Hi
I've been looking at some ways of wiring up voltage regulators while including diodes for protection. One of the ways to do so is shown in the image attached. This was found on:
I have a few questions about this configuration:
The diode(D1) is said to prevent damage if power is accidentally connected backwards, I gather that this is because it provides a path for the current to flow to ground. Notice that the two capacitors are electrolytic and therefore voltage should not be supplied in the reverse direction. Does the diode protect the capacitors C1 and C2 at all if the power is reversed? How so?
The RBBB board in the tutorial of the link above allows for programming via FTDI, the FTDI supplies 5V to the same rail as the output of the voltage regulator. That means, when the board is powered by the FTDI connection and not the power jack, there is 5V power being supplied to the output pin of the voltage regulator. By measuring the voltage between the input pin and common pin of the voltage regulator I found a voltage of 4.3V, which means that voltage does some how pass through the regulator in the reverse direction. Is this bad for the regulator?
I thought of adding a diode which can pass the current over the regulator as in Fig 2 attached, however, the current would still have now way to get to ground other than through the regulator. Unless maybe I add a LED to allow current to get to ground as in Fig 3, this would then also serve as a power indicator light. What would be the best way to do it.
And finally, if anyone could upload a schematic of the best way to wire a voltage regulator for applications such as an micro controller board I would appreciate it.
The purpose of D1 is to protect the regulator from reverse voltage connection by blowing the fuse in the power supply. Or if not enough current available from the supply, dropping the input to that across a diode which is 0.6v.
The diode across the regular is to protect against a higher voltage than the regulator input being applied to the output.
Look up the specification sheet of the regulator and it will show the circuit.
But what about a positive voltage applied to the output of the regulator? It is clear that when I apply 5V to the output pin (with ground connected to the centre pin also) and no voltage to the input pin then there is a voltage on the input pin. This means voltage passes through the regulator in the reverse direction. Is this bad for the regulator?
To my mind using a diode to pass excessive current in the hope that it will blow a fuse is just plain stupid. For starters the fuse takes time to blow and the diode itself may fail during that time.
Far better to instal the DIODE (edited) in-line with the supply to prevent the flow of current to the regulator in the event of reverse connection.
No fuse blows and once the defect has been established everything is quickly up and running.
Seems like the Arduino designer has the same policy as they also use the in-line diode
But what about a positive voltage applied to the output of the regulator? It is clear that when I apply 5V to the output pin (with ground connected to the centre pin also) and no voltage to the input pin then there is a voltage on the input pin. This means voltage passes through the regulator in the reverse direction. Is this bad for the regulator?
Bevan
There is no reverse current because the diode across the regulator passes the voltage to the input. As the same ( less 0.6v) voltage appears on the input and output there is no current flow.
jackrae:
Far better to instal the fuse in-line with the supply to prevent the flow of current to the regulator in the event of reverse connection.
I assume a typo and you mean instal the DIODE inline.
You are indeed correct, I meant diode to be in-line; and good on you to spot my (not) deliberate mistake.
Many, many years ago when at college we used to have lecturers who inserted deliberate mistakes intentionally to see if we were actually taking the subject in.... Cannot imagine how much trouble they must have caused in later years with some of their past pupils.
To my mind using a diode to pass excessive current in the hope that it will blow a fuse is just plain stupid. For starters the fuse takes time to blow and the diode itself may fail during that time.
You will find the majority of battery powered radios/transceivers all use the diode across the supply line to protect them.
A series diode would provide a volt drop for a start, and need to be rated higher than the continuous current of the circuit.
Using the shunt system the diode will only conduct for a short time, enough to blow a fuse, diodes have a peak current rating usually 5 or more times then its continuous rating.
In the system in question there is a voltage regulator which itself is designed to"'lose" voltage so losing an extra 0.7 across a series diode is no big deal. If that voltage loss is critical then a simple mosfet reverse polarity protection circuit will provide virtually no loss at whatever current the system demands.
Using a self-destruct circuit to provide reverse protection is just plain poor design. It may be simple but it certainly isn't foolproof and when "blown" is extremely inconvenient.
Hi, I regard the shunt form of protection extremely efficient when dealing with the public doing the installations and volt drop being a factor in performance. KISS
I agree that there are other devices around to do the job, and technology has made the problem easier to address.
Merry Christmas ...Tom......
(What no sandy claws smiley?)
weedpharma:
There is no reverse current because the diode across the regulator passes the voltage to the input. As the same ( less 0.6v) voltage appears on the input and output there is no current flow.
Weedpharma
To clear some confusion. I have built a board based on the RBBB described in the link in my original post. The voltage regulator circuit is exactly the same as the schematic in Fig1 of my original post, i.e. there is only a diode connected across the supply lines. A PCB fritzing image of the design and a photo is attached. The board has a 6 pin FTDI header on the left side for serial communication. When the FTDI cable is plugged in the board draws power from the FTDI cable. The power LED is on the INPUT side of the regulator whilst the FTDI header is on the OUTPUT side of the regulator. When the FTDI cable is plugged in, the power LED lights up, as shown in the figure. The path of the current is shown in red. This means that current does indeed flow through the voltage regulator in the reverse direction. Although the voltage and current passing through the regulator in the reverse direction is small, the regulator is obviously not designed for this. So coming back to one of my orginal questions: Is this bad for the voltage regulator? Or, will such small reverse current and voltage not affect the regulator?
The manufacturer recommends a diode in reverse across the regulator for this situation. He has many well paid engineers working for him and they make the recommendation. Who am I to question him?
