# Tag Info

0

The catch diode insures a proper return path for the inductors current. Without the diode there is risk of damage to the MOSFET switch and a greatly reduced output. There is also a possibility of incorrect polarity at the output, possibly causing damage or drawing excessive current from the source. The diode solves many problems on both the ON and OFF cycle ...

1

Pretty much all buck regulators I’ve seen have to have some form of compensation for the 180 degrees phase change incurred by the inductor and capacitor. It’s not a showstopper if you compensate your feedback circuit to add (typically) 10 to 30 degrees of phase up to the unity gain point in the error amplifier.

0

There are 100 V 5 W Zener diodes on market. You can construct a voltage limiter circuit. All you need to do is building the limiter circuit with less voltage level. For example consider a 5 kV DC with +250 V peak to peak ripples, a voltage limiter with 4.9 kV will kill all the ripples. This solution is likely cost you more because larger diodes that on ...

0

Knowing that you need 5V for the strips, you could segment them and supply them locally with individual 5V. So maybe five groups powered by a 60A 5V supply each. Then use an inverter to make 120V from your 12V battery bank to power the slip rings. Check this out: https://www.amazon.com/dp/B07G7S44CW/ And this: https://www.amazon.com/Power-TechOn-Inverter-...

0

Look for DC to DC converters that have a clock sync and/or parallelizable: Here is a list from analog of high current DC to DC converters. None of them got to the 300A you require, but some go to from 60A to 90A, in which you could parallel three or four of them. However DC to DC converters can be difficult to size components correctly and required a lot ...

2

Look at page 11 of that data sheet. There’s a table showing the pinout. In the same page, the mechanical drawings have the pin numbers listed.

0

If all else fails, and if you can reliably start the thing up with peripherals unplugged and then plug them in later, then you could intentionally stage the turn-on of the peripherals. Just have them on timed relays (or relays controlled from a microprocessor) that sequence them on in a way that doesn't challenge the power supply too much.

0

You could use an LDO to go from 24V to 3.3V, but you would also need heat sinking. How much heat sinking you need would depend on what current you are drawing. For a low current it would probably be okay, for a high current you would probably have trouble stopping it from over heating. A DC-DC converter is more efficient because it doesn't turn the excess ...

2

Q: the grid-side current feeding stage takes the source AC into consideration. How is this possible? A: the condition is a grid fault (0V sc.to gnd) with the voltage source now appearing as a current sink to the 400V charged microgrid capacitor. Q2: Also, the first and second stage is described as a natural response and the third stage is described as a ...

1

The fault characteristics can be divided into three stages as described because "the VSC...interfaces to the ac side through an inductor (Lac)." In addition, there is inductive impedance on the DC side in series with the fault. That impedance controls the timing of the response such that it can be analyzed in three stages.

0

So you want to measure the mean value and not the waveform. At first place you have to determine what are the capabilities of your MCU. You will neeed an analog low pass filter on input of ADC to elliminate aliasing effect, then you can oversample and do a digital low pass filter. For example: 1M sampling, 330kHz RC low pass filter, then a digital low pass ...

1

With a ratio of 5 or about 2 octaves with only a 2nd order filter you get -11 to -13 dB of ripple rejection at Fs and only half this at the Nyquist rate of Fs/2 so it will be noisy. If your ADC is say 10bit with -60dB quantization level and you want to reject PWM noise, then a higher order active Bessel or Chebychev filter with the group delay and rejection ...

0

The problem is a solar panel has very different output characteristics to a battery. Usually we would talk about 'max power point tracking' (MPPT) where you control the impedance to extract the most power for the conditions. At the very least you need to re-think your values. Sunnyskyguy has posted information on that. Though I think maybe a redesign with ...

0

You are wasting your time and money using supercaps because each tiny 18650 Li Ion cell has over 10 thousand Farads and you can use all of its Ah capacitance over a small voltage range of 3.7 to 3.0V unlike caps which must be up- converted to use all of its stored energy down to 0V. If you wanted more Jerk for about 100 milliseconds which won’t give you ...

1

The huge difference between a battery which is a voltage source and a PV panel which a current source is impedance. A charge needs a low impedance based on the need for a low voltage drop to rise in current ratio. Any time voltage is transformed up by N, impedance is also transformed down by N$^2$ thus stressing worse your design problem of impedance ...

1

There are multiple causes of improbability and impracticality, but they don't all reside in the amount of capacitance at a given voltage. You can calculate how many volts and how many farads would be needed, and come up with a value which would work. The math is relatively simple as the energy, in ioules, is equal to 1/2 C V^2, where C is in farads and V is ...

0

A motor would not need an isolated converter. If your using any electronics that would need to be referenced to ground, an isolated converter would be needed. If your only using a motor for a load, then the motor will not care what it's negative terminal is referenced to.

