Electronic – My shift register is working when I use LEDs, but not a solenoid

How is the shift-register wired? Do you have a 0.1uF bypass capacitor across the power leads close to the IC package?

It sounds to me like a noise issue, particularly since it's only triggered when you have a load on the relays. The fact that it results in the shift-register register-state getting reset makes me think it's a power issue.

Also, how are you wiring the shift-register.
With a 74HC595, you need to:

• Tie the two register clocks together (Pins 11 and 12)
• Pull the master-reset pin high (tie pin 10 to VCC)
• Pull the output-enable line low (tie pin 13 to ground)

Lastly, you need a 0.1 uF bypass capacitor between pin 16 (Vcc) and pin 8 (Gnd).

There may be a way of avoiding full H bridges providing you have some leniency with the power supply: -

The power supply is a split rail type that is normally used for producing +V, 0V and -V for op-amps and amplifiers etc.. Because the secondary is floating (i.e. not earthed) you can tie the negative rail to ground and have a half rail (formerly 0V) that the solenoid return current can use.

Now, you only need two FETs; one P type and one N type. You still need protection diodes of course.

When N type activated current flows thru solenoid from left to right. When p type activated current flows right to left.

If you want to melt things try activating both at the same time (this is a normal H bridge problem anyway but you have to be "solid" in how your control circuit works)!! With neither fet activated the solenoid draws no current.

There is some complication in that the P channel is referenced to the highest voltage supply and this will need an extra transistor circuit to make it gnd-logic referenced and, in the end you might favour going full integrated H bridge drive because it's simpler to build.