This is a placeholder topic for “3.3V + 5V Solderless Breadboard Power Supply Module Adaptor” comments.
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Are these things fail safe with a v12 plug pack input if they get a short?
Looks like there is no reverse polarity protection. Is there?
Normally in the device being powered. The power source has no way of telling which way around the powered device is connected except for possible overcurrent when something fails.
According to the only schematic I can find, which might not apply to this specific item, the only reverse polarity protection would be provided by the regulator. So whether or not there is protection would depend on which regulator is used (although typically it is not quoted as being provided).
However one supplier that appears to be stocking the same item mentions a reverse protection diode on the input, and it appears from the image as though there is something that could be a large diode (M7?) adjacent to the power socket. Diode protection would create a voltage drop which would affect the minimum dropout voltage of the device, which would be inconsistent with the quoted minimum supply voltage of 6.5 V. In the absence of confirmation I would be careful in selecting the correct plugpack.
Thank you. I hope the answers help the original questioner.
Purchased one of these sometime ago, managed to kill the 5V regulator, replaced with a large style one, not pretty but works.
The middle pin of the DC socket is connected to the Diode.
The other side of the diode goes to the 5V regulator input.
The output of the 5V regulator goes to the input of the 3V3 regulator.
It is useful for basic testing of 5V and 3V3 circuits but does not provide enough protection in my opinion.
It has a number of limitations.
Too easy to change the voltage and damage whatever circuit is being tested. I am forever making sure the jumpers are the correct way around.
The pins would not fit the breadboard I wanted to use, one side was removed and wires soldered on.
I think the 5V regulator failed because it must cater to the DC to DC input drop (12 - 5 = 7V) and also supply the 3V3 current.
You have got to watch this with linear regulators such as the LM7805 and the like. If you want to draw larger currents like 1A or so the best operation is with the input voltage no more than about 3V above the output. In your quoted case at 1A the regulator would have to dissipate or get rid of 7W which a 7805 might handle with its TO220 package but even then I would be thinking about a heat sink. Some of the physically smaller devices would struggle I think. The usual method of getting larger currents is to use one (or several in parallel) pass power transistors with the lower power unit providing the base current. The output here will be lower than the control regulator voltage by a value equal to the base/emitter voltage of the transistor(s).
Yeah I think I accidently shorted the 5v supply line and something died on this board and fried my circuit with 12v coming out of the 5v supply line.