Hot answers tagged

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 ...


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 ...


19

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 ...


14

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

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 ...


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

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 ...


10

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 ...


10

82B715 is a rather older device, with a limited specific function: It increases the bus current 10x. It does not however improve the threshold noise immunity (threshold levels remain the same). We used to have it in a product but dropped it - there were normally other options. 82B96 is a much more versatile device, and is able to split the bus into separate ...


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 ...


10

Large is when the processor cannot drive the various signals properly and that comes down to a number of things but primarily it is the number of devices on the bus. They all present a load to the drivers and this is not just for the processor - during the read process where the peripheral drives the bus the load is now from the peripheral perspective. ...


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 ...


9

A line driver is simply a buffer. What you put in one end comes out the other. However, it is typically able to sink and source much more current than, say, a normal GPIO pin on an MCU. The increased current is able to overcome the capacitance caused by having many devices on a bus. You can think of it as the logic gate equivalent of an amplifier.


9

If the only problem is that the amplifier stage you made inverts the signal, just flip the inputs of the opamp around to compensate. For another approach, use a different opamp, or do the whole thing with discrete parts in the first place. You said in your first sentence "I've chosen to work with an High Frequency Opamp that ...". You chose it, you can un-...


9

Even though this a a 'simple' circuit there are still lots of things that can go wrong. Finding a fault is quite an art. Let's start by breaking the circuit into two parts. The BJT switch and the MOSFET switch. Design issues The first thing is to look at the design - what's wrong? (if anything). Not every circuit you find on the internet actually ...


9

I'm Russian, so I'll help you out with this one. First of all, the first link google gives when I search for that is the museum of soviet electronics. Interesting choice of parts haha. Secondly, this part was later renamed into К264УМ1 K264UM1. Just for reference. Source: link (russian) As for schematic Going from top-left down and counter clockwise, name - ...


8

Do you really need to add a 6v regulator? To quote from the specs of your motors (my emphasis) - "These motors are intended for use at 6 V. In general, these kinds of motors can run at voltages above and below this nominal voltage, so they should comfortably operate in the 3 – 9 V range, though they can begin rotating at voltages as low as 1 V." Sounds like ...


8

Not much. It's just a part number, originally from the European company SGS-Ates. Perhaps at one time it indicated "Linear" (most of the chips SGS made were bipolar linear and started with an L or an T). With all the mergers over the years, any such meaning would be long gone from those 30-40 year old designs. When SGS made clones of then-popular parts ...


8

There are 2 approaches to driving a stepper motor. The simplest is just to connect DC to each winding in turn, via switches (FETs, driver ICs). And in that case, use 4.8V (5V - switch losses) as you confirmed from current and resistance. This is fine at low and medium speeds. If you need maximum performance, you'll find the motor's inductance attenuates ...


8

Use an H-bridge. It is cheap and straightforward to build. http://www.bristolwatch.com/k150/port4.htm I use this for high current switching: http://www.bristolwatch.com/k150/pics1/mosfet_hb2.png These N- and P-channel MOSFETs have a fairly low V_DS(ON) resistance, and don't run very hot at 2A. P = (I^2) x R = 4 x 0.8 = 3.2W for the IRF9630, less for the ...


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