# Tag Info

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 ...

4

An MPPT basically works by attempting to present a solar panel with the ideal "resistance". In the following image, the red line shows the current-vs-voltage curve of a typical solar panel. The basic idea is that for zero current (open-circuit), they provide maximum output voltage, and as you increase the current output (i.e. you go from infinite ...

4

The datasheet you linked to covers multiple variations of the switch. I count 14 different part numbers. There are multiple voltages and colours. The note under the table states there is a resistor included in the switch and special voltages can be made to order. It is probable that the LED will light up on the 12v version if you apply 5v, it will just be ...

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 ...

2

Since they permit both AC and DC power for the LED, I suspect there's some sort of electronics that is pre-conditioning power for the LED. I would say, try it. Apply 6VDC to the LED and see what happens. If it works, try applying 6V reverse polarity, which is unlikely to harm a 12V LED. If it lights in both polarities, that cinches it, there is a ...

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.

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 ...

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. ...

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 ...

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.

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 ...

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 ...

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 ...

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 ...

1

Whether you talk about a boost or a buck converter, it can be loaded by various types of load but let's assume a resistance $R$. In the literature, you often find $L$, $C$ and $R$ for the constitutive elements: Then, if you consider energy-storing elements parasitics like their equivalent series resistance or ESR, writers use lowercase to designate ...

1

if the drive is to be run at 50% speed for a long time, then this reduced power factor puts a strain on the AC Generators. That is not exactly true. At 50% speed, the power required from the generators will be 50% while the total RMS value (including harmonics) of the current remains constant. However the effect of the harmonics may be increased somewhat. ...

Only top voted, non community-wiki answers of a minimum length are eligible