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An LED requires a minimum voltage before it will turn on at all. This voltage varies with the type of LED, but is typically in the neighborhood of 1.5V - 4.4V. Once this voltage is reached, current will increase very rapidly with voltage, limited only by the LED's small resistance. Consequently, any voltage much higher than this will result in a very huge ...

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I would argue that there are fewer "gotcha's" with option A. I would recommend option A to people of unknown electronics skill because there's not a lot that can keep it from working. For option B to be viable, the following conditions must be true: $V_{CC_{LED}}$ must be equal to $V_{CC_{CONTROL}}$ $V_{CC}$ must be greater than $V_{f_{LED}} + V_{BE}\... 26 Converting to a "24 V signal" is missing the point. The real problem is driving a 24 V relay from a 5 V digital signal. Fortunately, that is easy. Here is one way: You didn't say how much coil current the relay draws at 24 V, so I picked an example part I had in my system (Zettler AZ8-1CH-5DSE). Figure the B-E drop of Q1 is about 700 mV. This means ... 22 An even better variation on your option "B" is to put the LED in series with the collector, while leaving the resistor in series with the emitter. simulate this circuit – Schematic created using CircuitLab This turns the transistor into a controlled current sink, where the current is determined by the base voltage, minus VBE, across the resistor. The ... 20 You should always tie unused inputs to a valid logic level. That could be tied to GND or to the VDD voltage rail. Never leave unused inputs floating in that it can cause excessive power dissipation in that IC package and introduce extra noise into the voltage / GND rails. It is common practice to use a pullup or pulldown on the unused inputs on unused ... 19 The answer is at the end, but, just in case you are not familiar with the concept of MOS capacitor, I'll do a quick review. MOS Capacitor: The Gate of MOSFET transistor is essentially a capacitor. When you apply any voltage to this capacitor, it responds by accumulating an electrical charge: The charge accumulated on the Gate electrode is useless, but the ... 18 There is one other way, much less commonly seen. Good for one LED, very simple, you can throw anything from about 4v to 20v at it, and it happily gives the LED a fairly constant current. Blue is the input voltage, 20v to 4v. Green is the current to the LED, about 12mA. Red is the power dissipated by the JFET, datasheet here. 18 The diodes in this application are not there to block current, but to allow a low-impedance path for the coils to discharge themselves through. If such a path is not provided, then when the coil's supply is stopped at each cycle, the stored magnetic energy must find a path for discharge. This results in the coil expressing an arbitrarily high reverse voltage ... 14 A microcontroller has a very low output current. You shouldn't drive directly more than an usual indicator LED with it. The motor draws a much higher current. Connecting directly will result in not working motor and destroying the microcontroller due to high currents. Drivers are not used only for motors. They are used for any device that usually draws ... 14 A LED/LCD display is constructed of multiple panels: 16x16, 64x64, etc depends on the supplier. These panels are mounted such to produce the 1920 x 1080 pixels Each panel will have its own interface IC and LED/LCD driver chipset like the tlc59283 How is each panel/module controlled? via a serial interface. If you look at a 320x240 panel such as this one ... 13 Assumption: LED Driver used is PDA012A-350S-R, from link provided in question. Assumption: A single LED is being driven, based on "flashing the LED about 3 times ..." The constant current driver is going into overload, due to being forced to dissipate power far over its design limits... From the datasheet of the LED, it has a forward voltage of 3.2 Volts. ... 13 Typically there are long skinny driver chips for row and column. They may be on the outside COF (Chip-On-Flex) or inside COG (Chip-On-Glass construction) the panel itself. Here's an HD TV panel from Dave Jones' blog showing the COF column drivers: Here you can find a datasheet for a Novatek driver chip showing the data interface for 384 outputs- 128 x 3 ... 13 Here is your circuit being discussed: R3 and R5 should not be getting "hot". Do the math. Even if Q1 were a perfect switch, there would only be 5 V across R3. (5 V)2(1 kΩ) = 25 mW. Unless this is a very tiny resistor, you wouldn't normally notice it getting warm. A 0805, for example, can usually dissipate about 150 mW safely in open air on a ... 13 Yes. All almost ICs need decoupling capacitors. Devices such as these '485 drivers especially need them due to the current surges the device experiences when switching the signal states due to the low value termination resistors used on the bus lines. 