What would be some of the benefits of doing this?
Power supply noise regulation is the main benefit - op-amps are good but they won't deal with power supply rail voltage variations without producing some distortion on the output signal. For an op-amp, it's called PSRR (power supply rejection ratio) and tells you (in the data sheet for the op-amp) what ...
I'd go straight for a single stage dual phase converter like this: -
One for each motor should do the trick.
Be warned though, this is not something to be undertaken without good tools and PCB design skills. You will certainly find more designs available if you split the power requirements by four and concentrate on a single regulator per motor load.
When the output finally turns off in your test, it follows a decay path that tells of an RC influence. There is some current being discharged through the high internal resistance (spec'ed at up to 2.6MOhm). The TLS715B0 datasheet also shows a fairly low threshold for EN, using the bandgap reference; a comparator may have some bias/leakage that ...
You can, but you don't have to. I would not because for isolation to be of much use, you must isolate all paths to the system. You can't leave direct connections for unpleasantness to go around your isolation. That means that every time your digital logic connects to something you want it isolated from, or connects to something connected to something you ...
You can if your laptop is using the traditional DC input. Many of the newer ones require intelligent at the power supply end and you'd better use an inverter for that (12V DC to 110V AC) like Solar Mike suggested.
If your laptop uses the traditional DC input, you'll have to note the followings:
The deep cycle battery shall have enough capacity to charge the ...
The casing maximum temperature is 105 C. (very hot)
At 90% worst case efficiency and full load it only loses 4 Watts.
At 60C Ta you can still do maximum load since with 10.8 C/W it's at 103.2 degrees celsius.
You can create a thermal pad with a via-stitching grid to all layers under the module and put some thermal transfer pads on it. Maybe transfer the heat ...
Input Voltage = 16V
Output Voltage = 5V
Load Current = 0.1A
That means you want to convert (16-5) V · 0.1 A = 1.1 W to heat inside the IC.
The thermal data says that on a two layer board, you'll be heating up your IC by 1.1 W · 78 K/W = 85 K. I don't think you want that.
So, reduce the supply voltage with an external voltage regulator. To avoid producing ...
Which cross-over frequency is this referring to with respect to the
DC-DC Converter Switching frequency
It's more than likely referring to the the LC resonant frequency (\$f_c\$) of the energy storage components within the DC-to-DC converter. See L and C below: -
Cross over frequency should not be higher than 1/8th of switching
The L and C form ...
would you still get some benefit in terms of cleanliness on the output
of the buck converter by having the input filter?
You would get reduced EMI and that could make a whole heap of difference if it allows your circuit//module/system to pass EMC regulations. If your battery has wires feeding your circuit then even more so because you don't want the battery ...
You can think of the first quadrant operation as being a buck converter down from the supply to the motor.
You can think of the second quadrant as being a boost converter from the motor's generated low voltage up to the supply voltage. This causes current generated by the motor to flow into the supply, braking the motor. The amount of braking can be chosen ...
If those converters have a built-in 10A constant-current limit they will share the load, if not they will quit one after the other.
the 10A fuse is too small. use thicker wire and a larger fuse and a 20A or larger relay.
with only 20A available the pump will be slow to start.
I watched enough of it to get what you're missing. It was obvious to me in a few seconds (but only because I'm already familiar with how those power supplies work)
That new resistor/pot combo is connected to an EXISTING potentiometer (the blue box). So what he's done is modify the resistance curve of the existing pot, not introduce a completely new element....
The type 3 compensator can be built around a variety of amplifiers like an operational amplifier (op-amp), an operational transconductance amplifier (OTA) and a TL431 to cite the most popular options. The type 3 offers the following characteristics:
a pole at the origin offers a high open-loop gain in dc (for \$s=0\$) and helps minimizing the static error (...
The description of loop compensation in the UC3842 datasheet is probably the best description you will find of how the closed loop control should work. The loop bandwidth is limited by the frequency.
Gain is controlled by R69, in conjunction with the CTR of the opto-isolator, gain of the TL431 error amplifier etc. See equation 58:
Based on the datasheet, I ...
The part was invented by Unitrode. TI now owns Unitrode, so I'd consider them the primary source of design information. The TI datasheet has detailed design guidance for each pin.
Your schematic has no output filter capacitor, and the voltage feedback input is shorted to GND. Each of these errors makes the circuit unworkable, so I'd start by trying to ...
The thing is, I have no idea how this thing works
The whole circuit is a flyback converter so there's your first term to start googling.
But, I have no idea how the jeezless thing is getting proper feedback
through the optoisolator
The TL431 acts like a voltage comparator - as soon as the "right" output voltage appears on the isolated output, it ...
Unless you expect to have solderability problems, have WIDE metal around that GROUND pin.
At least 3X wider.
That foil has 70 ° C thermal resistance per watt per square.
You have about 2 square of foil, out to the region with numerous vias.
At 2 squares, you have 2 * 70 == 140 ° C per watt.
Widen that 3:1, and drop the thermal resistance to 50 ° C per watt.
If your current out of the LM317 is Iout you are dissipating 3V*IoutA power at the LM317 as heat. Also, check what are your current ratings of your router and your ONU to calculate your power requirements and check if your supply can handle it. Measure the voltage and current at the 5V supply output to see if it is able to support the load.
At one time, a laptop charger simply supplied a fixed voltage. It varied a bit between manufacturers, but was usually about 19V.
Now some laptops use USB-C to charge. To get the correct voltage for the battery to charge, the laptop must communicate with the charger over the USB data lines to request the desired voltage. A simple DC-DC converter won'...
the disabled output is a bit of a mystery unless you exceeded the Absolute Maximum Ratings.
Pulse test with over 1.5W power drop and up to 150'C Functional range and Thermal resistance, you are exceeding 150'C which could easily occur in 100ms with your scope scale as 1s/div. The device would cool down slowly then recycle.
The mystery is the Absolute ...
Assuming an input voltage of 16V, a output voltage of 5V and a load current of 150mA implies that the LDO has to dissipate around 1.6W. Following the calculation done in section 6.3 of the datasheet we end up with something close to 33 K/W for the maximum allowed thermal resistance. If this is exceeded the thermal shutdown will be ...
A regulated charge pump can be more efficient at low currents; check out the LTC3246:
This type replaces the switching inductance with capacitance, which can make the overall circuit smaller and cheaper as well. Ripple will probably vary by input voltage and input stability; at lower input voltages, it just ...
As discussed in the comments, your question is somewhat theoretical and not caused by a concrete problem you are experiencing. As such, I'll need to make some assumptions to properly answer the question.
Assuming you are designing a stand-alone device, which plugs into an power source, the proper position to measure inrush current is the source power rail. ...