5V Regulator and Pickit2 Question

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gramo

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Just a quick question on the 7805 5V Regulator


Would anyone know if raising the output pin to 5V would be to much if there is 0V at the Input pin. The reason why I'm curious is because I am powering a circuit with my Pickit2, which also has a 7805 circuit. eg;



Now the problem lies in page 11 of the 7805 datasheet found here in the "RAISING THE OUTPUT VOLTAGE ABOVE THE INPUT
VOLTAGE" section.

It describes the damage would be similar to having a large capacitance on the Output of the Reg also found on that page (page 11, 2nd paragraph). Towards the end of that paragraph it implies that;
Generally no protection diode is required for values of output capacitance ≤ 10μF.
Click to expand...
Which kinda leads to suggest that small reverse leakage currents are "ok"

When I'm prototyping with software, I simply leave the Pickit2 connected and the whole circuit is powered from the Pickit2. This is fine for upto 100mA (25mA recommended), but will my 7805 bite the dust with no power connected to its Input pin because of the reverse leakage current?
 

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I thought it was a no no to do something like that, but I'm not sure and I'm sure others here can state for sure.

However a simple solution would be to use a DPST switch that removes both the input and output voltages from the regulator circuit.

Lefty
 
More of a convenience thing really. The board is a small protoboard that I made up for developing. When I'm prototyping and simming code, its left connected to the PicKit2 on the PC bench, and when I'm working on the hardware side of things it is on my work bench


Its handy having the PICKit2 and 5V reg, as when its on the work bench, I just plug it into the mains via a DC socket...


Well thats 'why' I have two supplies, I haven't blown a reg yet, but surely this is not good. Using Schottky diodes is out of the question, as even their low forward voltage drop will play havoc with ADC references

Perhaps a mod is in order and I'll put that switch in
 
Actually pins 2 & 3. The PICkit2 & Junebug both have a blocking diode. Your target can be powered and they'll keep their supplies off. They do need USB power but that's a given.

Ahh yes almost forgot, put a 1N4001 (in reverse) diode across the 7805 and it'll be fine. **broken link removed**
 
Last edited:
blueroomelectronics said:
Ahh yes almost forgot, put a 1N4001 (in reverse) diode across the 7805 and it'll be fine. **broken link removed**
Click to expand...

Doesn't the 7805 already have an internal diode like that?.
 
No, look at the schematic on the datasheet.

There is normally no need to have both a blocking diode and a diode in reverse parallel with the though.
 
It doesn't hurt. I recall the LM2940 doesn't need it but the old LM7805 might. I've used both diodes in the Inchworm design and have had no problems, it was there so the Inchworm could be target powered just like gramo wants to do.

 
Voltage regulators that are made for cars have protection built-in from reverse polarity.
I guess mechanics frequently connect the battery backwards.
 
audioguru said:
I guess mechanics frequently connect the battery backwards.
Click to expand...

You wouldn't believe how stupid mechanics can be! - mind you, most of them seem to be young kids!.
 
That's not true if there's a reverse polarity protection diode and a capacitor before the regulator. At the you disconnect the power, the voltage on the input capacitor which will gradually drop to the same voltage as the output. The LM7805 will not be harmed by the tiny capacitor and reverse bias leakage current.
 
The diode from the output to the input on a regulator is to prevent a large output capacitor from discharging with a high current into the regulator when the input is shorted to ground.
Inputs are rarely shorted to ground.
 
Well in a commercial product I'd add the diode, you have to plan for worst case and it's cheaper to add the diode then be servicing the stuff.
For the hobby bench I'd just leave it out.
 
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