I am trying to drive some small water pumps with a 74HC595, I added transistors to the output of the 595 in a common emmiter configuration, I used BD135 transistors and 700 ohm resistors for the base of the transistors.

This is an image from Eagle: enter image description here

The way I did the Resistor calculations is as follows, the expected output voltage from the 595 outputs are 5V, and the datasheet tells that a good current per pin output is 6mA, the transistor VBE on is 1V. Hence:

R = (5-1)/0.006 = 666 ohms -> 680 ohms resistor used.

BD135 has a minimum beta(HFE) of 25 , so should I expect to be able to handle at least 150 mA per pin output?

I have done a board with this schematic and it results on 595 heating until it is dead.

Can someone point me where do this went terribly wrong?

  • \$\begingroup\$ Please refer to the bottom of page 5 of this datasheet for an example of calculating dynamic load power consumption onsemi.com/pub/Collateral/MC74HC595-D.PDF. I don't know exactly which manufacturer and package you have, so you'll have to find the right figures for your part and compare to maximum power dissipation. Ensure you are within limits. If not, try clocking it slower. \$\endgroup\$
    – Void Star
    Commented Apr 8, 2020 at 19:38
  • \$\begingroup\$ Another thing to consider : run the logic from a different 5V supply to all those water pumps. At the moment those diodes re dumping all that switching toise onto that 5V supply ... it had better be VERY well regulated and decoupled, or noise on it won't do your logic any good. \$\endgroup\$
    – user16324
    Commented Apr 8, 2020 at 19:41
  • \$\begingroup\$ You're not exceeding the (TI's) datasheet's recommendation: Load currents should not exceed 35 mA per output and 70 mA total for the part So, it's likely one of the comments above. \$\endgroup\$
    – Huisman
    Commented Apr 8, 2020 at 19:42
  • \$\begingroup\$ In text you say you used common collector configuration but the diagram shows common emitter configuration. \$\endgroup\$
    – Curd
    Commented Apr 8, 2020 at 19:52
  • 1
    \$\begingroup\$ There is no bypass capacitor. The transistor is not in common collector configuration, it is used as a switch so it is common emitter configuration. Are the transistors mounted the right way around? Pins should be E-C-B order, and the TAB is connected to collector. The output high voltage driving at 6mA current would have 0.2V-0.5V drop already on chip output, so in reality the current is less than 6mA. Also, 700 ohm resistors do not exist, so please check which resistor you actually have there. But the transistors being mounted the opposite way could explain this. \$\endgroup\$
    – Justme
    Commented Apr 8, 2020 at 19:53

2 Answers 2


I’m thinking your problem has more to do with the inputs than the outputs.

It is essential that inputs remain within the GND to Vcc of the chip, and that definitely includes when power is first applied. You appear to have two 5V supplies and I don’t see where the ground to the MCU takes place.

If you cannot guarantee that condition, for example if the MCU power is applied before the 74HC595 power, then excessive current can flow between the MCU output and the corresponding input. When power is applied, the chip can conduct excessive supply current and fail.

The clean cure is to use voltage translators between the two power domains, the cheap approach is to limit the current with resistors of perhaps 1K.

Another possibility is that your layout is very bad and the switched high current causes the grounds to bounce around by more than a few hundred mV, so when the output is low, a negative bounce can cause excessive current to flow, which again causes latchup, and again the Vcc current kills the chip.

Latchup is basically the triggering of a parasitic SCR (across the power supply) that exists in junction-isolated CMOS ICs.


Your calculations are fine, the absolute max current is ±35mA in or out of the 74HC595 and 7mA is well under this assuming that the Vcc is actually 5V. If the Vcc is higher, then that could be a problem.

However there is another figure that you should be worried about and that is total Icc, which is ±75mA, even with all eight of the outputs on, and drawing 7mA, the design would still be under this figure (7mA*8 ports is 56mA, which leaves 19mA for the digital)

Make sure the digital ports (SER, SCK, RCLK, G) are at their proper voltages (either high or low)

Check Vcc, can't be more than 5V (or you'll need to redo your resistors)


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