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I used this to allow an ESP32 (3.3V) to control a Noctua 5V PWM fan and it worked great, even for the high 25KHz PWM frequency. Rob36 I too have been trying to use this as a level shifter for my 5v i2c sensor. But haven't been able to make it work. Should I keep the pullup resistors or get rid of them? When my sensor's connected to the arduino it requires a 4.4K pullup to function normally. I am trying to connect the sensor with electric imp. It works, BUT it only works at one value of load current and it wastes most of the power supplied. If the load value changes, the voltage will change, since there is no regulation. However, it will survive a short circuit at the output and protect the 12V source from shorting out. This logic level won’t work for anything lower than 1.8V. It won’t fully turn the MOSFET ON or OFF below that voltage level. We tried using it with 1V but the 5V logic side wouldn’t fully turn off. Do your lights include a led driver? LEDs typically need a constant current source. None of the converters above are constant current.
Next I measured the current used by the converter on the input side, for both a device load and a dummy load. I observed that a real load and a dummy load with same average current both have a similar efficiency. A difference could arise since the dummy load power draw is fixed while the device could draw power in bursts, but this didn’t affect the results significantly. I concluded that using dummy resistors is good enough approximation for this test. It works, BUT the device (or circuit) has to dissipate more power than is supplied to the load. It is even more inefficient than circuits 1 and 2, because the extra electronics take additional current. On the other hand, it would survive a short circuit and so is an improvement on circuits 2 and 3. It also limits the maximum current that would be taken under short circuit conditions, protecting the 12v source.And the winner is: “Fine”! This module is clearly the best for low power scenarios. When reaching currents of 100mA the differences between the 3 leading modules are minimal. It can handle input between 9V and 36V and put out between 5 and 5.3V depending on the power you put into the module. Circuit 2 is a series Zener diode (or you could use a number of ordinary diodes in series to make up the voltage drop - say 12 x silicon diodes)
Skill Level: Rookie - You may be required to know a bit more about the component, such as orientation, or how to hook it up, in addition to power requirements. You will need to understand polarized components. Voltage divider will do the job. If you are placing a resistor in the path of supply then it will only set the current not the voltage. Have you thought of making a shield to level shift between a 3.3V Arduino/shield and a 5V shield or the other way around. Then one simply uses the shield to make other shields compatible with shields of another voltage.
A resistor can only provided a fixed voltage drop if you send exactly the same current through it at all times. You would simply choose the resistor based on the amount of current so that it drops 7 V. The LM317 is an adjustable voltage regulator IC capable of supplying more than 1.0 Ampere of current with a wide range of o/p voltage from 1.25 Volts to 37 Volts. Its regulation is a little bit better than other fixed voltage-regulator IC like LM7805, 7806, 7808, 7810 … Do NOT waste your time with this, use the 74HCT245 like everyone who's actually tried it recommends. The module is pre-built. All you have to do is connect the positive and negative terminals to the connections on the module board. A simple screw mechanism secures them in place. You can then plug in a USB cable and any device you like to get a stable 5V output.
The input capacitor and an output capacitor is to be externally connected to the IC 7805, these capacitor acts as ripple reducer if it is present in the source power supply according to the datasheet. The heat sink is a must because the voltage drop of 7 volts is converted to heat though the Heat-Sink. What is the cut-on voltage for the BOB on the HV side if HV is set to 5v? I ask because sometimes the device I'm connecting to the HV side has 5v signals, and sometimes it has 3.3v signals (with HV set to 5v in both cases). It seems to work, but I wonder if I'm on the hairy edge. If cut-on is 2-2.5v then I think I'm ok, if it's around 3v then it might miss sometimes. I general you are right, but this article is titled “Efficient conversion for low-power electronics” not “… for phones”. I believe it is fair to perform a comparison for the 50mA range, nothing wrong with that.One 12v battery/ 12V adopter power source, 10uF capacitor, 1uF capacitor, IC LM7805, Heat-sink, some connecting wires, and a soldering iron (for soldering). This circuit is a voltage divider schematic. You can design it for the required ‘output voltage’ by using this formula: This depends very much on WHY you are trying to drop the voltage, and whether the LOAD is changing. To steal the picture from @Matthijs, To affect the measurement as little as possible, I used the ammeter on the input side (in series) and calculated the current on the output side using Ohm’s law I=V/R. This way there was no impact on the output side which could add a voltage drop and influence the results. Voltage V was measured in parallel and resistance R is known and depends on the dummy load used for each test. There are many different reasons why you may need a 12V to 5V converter. If you are looking to reduce the supply and create a USB port then the Uctronics converter module could be the perfect solution for you.
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