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I have a 74HC595 shift register that I shift bits into from an 5v Arduino. The pins on the shift register is connected to the base of 2N3904 BJT's through a 4.7kOhm resistor. I have noticed that if I include the base resistor, my current to the base is 12.8uA and if I don't the base current is 13.1uA.

Since the figures are so small, do I even need a base resistor?

I should mention that I am trying to build a circuit for one of those fancy LED cubes. On the schematics are shown 2 diodes, but that will be expanded to 8 LEDs for each shift register.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Could you perhaps provide a simple schematic? \$\endgroup\$ – Dzarda Oct 13 '14 at 12:27
  • \$\begingroup\$ Yes of course. I thought it would be a bit excessive, but I'll make one \$\endgroup\$ – Attaque Oct 13 '14 at 12:29
  • \$\begingroup\$ There is something wrong with your eeasurements. I would expect ~ 1mA with the ressistor, and ~ 30 mA without. A current in the uA range indicates that you either measured the wrong way, or the 595 output was low. \$\endgroup\$ – Wouter van Ooijen Oct 13 '14 at 12:31
  • \$\begingroup\$ The op must clearly have an emitter resistor given what he says for the current. \$\endgroup\$ – Andy aka Oct 13 '14 at 12:34
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    \$\begingroup\$ Q1 is an emitter follower, so R3 is not needed. If you want to scale the design to use a higher voltage source ie 12v instead of 5 for example you would want to use PNP transistors for the high side switching and than you would need a base resistor. \$\endgroup\$ – Mike Oct 13 '14 at 12:44
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The base resistors R3 and R4 are doing nothing useful. The circuit would be better with them replaced by wires.

You often see this kind of nonsense when someone designs a circuit by heresay and rules of thumb instead of actually undestanding the electronics. Whoever designed this heard somewhere that you're supposed to put a resistor in series with the base, but didn't bother to listen to the reason why and when this "rule" is appropriate.

Q1 and Q2 are used as emitter followers, so the base and emitter current will be nicely limited to safe values just because of the impedance of the load on the emitter. Adding base resistors will only make the output voltage less predictable.

Note that Q3 is used as a common emitter amplifier with the emitter tied to ground. In that case, something is needed to limit to the base current to a safe value. That is what R5 is doing.

So in summary, R3 and R4 are not needed and the circuit would be better off without them, but R5 is needed. Again, you have to undestand a circuit, not blindly apply rules of thumb, heresay, or any other type of silly superstition.

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  • \$\begingroup\$ Thank you for your answer Olin. I'm the one who designed this, and you are absolutely right. I applied a rule of thumb when adding the base resistors. Luckily I have you good people to answer my silly questions. Another common thing I have seen, is to add a pull-down resister to the base, to help the transistor switch on/off faster. Would it make sense to put pull-down resistors at Q1 and Q2? \$\endgroup\$ – Attaque Oct 13 '14 at 13:33
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    \$\begingroup\$ This circuit doesn't need pull-downs. The 74hc595 does that for you. \$\endgroup\$ – hoosierEE Oct 13 '14 at 13:51
  • \$\begingroup\$ One thing that should be mentioned: generally high side switching (Q1 and Q2) is implemented using PNP transistors to avoid unnecessary voltage drop. In your circuit, the voltage at the emitter of Q1 and Q2 will sit at about 4.3 volts due to the necessary Vbe of 0.7 volts. If you use PNP transistors, then this voltage could rise all the way to the rail. And if you use PNP transistors, the base resistors would be required to limit the current. \$\endgroup\$ – alex.forencich Oct 13 '14 at 19:26
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    \$\begingroup\$ The base resistors R3 and R4 would limit the current drawn from the 74HC595 if the supply of power to the emitters Q1 and Q2 were to fail. They could also protect the 74HC595 in the event that the transistors themselves were to fail. \$\endgroup\$ – supercat Oct 13 '14 at 21:38

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