Easy way is to connect a diode in series with your batteries (voltage drop across diode is around 0.7V). If you dont dissipate too much power you should be fine. For example: if you use a general purpose diode 1n400x (x = 1 to 7) You have to look in datasheet for voltage drop at max. current you want to draw.
3. These are motor capacitors, not electronic. 4.5 280v/ 5 250v/ 5 250v means the cap box has 3 separate capacitors inside, one that is the main Starting capacitor, likely the 4.5uF 280V one, and the other two are for speed changes. So low speed is both caps in series, middle speed is one of them, high speed is neither.
9 V – (1.8 V × 4) = 9 V – 7.2 V = 1.8 V. 1.8 V / 25 mA = 72 Ω (and we then round up to 75 Ω) Our generalized version of the formula with multiple LEDs in series is: [Power supply voltage – (LED voltage × number of LEDs)] / current = resistor value. We can even put a couple of these strings of four LEDs plus a resistor in parallel to
The chargers are "generic"; the same model is used for various laptops even with the OEM chargers. The 19.5V is close enough to 19V, it won't make a difference. The 4.62A is slightly less than 4.74A, but 4.74A was likely more than the laptop actually required. If your laptop is heavily expanded and was drawing the limit on the old adapter, the
Yes, that might make sense, again depending on your definition of "very clear 5.0V". You can get down to a ripple of a few mV with a DC/DC, so something like "5.0 +- 0.01 V". If you need a cleaner 5V, then an LDO as filter is probably necessary. Make sure, that the LDO has a high enough PSRR in the switching frequency range of your DC/DC.
EDIT: Since you did measure that the charger bridged the two 3.6 V packs in series, forming a 6S1P configuration during charging, the 9 V and 0.2 A makes perfect sense. It’s actually so low “overvoltage” (9-6*1,45=0.25 V) I’m suspecting the charger is delta V sensing or “smart charger”.
Now, use this voltage to calculate current, I= V/R, I= 0.01/0.0001=> 100 A. Q2. What is the primary advantage to using a current-sensor shunt instead of a Hall-effect sensor? Shunts have no measurement offset at zero current, so are good for avoiding drift in coulomb counting
Thanks for answering. @KyokoSasagava As long as the voltage is not more than 10% off and the power-supply can supply the same amount (or more) of current (Amps) this is in general fine. Do take care of the polarity of the plug. The + and - HAVE TO BE on the same side of the plug as on the original power-supply.
Step 1. Remove and count the batteries in the device you're adapting. Standard dry-cell round batteries such as AAA, AA, C or D are all 1.5 volts. Multiply 1.5 by the number of batteries. So, four batteries would equal 6 volts; six batteries would equal 9 volts and so on. Video of the Day.
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To estimate power supply needs, multiply the number of pixels by 20, then divide the result by 1,000 for the “rule of thumb” power supply rating in Amps. Or use 60 (instead of 20) if you want to guarantee an absolute margin of safety for all situations. For example: 60 NeoPixels × 20 mA ÷ 1,000 = 1.2 Amps minimum.
It acts like potentiometer and output's voltage varies between 0 (0 bar pressure) and 5 volts (2.5 bar pressure). I would like to make the simplest possible circuit to limit the output voltage from the sensor to 4.5 volts, instead of 5v: INPUT - OUTPUT. 0.5v - 0.5v. 3v - 3v. 4.5v - 4.5v. 5v - 4.5v.
According to Ward, “ [T]he incorrect detection circuitry on the Pi end of the USB connection” causes the Pi to be detected as an audio adapter accessory by more advanced cables, such as those used to charge laptops. As Ward explains, this is due to the fact that the Pi 4 uses a shared CC pull-down resistor on the USB Type-C connector rather
A simple fix would be to use diodes which drop about 0.7 V across them. simulate this circuit. Figure 2. A simple solution to drop about 3.5 V from the PSU. This will work reasonably well although you'll be wasting 70% of the power in the diodes and only using 30% in the motor. You might be able to find these in some junk if you don't have any
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can i use 4.5 v instead of 5v
