voltage regulator - What happens if you drive the output of a LM78CT05 when the input is not connected

I have a Fairchild LM78M05CT (datasheet) as a 5V regulator for a µC circuit that also has an ICSP port.

The actual circuit is the one in this question: Review of my first ever PCB design for a watering control robot.

I've seen some versions of the 7805 regulator data sheets where the application notes recommend a diode from Vout to Vin in order to protect the device from burning out if the output voltage goes higher than the input. However the datasheet to the Fairchild LM78M05CT doesn't show such a diode in the application notes. Hence I assumed it wasn't necessary with their implementation.

I went to program my device through the ICSP port so I disconnected it from the DC power (Vin to regulator unconnected) and applied +5V and GND to the ICSP header to power the target µC (which drove +5V to Vout of the regulator).

The programming failed, and the target device has a "electronic" smell not present before. Also a LED on the Vin side of the regulator shines indicating a current from Vout to Vin. When I next plugged in the DC cord I got a spark from the ground terminal (and there is a black residue from it on the solder mask) of the DC plug and immediately unplugged the DC cord.

Further I measure (in situ) only 500R or so between the Vout and GND pins on the regulator both ways which is much less than what I would have expected.

So my questions are:

1. Is my regulator shot?


2. Did I actually need the diode from Vout to Vin (would that have protected my device) in order to be able to program with ICSP?


3. Should I have had the DC power connected to the target? I would have to have had 5V and GND between the target and programmer (arduino working as ISP) to ensure matching signal levels so I would have had two +5V drivers.


4. Any other measurements I should make?

Status Update: Appears that I also bricked the DC adapter, LED doesn't come on and no voltage on terminals. Desoldered the regulator, all pin combinations with polarity measure within tolerance of a new one and it doesn't smell. Applied 5V over ICSP again with DC plug removed and no regulator mounted, smell came back and power LED lit up indicating RL1 is providing 5V on pin 11 in the circuit linked up top. Also the relay smells funny.

Answer

It seems rather unlikely to me that this is the cause. The output voltage has to be greater than the reverse breakdown of the B-E junctions in the regulator to cause damage, and 5V is not. There also has to be significant current flowing back (both conditions are necessary for damage to occur).

For example, the TI datasheet specifically says:

I suggest proceeding with caution and looking for something else that may have caused what appears to be significant damage. For example, reversed power connections.