Hot answers tagged

76

The two killer reasons are yield, and heat. Yield. Every time you do a process step, you get less than 100% perfection. Let's say you get 99% perfection per step. In a process with 20 steps, you would be down to 82%. In a process with 1000 steps, you would be down to 43 ppm, 43 successful builds for every million wafers started. Heat. Our existing designs ...


51

Properly designed, built and used, today's LEDs have incredibly long lives and the wearout mechanisms are not catastrophic in nature, so instead of using incandescent lamp MTTF statistics, a luminosity percentage (70%) is often used to define the lifetime- this doesn't mean that the LED burns out, it means that the light out put is only 70% of what it was ...


46

Your problem is the gate drive voltage. If you look at the datasheet for the STP16NF06, you'll see that the 0.08 Ω Rdson only applies for Vgs = 10 V, and you're driving it with only (a bit under) 5 V, so the resistance is much higher. Specifically, we can look at Figure 6 (Transfer Characteristics), which shows the behavior as Vgs varies. At Vgs = 4.75 V ...


40

When you say 120V @ 50Hz AC you are implicitly saying 120Vrms. The RMS voltage is qualitatively defined as the voltage which will give the same resistive heating (averaged out over time) as a DC voltage of the same number. Therefore, by the definition of RMS, the heating will be the same because the RMS voltages are the same. If you said 120Vpeak or ...


37

They aren't cooled from below because they have pins on the bottom, and FR4 below that. Due to having a much lower thermal conductivity, $$ \begin{array}{rrl} \text{Copper:} & 385\phantom{.25} & \frac{\mathrm{W}}{\mathrm{m}{\cdot}\mathrm{K}} \\ \text{Aluminum:} & 205\phantom{.25} & \frac{\mathrm{W}}{\mathrm{m}{\cdot}\mathrm{K}} \\ \text{FR4:...


36

There are two effects going on. The heat sinking effect of the connections and the temperature coefficient on the wire. Initially the wire is all at the same temperature. You turn the power on and it starts to heat up. The heating is determined by the electrical power dissipation in the wire, for any given section of the wire Power = Current * Voltage. ...


32

The issue with thermoelectric generators is they are horrendously inefficient. For a CPU you HAVE to get rid of the heat they produce or they melt down. You could hook up a peltier module and extract a small amount of electricity from them but you would still need to dissipate the remainder of the heat via a classical heat exchange method. The amount of ...


32

You can buy (or 3d print) fasteners made for such a case. Search PCB mounting feet. These include a standoff to help cooling and prevent shorts.


32

The electrode's resistance isn't what's heating things up – it's the resistance of the ionized air in the arc! Hence, things close to the arc get hot, and things farther away don't.


31

I haven't seen anybody else mention temperature. Perhaps you left the default 10 degree rise in the online calculator? That's pretty conservative. A 20 degree rise isn't that bad in a lot of situations. And if you aren't running the highest current continuously, it's quite possible even a higher temp rise would be acceptable, since it will have time to ...


30

From the datasheet: Figure 1. Absolute maximum forward current. and further on: Figure 2. Derating LED current at increased temperature. Current and temperature are your problems. You are running at absolute max current with no wriggle-room and you are allowing the temperature to rise. At 60° ambient the max current allowed falls off dramatically.


29

The power delivered to a resistor, all of which it converts to heat, is the voltage accross it times the current thru it:     P = IV Where P is power, I is current, and V is voltage. The current thru a resistor is related to the voltage accross it and the resistance:     I = V/R where R is the resistance. With this additional ...


29

The reason to use multiple MOSFETs is to lower power dissipation resulting in a cheaper design. Yes one MOSFET can handle the current but it will dissipate some power as it does have some resistance, typically 9 mohm for the IRFB3607. At 25 A that means 25 A * 9 m ohm = 225 mV drop At 25 A that means 25 A * 225 mV = 5.625 W of power dissipation A ...


28

You are getting your terminology all muddled up. Firstly, V = Volts which is a measure of potential difference (voltage), not power. So saying "extra 7V of power" is incorrect. Secondly, the resistor doesn't "resist voltage", it resists the flow of current. Thirdly, a device doesn't "draw 2V from the battery", it draws a current from the battery. So lets go ...


27

Cut a piece of pcb stock (or any sheet plastic) to fit properly in the housing, including screw holes to match those bosses. Then attach your components to this base plate either with double sided tape, velcro, or cable ties going through holes in the plate. Run the cable through a notch in the edge of the housing rather than a hole so it can be easily ...