1

You need an isolated converter when the device should be floating instead of referenced to the ground of the PoE injector. Since the injector may be in a different room or building, and thus may have a different ground than the one the device is placed in, no exposed parts may be connected to the circuit ground reference if you are not using an isolated ...

2

You need high and low buffer drivers. Using emitter followers adds 0.6V drop either way so you get about 3.8V swing and the Schottky takes 0.3V into the cap so you get 3.5V "pump" per stage. So ... 5 + 3.5 = 8.5V. Then 8.5 + 3.5 = 12V (Yeah. Right! - ie there will be losses.) Then 12 + 3.5 = 15.5 3 stages should then work even with some extra losses. ...

7

2mA sounds about right. The datasheet of the SN74HC14 says it has an output current of about 4mA. You are doubling the voltage. That means twice as much current must go in on the low side as you take out on the high side. 4mA available on the input gets you 2mA available at the output. You need to supply more current going in. I don't think a simple ...

-1

The LDO of specified type the output is 3.3V but input to LDO is 5V remaining excess power get dissipated in the form of heat since LDO type of linear voltage regulator leading to lesser efficiency Since the efficiency of is the major concern my suggest it will better if you proceed with switching regulators such as buck converters. Currently in market ...

0

how would C1 discharge since only I1 will be passing through the mosfet? Because it had been charged when the mosfet was not conducting as shown in your edited question, picture 6. In picture 2, when the mosfet is open, and C is being charged through D1. This is simplified in picture 6: the current $i_1$ is charging C1 and $i_1 = i_{C1}$. When ...

0

The low side sink current switch causes a high release voltage after clamp to 0V.

0

Open your eyes, son! The ones supposedly "teaching" you are really trying to sell you something (get you to buy their power conversion products), and they don't really have to give you the deep knowledge -- just enough so you know what you "need". Especially when the word "marketing" is in the URL, as is the case here, you can bet that they are not going ...

0

This is an inductor current graph, not an output capacitor voltage graph, so it's difficult to tell from it whether the output capacitor discharges through the load (it does). The current between $DT_s$ and $T_s$ is not constant, but the derivative of the current is proportional to the voltage difference between the inductor terminals, which during "off"...

0

Current limiting is based on the fact that I = U/R And the easiest way to reduce I (if it exceeds some limit) is to decrease U (voltage) which is what Great Scott is doing in his hacked video and explains it very well. It is also important to understand how op-amps works, since they are doing all the logic here. Typical simple DC-DC converter IC is an ...

0

We need a lot more information to give you good advice. For example, how high of a voltage makes a huge difference on the size limitation, what frequency are you thinking about, whatever info you can share makes this easier. The numbers can turn this from difficult to impossible very easily. The main thing to be worried about is straight up breakdown, you ...

3

I have the identical board. I connected it to a power supply set to 4.1V, adjusted the booster output to 5.60V and connected a 5.6Ω 10W resistor. The output stayed solid at 5.60V while the input current rose to 1.56A. After 6 minutes the chip temperature was stable at 39°C according to my infrared thermometer, 23°C above ambient. The inductor was ...

1

Yes, this is a really tiny power supply, yet very efficient if used as intended. It's the experience you gain from these failures about thermal resistance, power calculations, test methods with accurate measurements and thermal sensing, that's far worth more than the cost of this inexpensive board. Next time consider a thermometer and record hot spots and ...

1

Your load is 5.6V @ 1 A = 5.6W. If overall efficiency is 80%, that means an additional Power lost = (20%/80%) x 5.6W = 1.4 Watts is dissipated. That's more than an IC of that size can handle at sensible temperatures with PCB copper (if any) heatsinking plus a small amount of direct radiation and convection. Heatsinking can be obtained by adding a blob of ...

0

There is something funky about this circuit. The feedback voltage is 0. So there's that. I tried replacing the diode with another variant and landed on 4.7V as the output voltage .. If I were you, I would choose a new DCDC and try that :) I think I got the wrong pictures where I use 3.6V as Vin and tried putting a dummy load on the circuit :) Result is still ...

0

The error in the circuit is quite obvious once you see it. The values of C1, C2, C5, C8 and C9 are 0.1 which means 0.1 F. A huge bootstrap cap (C2), a shorted sensing circuit (C5), a very slow soft-start (C1) .... numerous reasons why the simulation is slow and gives weird results. The intended values probably are 0.1uF.

2

I like the cheapest solution US \$2.88 But provide room to add a tin-plated brass or steel foil Faraday shield with an RC filter to desireable breakpoint near 10 Hz. There may be noise from this to the source may need shielding, isolation and filtering. How much? depends on your specs. Also, when the output hits the rail, there is no PSRR. zero, 0dB nada, ...

1

I would start with a selection site where you input the supply voltages and output currents and it makes recommendations based on that. The former Linear technology site is good for that (now run by Analog Devices): - In the above picture I have pre-selected "External Power Switch Boost" (inside the red box) and added some values inside the purple box. The ...

Top 50 recent answers are included