC and corrupt the output. If you try to rectify the output without enclosing the diode in the opamp ... 0 My seat-of-the-pants analysis is that there's a fairly low impedance path from the opamp, through the diode, into C3. So I expect the attack time to follow the envelope pretty closely. But when the output falls below C3's voltage, the diode reverses and that just leaves C3 and R3 at the output. They may be in parallel from the standpoint of output and ... 0 The basic principle is simple enough. Each color represents a number. Black: 0, Brown: 1, Red: 2, Orange: 3: Yellow: 4, Green: 5, Blue: 6, Violet: 7, Grey: 8, White: 9. The normal pattern is. 2-3 bands for significant digits of the value (interpreed as an integer, e.g. brown-black-black is 100). 1 band for multiplier (10n for normal colors 0.1 for gold, 0.... 1 Presumably your input signal level is in the 1V+ range peak to get 5V out, so the diffusion isolation junctions in the op-amp front end are conducting on negative half cycles (there's a nominal 400$\Omega$resistor in series), which will shift the average input signal upwards (with a long time constant on decay due to R5*C1) and may cause some other weird ... 0 This is a poor solution to a peak and decay envelope detector. R3C3 determine the slow decay time. Look up a precision rectifier design that uses negative feedback after the diode and add a small cap for a hold time constant. Driving a large cap with an Op Amp also reduces phase margin which is current limited. 6 The order of the components does not make any difference. The voltage is diveded on the two components in the same way, in your case 2.1V on the LED and 1.2V on the resistor. This results in 1.2V/330Ohm = 3.6mA through your series circuit. So the power dissipation is also the same in both configurations. 6 What matters is the voltage across the terminals of the resistor (as you would measure with a meter connected directly across the resistor). It doesn't matter how the voltage gets there. So, no it doesn't matter which order they're connected when they're in series, because the magnitude of the voltage across the resistor will always be the supply voltage ... 1 10K 10K is often a good choice if you don't know what resistor to try. Simulate or think about the circuit with a 10K resistor. are the currents that will flow apropriate to the devices and the power-supply? will the voltages be good? if not adjust it. resistor choice for pull-ups and pull-downs is often not critical. 1 this looks like homework, so I'm only going to give hints Write the expression for the total complex impedance of the circuit for all three cases (R, L or C) Assume a fixed voltage as input, calculate the magnitude of the current for all three cases Identify the case, where the current actually goes up EDIT based on the work done. There nothing special ... 7 Ceramic resistors sometimes run pretty hot so something may have melted on it but my concern is to the right side of the yellow capacitors is visible corrosion that may be due to capacitor fluid leakage and may require replacement clean up may be possible with 97% alcohol 0 They are almost certainly not '3V LEDs'. LEDs can operate over a variety of voltages so long as the current through them is kept in check. A resistor is needed to do that. Your LEDs most likely have a forward voltage of 2.1V and want around 20mA to be at their brightest. Exceeding 30mA would probably cause them to burn out. So let's stick them in series ... 36 That's a cement block style, wirewound power resistor, sometimes called a square carbon power resistor. They look like this when new: The one in your circuit board has been mounted upside-down from usual, so you can see the guts of it through the opening slot in the bottom, but that's no big deal. Like all resistors, this component contains no fluid, so ... 8 No, it's fine (as far as we can tell visually). Maybe some foam got melted onto it or a bit of glue was on there and was heated by the resistor. Those resistors have a ceramic housing, resistance wire winding around a core, and are held in by a type of ceramic cement. No liquids at all. I wonder why they put the ugly side up. Whatever their reason, it's ... 0 Do you have an ammeter to measure precisely how much current the tig welder outputs at its lowest 15A settings? Generally welding machines output current is not so well regulated in tight margins especially at such low settings. To answer your question, YES, you can use nichrome coil from stove heaters to get the current you want if you follow these ... 0 TIG (Tungsten-Inert-Gas) supplies are CC (Constant Current) more or less, anyway. Series or parallel resistor won't be much help because the voltage has to increase as the anodized layer forms. Maybe you could parallel the anodized part with a dummy part to use the excess current. 1 s-domain analysis is unitless, so you are free to describe any transfer function. You may simply provide your instructor with a resistor with a resistance greater than$ 1 \; \Omega$, along with the current to voltage transfer function of your resistor. You could also employ an ideal transformer with secondary:primary turns ratio greater than 1. 1 Try doing the problem again with a much lower restance for the IC input, you'll find you get a very different answer. That is why it matters. Most (modern) IC inputs (not power supply pins) can be modeled as a large resistor. Again, this problem is talking about an input pin, not a power supply pin. The input pin could be one of the sense terminals of an op-... 0 What you want to do (a passive network with gain > 1 independent of frequency) is not possible if you use only passive components. If you use resistors, one of them will have to be negative. The last time I checked, negative resistors were not yet invented. It is possible to simulate negative resistances, using a device called a gyrator. A gyrator always ... -1 We can apply voltage divider rule only when same current is running through these resistors. This means, if the IC is supposed to draw current, sum of two currents: one towards the IC and another towards the R2 passes through the resistor R1. So there is difference in the currents flowing throught R1 and R2. Look at the schematic I have added. I have just ... 2 The formula for the voltage divider's output$V_{out} = V_{in}\cdot \frac{R2}{R1+R2}\\$ only applies if you're not drawing any current of the output node. This because the formula above presumes that the same current flows through both the resistors R1 and R2. In other words, to have a stable output voltage Vout you either have to take the required output ...

5

Not every gate needs current for it to operate. MOSFETs are an example of voltage controlled devices whereas BJTs are current controlled devices. A high input impedance means that no significant current will be drawn and as such components can be chosen accordingly. It also means that the voltage between the divider won't change according to current draw ...

4

Firstly the Voltage isn't all that you need to know. The Amperage that can be drawn at that voltage is also important to know. In your Aurduino documentation it will tell you what the maximum current that can be sourced is. As already said in the comments, it will be around 400mA. Current or Amps is how much power you have available, not Volts. We don't ...

2

simulate this circuit – Schematic created using CircuitLab Consider this 100mA layout for 2S5P array. You can always recompute R1 if this is blinding too bright ! Some are > 10 cd such as message signs. You better read a datasheet so you know how not to handle them and how not to solder them. (ESD and 5sec max solder time >5mm from base.

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To answer your question about the E12 and E24 series, you have to go back to the E3 series. There's no escaping that fact. You cannot derive the values for E12 or E24, without retracing backward to E3. Historical Context The history of this goes back at least to Charles Renard, who proposed specific ways of arranging numbers to divide (decimal) intervals. ...

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