Electronics > Beginners
Replace 1N4848 with 1N4148?
groinksan:
It appears to be a very simple circuit. The purpose of the diode is to prevent VCC being provided by another source from being fed into an Arduino Nano's 5V pin when the Nano is being powered by the USB port, and when a jumper is mistakenly left in-place. I believe on the Arduino Nano, the 5V pin is an output pin, which puts out 5V. So it wouldn't be a good idea to have VCC coming from another source colliding with the 5V pin. Therefore, a diode is put into the circuit to prevent something like this from happening.
The schematic diagram shows a 1N4848 being used here. I read the data sheet, and it seems to be quite beefy for the actual application. I tried looking for a 1N4848, but I can't find any at the local electronics store. I have a ton of 1N4148. Since both the VCC and 5V falls within the 1N4148, I feel that I can use this in-place of the 1N4848. Other things in the data sheet such as maximum forward current, maximum reverse voltage, although all important, I don't think they are in play for this particular application. And, the power supply I'm using is well regulated at 5VDC.
Thanks!
PA0PBZ:
Since the 1N4848 is a 47V zener diode I'd say it is a typo and it should be 1N4148.
rsjsouza:
If the description of the 5V pin is accurate (I am not familiar with the board), the replacement could be done if the output current is well below 75mA (the maximum forward current of the 1N4148)
I would instead replace it wth a regular 1A rectifier such as the 1N4001 as a safety margin.
Ian.M:
Look at the Arduino Nano schematic - there is already a diode between USB Vbus and Nano +5V to protect Vbus from being back-fed, so, as the on-board LM1117 5V regulator will tolerate being back-fed as long as Vin isn't grounded, back-feeding the +5V pin will have no ill effect. If you are using the Vin pin, and have other loads on the same supply, a series diode there may be advisable.
If you use a diode in series with the +5V pin, the rest of your circuit could be powered up and driving inputs of the ATmega328P when it isn't powered. That is *NOT* a good idea.
Also if Vbus is on the low side of the permitted tolerance range under load, and you add another diode drop, you could, when USB powered, end up with less than 4V on your Vcc rail, which may be problematic for your circuit.
David Hess:
The +5V pin is directly connected to the +5 volt output from the Arduino Nano's built in 5 volt regulator. VUSB connects through an MBR0520 0.5 amp schottky diode directly to the +5 volt supply. So the diode as shown prevents any backfeeding of the +5 volt supply which seems a little weird because the forward voltage drop will be excessive for many applications and they included a jumper.
--- Quote from: rsjsouza on December 14, 2019, 09:48:52 am ---If the description of the 5V pin is accurate (I am not familiar with the board), the replacement could be done if the output current is well below 75mA (the maximum forward current of the 1N4148)
--- End quote ---
The 1N4148 is 75 volts and 300 milliamps continuous current but voltage drop becomes excessive above a couple 10s of milliamps. They might have intended it to be a 1N4448 which is essentially a 500 milliamp 1N4148 but a schottky rectifier would be more suitable.
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