13 Schematics like the one that you show should be regarded as simplified schematics - the actual implementation is likely more complex. Anything that is not specified in the tables, graphs or text --- generally with tolerances --- is not a useable design property. When reading the NUD3160 specification, I do find a property related to the value of the ... 12 Here is a collection of LED driver options you can play with. simulate this circuit – Schematic created using CircuitLab 12 Option B requires the control signal to be raised to a higher voltage than the LED drop voltage plus the base/emitter drop voltage. If your control driver is able to operate at a higher voltage than the LED drop voltage plus the transistor base/emitter drop voltage, then Option B would be valid. Option A on the other hand can easily drive any LED drop ... 12 Try probing on the power supply rail. I bet you see those spikes on there. It'll be due to the lead length between your bench supply and the MOSFETs. Clearly you won't see it on the lower FET side because your scope is referenced to that rail but, if you probed back at the power supply I bet you would. Try a 1uF or 10uF ceramic across the power rails close ... 12 Figure 1. When Q1 is off both Q2 and Q3 are forward biased (green) and will turn on. The result will be shoot-through the two transistors (red). If the supply voltage doesn't collapse (1) will be at about 4.3 V, (2) at 2.5 V and (3) at 0.7 V. Ignoring R3 the current through Q2, R4, R5 and Q3 will be about$ \frac {5 - 0.7=0.7}{2k} = 1.8 \ \text {mA} $. I ... 11 This is a very old topic, but it has no answer yet. This is my attempt: Why are today's implementations not capable of streaming at 53 MB/s? The calculations are nearly fine, but you are forgetting a couple of things in the available number of bytes between frame markers: Each microframe has two thresholds called EOF1 and EOF2. No bus acivity must occur ... 11 The datasheet says Low input voltage noise: 2.2 nV/√Hz. The parameter is really just input voltage noise, and they are saying that it's low to make the part sound more awesome. The noise is usually modelled as having a constant spectral density. Thus, the spectral density of noise power would be specified in$\mathrm{W/Hz}$. The higher the bandwidth (for ... 10$R_{\theta JA}$is one small part of the story. The main part of the story is$R_{\theta JC}$(at 1.5 °C/W), because this is the junction-to-case thermal resistance and it adds to the heatsink thermal resistance to give the lowest (normally) path for heat. So, if the thermal resistance of the heatsink is (say) 6 °C/W then the total thermal resistance ... 10 With any CMOS logic IC, you MUST connect unused logic inputs to a known logic level. You may connect unused inputs to either High or Low, whicever is convenient (or whichever is necessary to make the part work as desired). An unconnected CMOS input may take on any level - if it sits at a "maybe" state, the input circuit will draw excessive current, and ... 9 Option A is a neat ON/ OFF switch. When BJT is saturated, LED current depends basically on Vcc and R3, so LED will have a constant brightness. Option B is an "emitter follower" and makes LED current to depend on input voltage, as VE would be Vin -0.7. Option B is good if you want to control LED current and brightness. But most of the times, it's better ... 9 An H-Bridge contains 4 transistors. Often these are FETs. The gate of a FET is not as simple to drive as you might think, for two reasons: FET gates have some capacitance, which takes a little while to charge up to the threshold voltage. This in turn means the FET take a little to switch on. While the gate capacitor is charging up, the FET is only partially ... 9 To supplement Anindo's fine answer, specifically, the voltage across the inductor (i.e., in this case the motor) is  L \frac{di}{dt}  Thus, when current is cut off suddenly (exactly what an h-bridge wants to do, especially when controlled as a PWM),$\frac{di}{dt}\$gets extremely large and there is an associated very large spike in voltage. The ... 9 A charge pump will give you more flexibility. Bootstrapping is essentially a charge pump operated by the main switching devices. The advantage of bootstrapping is low cost - usually a single external capacitor and a diode. The diodes is internal to the driving IC if an integrated driver IC is used. A separate charge pump requires a clock and typically ... 9 The cheapest solution may be a microcontroller programmed for that purpose. This is all you need: displays, a couple transistors and resistors. Should be possible for about$2. If you really want to control the colon you need a fifth transistor. (If it's always on you can leave it out of the microcontroller's way, and simply connect current-limiting ...

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The reasons are different depending on if it is a BJT or a MOSFET. The effect, though, is the same - it's there to reduce current and protect the IO pins on the controller that are driving the transistors. On the BJT the base -> emitter junction is essentially like a diode, and without a resistor would be like a near short circuit. The resistor stops the ...

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