# help understanding relay circuit

#### minkey01

Joined Jul 23, 2014
185
Could someone help me understand what this circuit is doing?

I believe the switch (S1) is off in the picture and when turned on is suppose to activate that relay. The power is DC at the + & -.

What are all the resistors and capacitors doing? Is this a timing thing? Could someone maybe run a simulation for me? Also, last question, why is the initial switch position (off) connected to R1 and C2?

Thanks!

#### minkey01

Joined Jul 23, 2014
185
initial switch position may be ON in picture. not sure. need to figure that out too.

#### ericgibbs

Joined Jan 29, 2010
9,560
hi,
The initial condition with the S1 as shown, C3 will be held in a discharged state via R1

Switching S1 over will allow C3 to charge up via R2.
When the voltage on C3 reaches the relays pull in voltage it will energise and remain so.

Switching S1 back to its initial position will allow C3 to discharge via R1, when the voltage on C3 falls to the relays drop out voltage it will de-energise.

C1 and C2 are for switch contact snubbing

Its a relay delay On and Off

E

#### TheButtonThief

Joined Feb 26, 2011
237
When in the "off" position, S1 shorts C2, meaning C3 discharges through R1. If C2 wasn't shorted, it would behave as a DC filter and stop the discharge of C3 through R1.

#### minkey01

Joined Jul 23, 2014
185
Awesome explanation!! Thanks, Eric! (&ButtonThief)

Ok. Could you help me understand the second part of my problem please? It has to deal with what that relay is controlling. See my new uploaded jpg.

When I throw S1 this is going to turn on a solenoid. It looks like the relay keeps that K9 path (lower left corner in pic) connected until C3 charges like you were saying. When C3 is finished charging the relay is activated and turns off the K9 path. I believe this is to have higher voltage from the power supply in the beginning to pull in the solenoid and then when K9 turns off the hold-in solenoid voltage power is lower.

Could you explain the volts from the power supply? It looks like power from D16 and D17 are added to D7 for the initial solenoid pull-in time. And then after for the hold time is just power from D7 alone. I thought for volts to be added it has to be in series? It looks like the D16, D17, & D7 are in parallel. Thoughts?

Thanks so much for the help!!

#### minkey01

Joined Jul 23, 2014
185
oh... in case it's not clear: orange-white-orange from the transformer is 11.5Vx2 AC

#### strantor

Joined Oct 3, 2010
5,225
When I throw S1 this is going to turn on a solenoid. It looks like the relay keeps that K9 path (lower left corner in pic) connected until C3 charges like you were saying. When C3 is finished charging the relay is activated and turns off the K9 path. I believe this is to have higher voltage from the power supply in the beginning to pull in the solenoid and then when K9 turns off the hold-in solenoid voltage power is lower.
I agree
Could you explain the volts from the power supply? It looks like power from D16 and D17 are added to D7 for the initial solenoid pull-in time. And then after for the hold time is just power from D7 alone. I thought for volts to be added it has to be in series? It looks like the D16, D17, & D7 are in parallel. Thoughts?
D16 and D17 are actually in parallel with D3 and D4. D3 and D4 form the top half of a full wave bridge rectifier, and D5 and D6 form the bottom.
D16 and D17 form an additional (redundant) top half of the same bridge rectifier, and I can't figure why they are there.... Unless they're installed for fear high solenoid current of blowing up D3 and D4, but that would be silly since they still utilize D5 and D6 for the bottom half. From what I can see and understand, you could eliminate D16 and D17 and get the same voltage from the junction of D3 and D4.

D7 is not in parallel with anything. You are mistaken.

With K9 de-energized, when S1 is depressed, 22VDC (rectified from both transformer halves in series, thru D16 & D17) is applied to the solenoid via the NC contact of K9.
After K9 energizes, the full-voltage path (22VDC thru 16 & D17) is removed, and instead half-voltage (11VDC) comes from D7. The output from D7 is half, because it comes from the center tap. The output from D7 is still full-wave rectified despite only using 3 diodes (D7, D5, and D6), since it comes from the center tap.

#### minkey01

Joined Jul 23, 2014
185
Perfect!! Thank you, Strantor. This is exactly the functions I am looking for. It is finally figured out.

I guess I don't really understand series and parallel with AC circuits too well yet. I'll have to study more. But thank you for confirming that the rectifying area will indeed work.

It is ok to keep the redundant D16 & D17? It came from an existing board. If it doesn't hurt anything, I'd just like to keep it.

Have a great day.

#### strantor

Joined Oct 3, 2010
5,225
It is ok to keep the redundant D16 & D17? It came from an existing board. If it doesn't hurt anything, I'd just like to keep it.
Sure, no problem there. I guess it even gives you the bonus capacitor C5. Maybe that's why they put D16 and D17; so that they could put in an additional capacitor, so that the solenoid energizing wouldn't suck C6 dry and screw up the rest of the circuits.

#### minkey01

Joined Jul 23, 2014
185
A few last detail questions.

Should I put another contact snubbing capacitor along the second pole of S1, similar to C1? Probably should?

And...just curious, how many volts are turning the relay on?