Voltage drop diode question

[panalog9]

dropping diode

Hi

I am building a circuit to be powered by 3.6V. Except one part can only take 2.7-3.3V.

Can I use a diode from the +3.6V rail to the VCC of that part ? Diode would have say 0.7V drop, so would provide 2.9V to part.

Part is SPI interface. I could also use same diodes for the SCK and MOSI lines. MISO would go from part to controller, so that would not need dropping diode.

That would work ?

Thanks.

 

[jzaghal]

diode spi miso

Hi,

I would use a 2V7 and a small resistor for
better result.

Good Luck.

 

[House_Cat]

For low current applications, a zener with a dropping resistor from 3.6V would be a more reliable and safer solution.

For high current, I wouldn't skimp - use a separate regulated source for each rail.

The dropping resistor in series with a zener will limit current in either failure mode for the zener. The series diode idea for dropping voltage will take out the lower voltage device if the junction shorts.

The series dropping diode also suffers from the tendency to pass-thru fast transients via the junction capacitance. It can be filtered, but it is touchy compared with the simple zener source.

 

[panalog9]

It's only one part at 2.7V-3.3V, and low current. Peak is 60mA for < 1uS at startup. Average m@x current is 37mA.

**broken link removed**

So use a 3.3V zener diode with a dropping resistor to feed the 3.3V VCC pin. How about the SPI interface pins ? Current there will be extremely small, so the simple voltage dropping diodes should be OK there.

Should they be zener diodes as well ?

Thanks for the quick replies.

 

[jetmarc]

Can I use a diode from the +3.6V rail to the VCC of that part ? Diode would have say 0.7V drop, so would provide 2.9V to part.

Part is SPI interface. I could also use same diodes for the SCK and MOSI lines. MISO would go from part to controller, so that would not need dropping diode.

For the VCC supply, you can use the series diode. It will work, except for the fact that the voltage drop will not be exactly 0.7v. The lower the power consumption, the smaller the voltage drop. To be on the safe side, add a pulldown resistor from VCC_2.9v to GROUND. Something that always consumes about 1-2ma.

For the data signals however, you cannot simply add series diodes. Here you need pulldown resisters (not only for a current bias). The host cannot drive a logic LOW through the diode. You need resistors similar to the optional VCC resistor. The lower the value, the faster your SPI can go. Check with a scope on your final PCB layout to see what value gives best results.

All this stuff is very power consuming. If you are out for low-power stuff (battery operated), you should really check if you can replace the SPI component with a 3.6v compatible one.

Another option is to use a dedicated voltage regulator and logic level converter. This is more expensive than diodes + resistors, but saves a lot of power. Use a 74VHC bus driver to convert the 3.6v logic levels down to 3.0v.

If you can't afford this luxury, at least make sure that your SPI signals are LOW when inactive. This avoids current flow through the pulldown resistors.

jetmarc

 

[panalog9]

Thank you very much for the information. I just discovered good news - the part is a SanDisk MultiMediaCard - the specs quote all electrical characteristics at 2.7V and 3.3V, so I assumed that 3.3V was the m@x.

I just noticed (in very small type) that the timing values are specified as "only in the voltage range 2.7V - 3.6V"

Hah - 3.6V - so the whole question is moot. The whole circuit can run at 3.6V.

Thanks again for your time and help.