-1
\$\begingroup\$

So I have a 240 VAC (single-phase) mains source, and I'd like to design a circuit that switches between that and a DC power supply upon receiving a 5 V TTL signal from a delay box.

Basically this fast-switching will be used to open a solenoid valve rapidly (using the DC pulse) then hold it open for a known time interval (using the AC signal) then close it, and repeat. I do know the in-rush and holding current of the valve (1.25 A and 100 mA, respectively) but that and the amplitude of the DC signal aren't important right now. What I'm unsure about is how to safely wire this, and what circuit elements are appropriate for this application.

I appreciate the help.

Edit #1

the valve data sheet can be found here: http://clarkcooper.com/hpv/EH30.html#HiPresstabs=0

"Delay box" refers to a signal delay generator.

Clarification

Sorry for all the confusion, this is (obviously) my first time here. What I'm asking is can someone draw a circuit diagram, with the exact circuit elements I need, to switch between an AC and DC source upon receiving a 5 volt signal from the signal delay generator? I can connect the delay generator with a solid-state relay, but from there I don't know what circuit will allow me to rapidly switch between an AC and DC source. The AC source will, of course, be from the building (mains) while the DC source will be from a standard power supply.

\$\endgroup\$
7
  • 3
    \$\begingroup\$ This is very unclear. I get that the SSR controls the 240 VAC to the valve -- but where does the DC power supply fit into this? \$\endgroup\$
    – Dave Tweed
    Jun 28, 2017 at 18:42
  • 2
    \$\begingroup\$ (1) Your schematic doesn't show a complete circuit. There is/are no power supply/supplies. If #1 and #2 refer to 240 V AC and N then label them that way. (2) What are you showing the ground wire for? (3) What is a "delay box"? (4) Why are you using DC? (5) Where is the link to the valve datasheet? Details in your question, please. \$\endgroup\$
    – Transistor
    Jun 28, 2017 at 18:42
  • \$\begingroup\$ (1) I apologize. Yes, the #1 and #2 are the 240 VAC and N wires. (2) Just for completeness. (3) The "delay box" is a pulse delay generator. (4) We want to use a DC power supply to drive the valve and reduce the rise-time of the gas. Basically we want to go from a "no gas" to "gas" state as fast as possible. (5) The valve data sheet can be found here: clarkcooper.com/hpv/EH30.html#HiPresstabs=0 \$\endgroup\$
    – knotwhirl
    Jun 28, 2017 at 18:45
  • 2
    \$\begingroup\$ This is starting to sound like an X-Y problem (and a non-EE one at that). Are you sure that the mechanical speed of the solenoid plunger is the limiting factor on the risetime of the gas pressure? It seems to me that even if the plunger moved instantaneously, the tiny orifice (0.019" or 0.032") and the volume of the output connection will define how fast the gas pressure can rise. \$\endgroup\$
    – Dave Tweed
    Jun 28, 2017 at 19:17
  • 3
    \$\begingroup\$ So what kind of risetime do you get when you drive the valve with DC? Surely you tried that as well, right? Switching between DC and AC just seems silly. It would make more sense to use DC only, but limit the current after the first few tens of ms. This could be nothing more than a power resistor and a largish "reservoir" capacitor that provides the initial current spike. \$\endgroup\$
    – Dave Tweed
    Jun 28, 2017 at 19:36

2 Answers 2

Reset to default
1
\$\begingroup\$

You will only achieve your rapid switching with predictable timing using a normally closed valve.
You should use a DC valve and not an AC one. If you use a DC activated valve you will have more predictable hold voltage (current). If you use a 12 V DC valve for example it might have a hold voltage of say 5 V.
You can then hit the valve with a high voltage (say 36-40 V) DC pulse for say 20-30 mS. With a normally closed valve you only have to raise the needle from the seat by a fraction of a mm and the valve will essentially auto open at high pressure.

If you want a suggested circuit, add a request in comments.

\$\endgroup\$
1
  • \$\begingroup\$ That makes perfect sense, thank you. Please do insert a suggested circuit. \$\endgroup\$
    – knotwhirl
    Jun 30, 2017 at 15:14
0
\$\begingroup\$

I am unable to find any reference in the linked web pages to running the one coil on both AC and DC. It appears that you have an AC option or a DC option.

The AC specification states:

Power: 22 Watts
AC Inrush:  2.5 amp @ 120V AC
AC Holding: 0.2 amp @ 120V AC

This is normal for any AC electro-magnetic device such as a contactor or solenoid and is one of the advantages of AC coils.

  • When the coil is de-energised the armature is sprung out of the coil and the inductance drops so that the coil is largely resistive. In your case the impedance drops to \$ Z = \frac {V}{I} = \frac {120}{2.5} \approx 40 \; \Omega \$.
  • When the coil is energised the armature pulls into the coil, replaces some of the air-core with iron and the impedance increases to \$ \frac {120}{0.2} = 600 \; \Omega \$.

The initial inrush results in a very fast pull in with low powered holding without any coil switching required as would be the case with DC. The downside is that if the armature sticks open the impedance will remain low and the coil overheat.

\$\endgroup\$

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