MPU-9150 (IMU) AD0 address change

Question

I have an MPU-9150 which uses I2C address 0x68 by default, but I'd like it to use address 0x69. This is possible by desoldering the 3-way AD0 "Jumper" on the breakout board. By default there are 3 solder pads and pads 2+3 are connected, so that pin AD0 is defaulting to "ground".

So as I understand it, I need to "desolder" the connection from pads 2+3 and then I can connect VCC to pin AD0 which will switch the I2C address to 0x69.

I tried desoldering, but it looks like this won't "break" the connection even if I get rid of all the solder. It looks like it'll still be connected due to the little "circle" and the edges around it (see images). I haven't done such tiny connections before yet either - so any tips on how to best proceed ?

Also, I'd like to understand better how this 3-way jumper works. So assuming pads 2+3 are desoldered, can I "manually regulate" which address is going to be used by putting "high" or "low" on pin AD0 (or is it then going to be fixed to 0x69)? ... and not connecting anything to the pin now would probably be "bad"?

What happens if I'd solder AD0 jumper pads 1+2 together?

Will it then always be using the higher I2C address (0x69)?

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For future reference: Following the suggestions of SamGibson, cutting the connection between pads 2+3 using a hobbyknife + loupe - and then connecting 3.3V to AD0, it now indeed reports I2C address 0x69, see image :

Answer 1

(For consistency with your numbering, I'll count the 3 pads of the AD0 jumper starting with pad 1 on the left-hand side.)

as I understand it I need to "desolder" the connection from 2+3 and then I can connect VCC to pin AD0 which will switch the address to 0x69. I tried desoldering, but it looks like this won't "break" the connection even if I get rid of all the solder, it looks like it'll be still connected due to the little "circle" and the edges around it

Agreed. That breakout board for the MPU9150 seems to have a design defect and some kind of square-ish through-hole pad is bridging the right-hand pair (pads 2-3) of the 3-pad AD0 jumper. I've marked it in green on this enlargement from your image, for readers to see what I'm referring to:

People with the necessary experience, tools & perhaps some magnification, could bypass those pads, cut the PCB track leading from pad 2 to the AD0 pin on the device, and connect the track however is required (logic high or low).

If you want to use those 3 pads of the AD0 jumper, you'll need to investigate the purpose of that square-ish through-hole pad - does it have any connections, top or bottom, except to AD0 jumper pads 2-3? I expect the square-ish through-hole pad has a connection to Gnd. If so, you'll need to use a sharp knife (scalpel) to cut away the copper from that square-ish pad which connects it to pad 2 on the AD0 jumper, leaving it only connected to pad 3.

What happens if I'd solder AD0-jumper 1+2 together?

Do not connect pads 1-2 of the AD0 jumper until pads 2-3 are definitely disconnected, otherwise you will short pad 1 (VLOGIC) to pad 3 (Gnd).

You should really get support for this PCB problem from the vendor of the breakout board. Unfortunately a low-cost supplier may not care, and in that case, you would have the challenge of solving this yourself.

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After the incorrect copper "bridge" has been removed between pads 2-3, then the following options are possible:

If you want to use the AD0 pin on the 0.1" header to select the I2C address, then the AD0 solder jumper must not connect the centre pad (pad 2) to either of the other pads.

Alternatively, if you want to fix the I2C address on the board without using an external connection to the AD0 pin on the 0.1" header, then:

• connecting AD0 solder jumper pads 1+2 (pad 3 unconnected) sets AD0 high (I2C address 0x69), or
• connecting AD0 solder jumper pads 2+3 (pad 1 unconnected) sets AD0 low (I2C address 0x68)