0
\$\begingroup\$

I am using a Bluetooth module (RN-41 SM) that is spec'd to run at 3.3(V).

I can operate this module at 5(V) however, due to another part of the circuit, I need to run the micro controller at 5(V)

The issue I have is that the only rail I have available is a 5(V) rail. My idea was to use a 3.3(V) zener diode and buffer it through an emitter follower, however when I tried this, it failed on me. I have to supply at max 100 mA to the circuit.

I am using a garden variety transistor (2N3904).

I think my issue may have been related to biasing the transistor properly. Could anyone guide me on how to bias a circuit like this or suggest any other method I could use to supply the 3.3 (V) rail.

Thank you for your time.

\$\endgroup\$
2
  • \$\begingroup\$ Do you need to go with parts you already have? If not, I'd suggest a voltage regulator. An LDO would be easy to setup and at 100 mA, wouldn't waste much power. \$\endgroup\$
    – exscape
    Oct 6, 2012 at 18:01
  • \$\begingroup\$ yes... thank you... i have realized that an LDO is the way to go here... thanks for your help \$\endgroup\$
    – Edwin
    Oct 6, 2012 at 19:36

1 Answer 1

Reset to default
10
\$\begingroup\$

Why not an LDO (Low Drop-Out) regulator? They're easy to use: capacitor on the input, and one on the output, and have much better performance than any zener circuit, load regulation-wise and line regulation-wise.

The TC1269 (just a random pick) has a 2.5 % output voltage accuracy, can supply 300 mA, and has dropout voltage of maximum 160 mV at 100 mA. Dissipation will be less than a shunt regulator as well.

The TC1269 is only 31 cents in 1s at Digikey. A zener + transistor may cost as much.

Digikey lists 3600 types which can supply at least 100 mA.

If you really want a voltage follower then remember that the output voltage is about 0.7 V ( the base-emitter junction voltage drop) lower than the base voltage, so you'll want a 4 V zener to get 3.3 V out.

\$\endgroup\$
5
  • \$\begingroup\$ A zener + transistor should not cost as much, but that is probably the only advantage, and I would rather have a LDO any day for something sensitive. \$\endgroup\$
    – Kortuk
    Oct 6, 2012 at 18:33
  • 1
    \$\begingroup\$ @Kortuk - if \$\Delta\$ cost < 10 cent then cost = equal :-). \$\endgroup\$
    – stevenvh
    Oct 6, 2012 at 18:46
  • \$\begingroup\$ thanks for the answer Steven. I will check out both options. I was trying to avoid another regulator to avoid redundancy, but I have gotten that suggestions from many people. \$\endgroup\$
    – Edwin
    Oct 6, 2012 at 19:29
  • \$\begingroup\$ @Edwin A regulator like an LDO is a transistor with feedback, you are using a zener and a transistor, you are building a regulator, just not as great of one. \$\endgroup\$
    – Kortuk
    Oct 6, 2012 at 21:17
  • \$\begingroup\$ @stevenvh not if 10 cents is 1/3 the price and you make millions. But honestly, I consider the price close enough and the performance difference great enough to be worth it. \$\endgroup\$
    – Kortuk
    Oct 6, 2012 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.