2
\$\begingroup\$

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.

30s timer circuit diagram

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?

Improve this question
\$\endgroup\$
7
  • \$\begingroup\$ You need to define if it is retriggerable or not. But your current drive is far too weak. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 15 '19 at 4:55
  • \$\begingroup\$ Not sure about the TRIG input not needing a pull-up resistor, I would double check that one. D1 is in the wrong location, it should be in parallel to the solenoid to clamp inductive kick-back. \$\endgroup\$ – sstobbe Feb 15 '19 at 17:07
  • \$\begingroup\$ @sstobbe Thanks, will have a look at a pull-up resistor on the TRIG input. When solenoid collapses, D2 is the diode to allow the spike to dissipate. D1 is a power diode to protect the transistors. \$\endgroup\$ – Austenite Feb 16 '19 at 5:15
  • \$\begingroup\$ The solenoid will kick-back negative in this case and D1 will pull the collector of Q2 past it's VCE rating. \$\endgroup\$ – sstobbe Feb 16 '19 at 16:01
  • 1
    \$\begingroup\$ Thanks again, I searched last night and the Wikipedia article link was helpful for me, explaining it clear as day, now that you alerted me to it being a negative voltage spike. I had always thought it was a positive voltage spike. \$\endgroup\$ – Austenite Feb 19 '19 at 0:12
1
\$\begingroup\$

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.

schematic

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.

Improve this answer
\$\endgroup\$
13
  • \$\begingroup\$ Thanks for the advice about voltage drop @SunnyskyguyEE75 . Solenoid seems to work fine at that voltage but would be better to avoid the voltage drop, so will look into P channel mosfet at the next opportunity. Ic/Ib for the PNP darlington is somewhere between 2,000 and 3,000 according to the datasheet, that's why R4 is 10k resistor. \$\endgroup\$ – Austenite Feb 16 '19 at 5:21
  • \$\begingroup\$ That is the linear hFE, but when working as a saturated switch, it reduces to 10% as a rule of thumb but in this case is rated at Ic/Ib=50 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 16 '19 at 5:27
  • \$\begingroup\$ Thanks again. I did not know that! I'll have a closer look at the data sheet and learn how to read for that parameter. \$\endgroup\$ – Austenite Feb 18 '19 at 4:47
  • \$\begingroup\$ 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, so adding a series R in the diode 25x the DCR of the solenoid restores the solenoid response time with a compromise on reverse voltage. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 26 '19 at 1:34
  • \$\begingroup\$ Once again, thank you everyone for your input. I'm continuing to work away at this in spare moments. The more I study the diagrams above, the more questions I have. Understanding that you're offering assistance entirely out of goodwill, I thought I would ask them here so that others who find this post later may also learn. Question 1: In the diagrams above, what's the purpose of the C1, the low ESR capacitor? Question 2: In the left hand diagram with the mosfet, why is D1 in parallel with the mosfet and not the solenoid? \$\endgroup\$ – Austenite Feb 26 '19 at 1:37
0
\$\begingroup\$

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.

Improve this answer
\$\endgroup\$
1
  • \$\begingroup\$ Actually, turns out the issues with the voltage divider and 6.3 volt capacitor may have been confounding issues but not the actual root cause. The real root cause was R6, in the diagram above. It should be a pull-up resistor connected to the 12v rail, keeping pin 2 high except when the switch is pressed. \$\endgroup\$ – Austenite Apr 2 '19 at 1:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.