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L,H on the diagram below represent processor pins in LOW/HIGH states. LOW is 0v, HIGH -- 5V. Difference in voltage levels (5V vs 12V) leads to problems -- current flows on paths where it shouldn't. If I set the HIGH level to 12V, pins at the top of the diagram select the LED group and pins on the bottom the LED column, as expected.

How do I solve this incompatibility? One idea I have is to use multiple transistors at the top and at the bottom, but it requires 6 additional parts. Is there a more efficient solution?

enter image description here

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1 Answer 1

Reset to default
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You need to do something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The digit select line must be High to select the digit - thia may require a software change. Otherwise, you would need another NPN to invert the signal.

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  • \$\begingroup\$ Why do we need R2? What would happen if R2 was removed and the only connection between 12V and R1 was through Q1? \$\endgroup\$
    – pedro
    Jan 26, 2015 at 12:02
  • \$\begingroup\$ R2 ensures that Q1 will be turned off when the Select input is low. Theory may say that the circuit should work without R2, but having R2 makes certain that it will work, even if Q2 is a bit leaky. \$\endgroup\$
    – Peter Bennett
    Jan 26, 2015 at 16:38

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