26

A transistor (FET, in modern ICs) never switches instantly from full OFF to full ON. There is a period while it's turning on or off where the FET acts like a resistor (even when fully ON it still has a resistance). As you know, passing a current through a resistor generates heat (\$P=I^2R\$ or \$P=\frac{V^2}{R}\$). The more the transistors switch the more ...


26

The normal cable ratings assume that the wire can adequately disperse heat generated in the cable due to the current flowing. If you coil it up and use close to the maximum rating then it stands a good chance of melting the plastic insulation and then causing a short.


25

From the driver to the gates, the wires are ~15cm. Does this cause rining? Almost certainly, and it's a fair bet that this is destroying your MOSFETs, by one or more of these mechanisms: exceeding \$V_{G(max)}\$ even for the briefest instant exceeding \$V_{DS(max)}\$ simple overheating due to slow switching and unintended conduction #3 should be pretty ...


24

Best guesstimate I have for an indicator LED life span is more than 5,000,000 hours. I don't think anyone would be surprised if a micro controller lasted 10-15 years. More likely one would EXPECT one to last that long. Think about how many pn junctions could fail in a microcontroller. An LED has only one pn junction. An indicator LED will last longer ...


23

All current flow in anything that isn't a superconductor generates heat. In chips, it's mostly flowing in aluminium "metal" layers (why not copper? Nasty chemical interaction with other parts of the silicon, it turns out). What causes current to flow? Every time a transistor changes state, this can be modeled as a capacitor (the FET gate of the driven logic ...


23

You have already hit on the answer: your LEDs are getting hot. 15 watts may not sound like much, but it's building up and killing your LEDs. I suggest you get a thermistor and attach it to the center of the board, then monitor the temperature as the system operates. Even better, attach it to the body of one of the LEDs. Because you're using this as a ...


22

Newtons Law of Cooling - Scala graduum caloris the heat which hot iron, in a determinate time, communicates to cold bodies near it, that is, the heat which the iron loses in a certain time is as the whole heat of the iron; and therefore, if equal time of cooling be taken, the degrees of heat will be in geometrical proportion \$ \frac{dQ}{dt} = h\...


22

First let's do a quick number crunch: 6.528W/10W = 65% (of 10W) Referring to the datasheet: There is about a 165C rise in temp. Do not touch!. As for "Is it a safe temperature for the resistor?", refer to the next figure: I'll admit that the Derating Curve Graph kinda hurts my head. But, if you follow the 10W curve over to 25C (about room temperautre), ...


22

Cooling isn't important, it's crucial. A modern CPU can easily put out something between 15 W and 200 W, from a die that's a few cm². If you're not transporting that heat away, that chip has to stop operating, slow down, or: just burn up. With that out the way: Where do you put your heat from there? The cooling surface of a motherboard is very limited ...


22

Yes, electric heaters are basically a resistor. A resistor converts electric energy into heat, it does that with 100% efficiency. That might sound weird but think about it this way: if a resistor was 90% efficient, where would the 10% "lost" power go? Nearly all not 100% efficient devices lose the wasted energy as heat. Generating heat is the sole purpose ...


21

No, current is not lost due to heat. The same current flows in one end of a wire that flows out the other end, regardless of how much heat is dissipated in the wire (or any other resistor). Energy is conserved because the resistance times the current causes a voltage drop. This voltage times current is power, which is taken from the circuit and heats the ...


21

This can be easily calculated. The power supply is 5 V and the LED drops 2 V. That leaves 3 V accross the resistor. 3 V / 150 Ω = 20 mA, which is a typical max current for small LEDs. That means the LED is driven correctly. Now look at the power dissipation. 20 mA x 3 V = 60 mW. That's well within the capability of what looks to be a "1/4 W" ...


20

You need to measure the heat loss in your mug around 50°C. Measure the mug capacity. Fill it with hot water or coffee. Insert a temperature probe and close the lid as much as possible. Use tape to stop any heat leaks. Record the temperature drop over time in the zone of interest. Power loss (and power required to maintain temperature) will be given by $$ ...


20

Yes, CMOS circuits can get hot when there are floating inputs. You should always connect unused CMOS input pins to a defined voltage, usually GND or Vdd, unless the datasheet tells you otherwise (see also the end of this answer and Michael's answer). If a pin could be configured as either input or output and you are not sure which it will be, then you could ...


19

The hob power is typically controlled using a thermo-mechanical duty-cycle controller. Figure 1. Part of a hob power regulator. There are three parts to the control. A small heater element that turns on with the hob. A switch contact containing a bi-metallic strip. This is designed to suddenly toggle over at a certain temperature to give a fast contact ...


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