A transistor gives a higher voltage than its source?

[jimjam]

The input voltage is 12.24V connected to the emitter
When I measured the voltage at the Base (yes it's floating), it reads 13.24V. How come?

[rob77]

i think you should buy a new voltmeter that's not possible with the given circuit

[Rick Law]

The input voltage is 12.24V connected to the emitter
When I measured the voltage at the Base (yes it's floating), it reads 13.24V. How come?

JimJam, that happens to be at times and it is often because I forgot to ground it properly.

I would have one part of the circuit on a breadboard and the other part on a prototype (soldered) board and the connection ground wire got unplugged between the two.

Second common (common to me) source is depending on length of connections, the wire could also pick up some AC noise.    If possible, put a 1Meg between your measuring points to do a sanity check.  That tiny load would make the noise disappear (in proportion to the 10volt-ish you are measuring.)

[mzacharias]

Don't rule out oscillation...

[jimjam]

i think you should buy a new voltmeter that's not possible with the given circuit

Good point there. I thought it shouldn't be possible either.

Here's some more information:
The power came from a wall wart 12V 2A adapter (switching power supply?)
I set this up on a breadboard
The multimeter is Fluke 87V and I have no reason to believe that it's malfunctioning.

What's more bizarre...

if I disconnected the positive from the breadboard, so that only the negative from the wallwart is connected to the breadboard, things get more interesting. In theory, I should not read any voltage on the circuit. Indeed when I measured either side of the resistor, it reads as 0V.

However, the base reads 5.7V and Emitter 3.4V

This is with the + from the wall wart supply disconnected from the breadboard. Only one wire - the negative, goes to the breadboard.

[rob77]

Don't rule out oscillation...

reverse biased C-E with enough capacitance would oscillate, but the circuit above.....

[rob77]

i think you should buy a new voltmeter that's not possible with the given circuit

Good point there. I thought it shouldn't be possible either.

Here's some more information:
The power came from a wall wart 12V 2A adapter (switching power supply?)
I set this up on a breadboard
The multimeter is Fluke 87V and I have no reason to believe that it's malfunctioning.

What's more bizarre...

if I disconnected the positive from the breadboard, so that only the negative from the wallwart is connected to the breadboard, things get more interesting. In theory, I should not read any voltage on the circuit. Indeed when I measured either side of the resistor, it reads as 0V.

However, the base reads 5.7V and Emitter 3.4V

This is with the + from the wall wart supply disconnected from the breadboard. Only one wire - the negative, goes to the breadboard.

If you have acceess to an sciloscope, then ccheck with scope. Must be some kind of RF noise. Probably the smps adapter  gone crazy ?

[jimjam]

I tried another wall wart adapter - different brand. 12V 2A also. Similar results. The base voltage reads about 1V higher than the supply voltage.

No I don't have a scope - I wish I did.

[rob77]

I tried another wall wart adapter - different brand. 12V 2A also. Similar results. The base voltage reads about 1V higher than the supply voltage.

No I don't have a scope - I wish I did.

Are you sure the circuit is exactly as the schematic you posted ? Isn't the C-E swapped ? as i mentioned above - a reverse biased C-E with a floating base is a nice oscillator - probably the breadboard's capacitance is enough to oscillate.

[Andy Watson]

If the output of the wall wart is not grounded there will be significant AC leakage present. Measuring the supply is a low source impedance - you read the supply voltage. Measuring the floating base you have one lead that has capacitance to ground and is being charged from the AC leakage by the rectifying action of the emitter-base junction.

[jimjam]

OK I tried a third adapter. The "12V" reads 16.64V
The base voltage is 15.56V - at least this time it's lower.

The transistor is 2N3906. When looking at it onto the flat part of the transistor, I am assuming it's EBC from left to right, which corresponds to all the datasheets I can find, e.g.

[jimjam]

If the output of the wall wart is not grounded there will be significant AC leakage present. Measuring the supply is a low source impedance - you read the supply voltage. Measuring the floating base you have one lead that has capacitance to ground and is being charged from the AC leakage by the rectifying action of the emitter-base junction.

This seems to be the most plausible explanation... although I still don't fully understand where this capacitance exists... is it inside the meter? Should it be there?

[madires]

There's a small cap (class Y) between the primary and secondary side of the SMPS wall wart to reduce RF emission.