Can I use a Zener for 556 protection from relay coil back EMF?

Ron H

Joined Apr 14, 2005
7,063
I was just objecting to the 470 ohm resistor. You don't need it, and it only gives you about 9V across the coil. I prefer to keep relay coil drive as close to rated voltage as possible. Get rid of it, and you'll get about 12V, with a 14.5V supply (again, assuming a CMOS 556).
There is no way a 240k gate pulldown will speed turn-off. It is good to have, to keep the gate from floating if you remove the 556.
 

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mrpict

Joined Nov 4, 2012
26
OK I understand your objection. My reasoning behind putting it there was that without it when 12.5V (556 output with a 14V supply) is applied to the gate and the MOSFET turns on the voltage at the source will jump to 14V. I thought that then because Vs is now greater than Vg that the device would promptly turn off again. With the resistor there I thought I was ensuring that Vs would always be 3V below Vg. I am not getting something obviously. Are you saying it will still work if I remove the 470R resistor and connect drain directly to the 14V rail? If so I don't understand how the MOSFET stays on if Vs is greater than than Vg. What am I missing?
 

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mrpict

Joined Nov 4, 2012
26
I was just objecting to the 470 ohm resistor. You don't need it, and it only gives you about 9V across the coil. I prefer to keep relay coil drive as close to rated voltage as possible. Get rid of it, and you'll get about 12V, with a 14.5V supply (again, assuming a CMOS 556).
With the supply turned down to 12V and the drain connected directly to it (no 470R resistor) I get 10.6V at the coil (source). I love that this is the case but still don't get why this works per my previous post.

There is no way a 240k gate pulldown will speed turn-off. It is good to have, to keep the gate from floating if you remove the 556.
With the gate connected directly to the 556 output there is a short but noticeable period of dimming when the LED/resistor across source/ground deilluminates. With the gate pulldown resistor it snaps off abruptly??
Don't think I am imagining thins but you never know.
 

Ron H

Joined Apr 14, 2005
7,063
With the supply turned down to 12V and the drain connected directly to it (no 470R resistor) I get 10.6V at the coil (source). I love that this is the case but still don't get why this works per my previous post.



With the gate connected directly to the 556 output there is a short but noticeable period of dimming when the LED/resistor across source/ground deilluminates. With the gate pulldown resistor it snaps off abruptly??
Don't think I am imagining thins but you never know.
It's because you have that damned diode between the 556 and the gate.
 

Thread Starter

mrpict

Joined Nov 4, 2012
26
The diodes are there because I have the outputs of two monostables feeding a common output connected to a single relay coil. This transpired as it was the only way I could think of to engage a latching relay without simply using a battery draining constant current solution. To latch the relay I only had a single SPDT left on a 3PDT footswitch. The other two switches are toggling the status LED and the target set/reset coil for the latching pulse.

If I don't have the diodes there then the voltage pulse from one monostable will drain to ground through the other one's low output, instead of firing the coil.

Please don't diss the diode. He's doing the job and we've bonded. He's on my Christmas list. Thank you for the insight.

Regards,
John
 

Ron H

Joined Apr 14, 2005
7,063
The diodes are there because I have the outputs of two monostables feeding a common output connected to a single relay coil. This transpired as it was the only way I could think of to engage a latching relay without simply using a battery draining constant current solution. To latch the relay I only had a single SPDT left on a 3PDT footswitch. The other two switches are toggling the status LED and the target set/reset coil for the latching pulse.

If I don't have the diodes there then the voltage pulse from one monostable will drain to ground through the other one's low output, instead of firing the coil.

Please don't diss the diode. He's doing the job and we've bonded. He's on my Christmas list. Thank you for the insight.

Regards,
John
OK, so the diodes are an OR gate. There is a compromise between MOSFET turn-off speed and power dissipation.
I would trade in the two diodes for one more MOSFET, in parallel with the existing one except for the gates. Then you get more speed, higher coil drive voltage, and less wasted power.
 

Thread Starter

mrpict

Joined Nov 4, 2012
26
That's a really good idea. It would be a bit messy to replace the diodes with individual MOSFET drivers on each 556 output pin cause of the current board layout, so I'll probably embrace the less elegant solution and stick with the single source follower cause I can add that on the board without lengthy jumping wires. Next time around I'd do it your way.
Cheers,
John
 

Ron H

Joined Apr 14, 2005
7,063
That's a really good idea. It would be a bit messy to replace the diodes with individual MOSFET drivers on each 556 output pin cause of the current board layout, so I'll probably embrace the less elegant solution and stick with the single source follower cause I can add that on the board without lengthy jumping wires. Next time around I'd do it your way.
Cheers,
John
And next time around, use low-side switches.:)
 

Thread Starter

mrpict

Joined Nov 4, 2012
26
Well I got all the relay logic switching like I want it. Very tight getting everything inside the housing but no chance to switch from PP3 to AA. Was toying with the idea of trying to reduce the size of the board with the 6 x 556 dual monostables. Also was hoping to reduce the ambient current drain (20mA). This might allow me to fit in the required 12 AA cells. I was wondering if I could lose the 556 and associated caps and resistors and just use a capacitor to bias the 2N7000 gate? This might look pretty ugly to someone who knows what they are doing but I'm wondering if there is mileage in such a solution as this:


Must switch criteria for the relay is a coil voltage of >8.4V for 6mS. I'd need the circuit to re-arm itself within 500mS ideally.

