555 timer circuit switching automotive solenoid with PNP & NPN switching transistors

Question

I am trying to design a 30 second timer circuit to switch an automotive solenoid that draws about 0.8 amps, and need some help please.

StackExchange posts have been very very helpful so far, and by reading older posts it has helped improve the design of the circuit with protection diodes D1 and D2, and decoupling capacitors C3 and C4, placed very close to pins 1 and 8 of the 555.

Due to the configuration my control circuit must be above the load, so I am using a PNP transistor (Q2) to switch the solenoid. The Base of the PNP is connected to ground through an NPN transistor (Q1), which is switched by the output of a 555 chip.

My problem is that the solenoid switches on as soon as the NPN transistor Q1 is connected. Disconnecting either the 555 output or the connection from the transistor to earth switches the solenoid off.

When prototyping, it worked previously when using BD682 as the PNP transistor and BD681 as the NPN transistor with appropriately sized base resistors, but have now switched to surface mount parts and is broken. Any advice gratefully received!

In this thread: PNP to NPN Transistor switch, some of the replies discuss adding transistors between the Base and Emitter of the PNP transistor due to leakage of the NPN transistor, but R5 does that in my circuit, I think?

Answer 1

If solenoid is 12V / 0.8A = 15 Ω =Rc and Vce(sat) is rated for Ic/Ib= 50= 1.5A/30mA then Rb = (12-Vbe1-Vce2)/30mA = (12-5V)/30mA= 266 Ω @12V and 307 Ohms at 14.2V or nearest, your choice.

But your Darlington drops 2.5V for both Vbe and Vce @1.5A so the base drive changes, if the engine has started which affects Vce(sat).

The solenoid only gets 9.5V from 12V but 11.7V from 14.2V.

Using a Pch FET with an RdsOn of say <1% of load = 100 to 150mΩ will supply 14V to the Solenoid.

^{simulate this circuit – Schematic created using CircuitLab}

The Diode protects the output from Excess voltage =LdI/dt, but also slows down the solenoid release due to L/R=T where R is the diode resistance ~1 ohm decay time.

Turn on Speed = T= L/DCR of the solenoid

Turn Off Damping time constant = T=L/R where R = 1 to 20 x DCR by choice with the diode to compromise overvoltage. V=IR =0.8 * (16+16)= 28 V or V= 0.8A x 320= 256V

Reducing the series R in the Diode increase the power dissipation and slows the response of the solenoid release as the energy must be dissipated E=1/LI^2 which can be in Watts. Choosing series R equal to the DCR of the solenoid may be a good compromise depending on your solenoid spring and release time design specs.

Measure coil DCR with DMM and then choose 1/4W or so depending on the rep. rate of the actuator as energy is dumped in the R with low damping factor L/R.

Keep wire pairs tight for solenoid to minimize area of current loop incl diode to Vcc cap. This will reduce radiated EMI.

Answer 2

To answer the original question, I think i've found the root cause of my problem with the solenoid being always on:

I originally used a voltage divider to provide 5 volts to the 555 timer and associated capacitor/resistors. This is because I didn't know what I was doing - the voltage divider has a very limited ability to supply current, so would operate the 555 just fine but as soon as the connected transistors drew a small amount of current, the supply voltage to the 555 would drop and cause issues with timing and signals.

Since the 555 is rated up to 18 volts, I deleted the resistor creating the voltage divider, and operated the circuit on 12 to 14.2 volts (automotive voltage).

However, the timing capacitor I had on my prototype PCB was only rated to 6.3 volts. It was the only component not rated sufficiently, and was causing issues with triggering and timing. Replacing the capacitor made my circuit work as expected.