I want to drive 120 solenoids froenter image description herem 120 shift register outputs. I am driving solenoid by using TIP120 Transistor. When i connect all the solenoids Shift register circuitry heating.

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    \$\begingroup\$ Can you add a partial schematic? Not all 120 outputs are needed, just the general idea. \$\endgroup\$ – winny Sep 1 '16 at 13:22
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    \$\begingroup\$ Have you used a base resistor? I bet if you have it's too small. \$\endgroup\$ – Andy aka Sep 1 '16 at 13:23
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    \$\begingroup\$ You must add a flyback diode, see: brenamanf.files.wordpress.com/2014/08/flyback_diode.gif or you will destroy the transistor when the solenoid is deactivated. Consider using an NMOS instead of the NPN transistor as an NMOS does not require a drive current so the 74HC595 has an easier job. Most NMOS already have a diode so you might leave out the flyback diode if the one in the NMOS is suitable. \$\endgroup\$ – Bimpelrekkie Sep 1 '16 at 13:42
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    \$\begingroup\$ @FakeMoustache The body diode is the wrong way in a MOSFET. Unless it's avalanche rated for the solenoid energy you must add the flyback diode. \$\endgroup\$ – Spehro Pefhany Sep 1 '16 at 13:48
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    \$\begingroup\$ @SuryaPrakash You can't supply that much current from it. Please see FakeMostache comment and replace Q1 with a MOSFET with low enough Vgs and add a flyback diode and you should be fine. \$\endgroup\$ – winny Sep 1 '16 at 13:53

First, you do need flyback diodes across the coils, or zeners + diodes if you need the solenoids to switch as fast as possible.

If the chips are getting extremely hot, that's a strong indication of latch-up.

This is not an uncommon situation when you combine large currents and sub-optimal layout. Chips that are connected to each other have protection diodes on the inputs and if one chip's ground bounces by more than a diode drop or two you can have large currents flowing between CMOS output and input, through the protection diodes. If that current exceeds some limit (probably in the range of 100mA for modern parts) the CMOS chip will latch up and may be destroyed if there is enough power supply current available.

The solution is to separate the high and low current paths so that the solenoid return current from the TIP emitters does not flow through the grounds on the shift registers. Connect them together at one point, usually the power supply return. The 1K resistors will prevent anything too dastardly happening if your layout isn't the greatest, but try to keep the return on the transistors physically short (low inductance). Fortunately the TIP transistors are very slow to switch so the problem isn't magnified.

If you do try MOSFETs, keep the resistors, in fact you can increase them to a few K to slow the switching.


The shift registers should not be getting hot. Even with a 5 V supply, each output only sources about 4.3 mA (check to make sure this is within the capability of these digital outputs). Times 8 outputs that's about 35 mA, which is 175 mW total with a 5 V supply. Most of that total will be dissipated in the base resistors, not the chip. Even if it was, that would only make the chips get warm.

One glaring problem with your circuit is lack of kickback catch diodes across the solenoids. When a transistor shuts off, the stored energy in the solenoid ends up going somewhere. If you don't provide it a nice and safe path, like thru a diode, then it will find a path somewhere, producing whatever voltage it takes. In this case that means creating a high enough voltage so that the transistor conducts anyway. Such a high voltage is out of spec, can damage the transistor. Doing this repeatedly is pretty much guaranteed to fry the transistor.

If you fried some of the transistors, they may be shorting the supply voltage thru the solenoid and thru the base resistor into the protection diodes of the shift register outputs. That could make them draw much more current than intended, and getting hot is quite likely.

  • \$\begingroup\$ You might want to check the data sheet on the TIP122, which is a darlington. So the output currents are smaller than you're calculating. Of course, this just reinforces your conclusion, but still... \$\endgroup\$ – WhatRoughBeast Sep 1 '16 at 17:51

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