Does the following circuit boost the power of an arduino output (Arduino Mega 2560) from the 5V of the Pin to the 12V that I need to control a Wireless Transmitter Module without Inverting it?

Can I use 1kOhm for R3? I have no 2k2 laying around.

TX_DATA_5V is the Output of the µC, TX_DATA_12V will be connected with TX Module.

Circuit to Boost from 5V to 12V

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    \$\begingroup\$ you can connect two 1K resistors in series to make 2K ohms. \$\endgroup\$ – DoxyLover Mar 27 '14 at 22:07
  • \$\begingroup\$ what is TX_DATA_12V? is it pulled up some where else or a high impedance input? what is it pulled up to? What rise time do you need? \$\endgroup\$ – smashtastic Mar 27 '14 at 22:25
  • \$\begingroup\$ I want to transmit data with a cheap 12V 433MHz sender from eBay. TX_DATA_12V is like the µC-Pin, just at 12V and I do not think that it is pulled up or down inside of the tx module. \$\endgroup\$ – Tobi Mar 28 '14 at 1:37

It does look like it should work. With a 5V input at the emitter, T1 base will be reverse biased by about 3V (no problem here for most transistors) and there should be no collector current and TX out will be at 12V.

With 0V at the input, the base is now conducting forward current and the BJT shouls turn on and the collector will drop to 200mV above the input (if input at 0V then that's 200mV above ground).

Things to watch for - rise time of the 12V out is subject to the speed at which the transistor turns off so if you expect to pass data thru at 1Mbps forget it but at 10kbps (some hand-waving going on) it should be OK.

Can I use 1kOhm for R3? I have no 2k2 laying around

It should be OK but you are starting to approach the maximum reverse voltage for a B-E junction on a BJT - check the data sheet to see that -4V reverse bias on base-to-emitter is well-within the spec.

  • \$\begingroup\$ +1 for good answer. but I curious, Why isn't it suitable for 1Mbps? \$\endgroup\$ – Roh Mar 28 '14 at 4:26
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    \$\begingroup\$ @roh the 10k resistor is too high to adequately produce fast enough rise times. \$\endgroup\$ – Andy aka Mar 28 '14 at 9:53

Yes, your design looks good, assuming 10 kΩ is a good pullup value for it. When the digital signal is high, the only thing driving the output high will be that pullup resistor.

No, I would not use 1 kΩ for R3. Actually, 1 kΩ should still work, but 2.2 kΩ is already on the low side. There is little harm in making the threshold voltage higher. For best noise immunity, you want the threshold where the circuit is in the middle between a high and low to be near the middle of the digital signal's range, in other words about 2.5 V. Figure the B-E drop of T1 is 700 mV, so you want to aim the R2-R3 voltage divider to make about 3.2 V. Assuming you leave R2 alone, that would make R3 3.6 kΩ. Do you have one of those lying around?

On a separate note, get a resistor assortment already. 20 each of all the 5% values from 1 Ω to 1 MΩ doesn't cost that much.


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