Will this work as drawn? Is there a better way to attempt this? Many thanks for any thoughts,
John

PS. The 9V battery should be a 12V one.
 
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Ron H

Joined Apr 14, 2005
7,063
Tell us what the circuit is supposed to do. This doesn't seem to relate to your original goal of driving a latching relay
 

Thread Starter

mrpict

Joined Nov 4, 2012
26
K1 is latching relay. I'm thinking the circuit should behave like this. Reality may well be different:

In the switch position shown, C1 is charged up via R1 to 12V. Gate of Q1 held at 1V by resistor divider R2/R3. I presume from the datasheet plots that this is low enough to prevent any significant source/drain current.

When switch is flipped to the left. C1 presents 12V to the gate turning it on and latching the relay while C1 discharges towards 1V via R3. After voltage at C1/R3/Q1g declines to the gate threshold (say 3V), Q1 turns off, and relay coil current is snuffed.

Meanwhile (when switch is flipped to the left), C2 charges up to 12V via R2 ready to pulse the relay when the switch is flipped to the right again.

That was the idea anyway.

As drawn this will only latch the relay on (once), I'll add the rest of the coil switching wiring and post in case that makes things clearer.
 

Thread Starter

mrpict

Joined Nov 4, 2012
26
Here it is redrawn with the intention of the driver doing something sensible with the relay coils. SW1 is DPDT, with SW1A triggering the pulse (hopefully) and SW1B (labelling it SW2B is a typo) targeting the coil to latch the relay on and off.

 

Thread Starter

mrpict

Joined Nov 4, 2012
26
One second might be a bit much. Intolerable even. If the MOSFET stays off (insignificant leakage) at 2V which looks possible provided I check, then I might be able to lower RI/R2 to 75K and R3 to 15K. These values would maintain the gate at 2V under static conditions.

Retaining the 0.47uF capacitor, C1 would charge from 2V to 11.4V in 99mS (via 75K).

When triggering the MOSFET, C1 would discharge from 11.4V to 4V in 11mS (via 15K).

Since I need the MOSFET on for >6mS this should work in theory. At the very least assuming it does work there should be significant wiggle room beneath the conservative values chose initially.

100mS recovery would be ideal. No way you could push a footswitch twice that quick. I'll bread board it when I get a chance and play with the values.

I can check the conduction behaviour of a particular MOSFET using a couple of those variable DC-DC buck converters I have. One to vary Vgs and one to put 12V across the source/drain/relay in series.

Thanks again for taking a look at this.
 
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Thread Starter

mrpict

Joined Nov 4, 2012
26
As an aside I dispensed with the 36V zeners. I had been reading stuff about the downside of snubbing using only a diode which suggested that the armature would move in a less snappier fashion than unsnubbed due to the slower collapse of the magnetic field. Now that might be applicable in a plain vanilla relay where the relay opens when power comes off, but in my case I am using latching relays where the armature does not move when the power comes off. Only concern is to protect the MOSFET as far as I can see.
 

Ron H

Joined Apr 14, 2005
7,063
Your circuit should work if you choose a mosfet with Vgs(off)>2V.

You can get the gate voltage down to around 100mV when off if you add a couple of P-channel transistors and 4 resistors. They switch the charging current off when not needed.
I realize you are trying to save space, so this may not be of interest to you.
The voltage-controlled switches are equivalent to a DPDT switch.
 

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mrpict

Joined Nov 4, 2012
26
I think I get the operation, but I am assuming V1 is intended to be a SPDT switch connecting sw_ctrl to either Vcc or GND, either S1/S3 or S2/S4 are closed with the other pair open. The 100n caps are supposed to turn on M1, fire the selected relay coil, and turn off M2/M3 until M1 gate falls below threshold again. If I'm seeing it right and either S1/S3 or S2/S4 are closed all the time is the biasing on M1 gate right? With either S1 or S2 closed you have the gate of M1 biased on via R1/R2 at 47K (Vcc to gate) and R3 at 100K (gate to ground). M1 is never off? Must be missing something. What is the nature of V1?

Ambient current with the circuit I proposed is under 1mA for the six instances I need, so I'm happy with that and the reduced component count, but I'm interested in how your solution works. Thanks.
 

Ron H

Joined Apr 14, 2005
7,063
V1 is the foot that controls the DPDT switch. When V1 is low, S1 and S3 are closed. When V1 is high, S2 and S4 are closed.
The sim starts with S1 and S3 closed. M2 and M3 are both off, because M1 is off. This makes V(g1)≈100mV (R3/R5 divider). Node a is high, being pulled up by R7 (S2 is open).
Now, when V1 goes high, the DPDT toggles.The charge on C1 dumps onto node g1, turning on M1 briefly. M1's current is directed through S4, which is now closed, so M2 stays off, while M3 turns on briefly, long enough to recharge C2.
On subsequent switch toggles, M2 and M3 alternately deliver short charging pulses to their respective caps. The result is that the cap which has been discharged charges rapidly, while the charged cap dumps onto node g1 and gets discharged rapidly.

The attachment may explain it better.
I have to say that I'm a little concerned about what would happen if the DPDT switching is not simultaneous, as it is in the sim.
 

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