Zener voltage limiter without voltage drop to the load

Schematic at link below.

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Input voltage varies from 8-18V. The voltage to relay shouldn't drop as long as it is less than 12V and remain at 12V when the input voltage raises above 12V. Will the above circuit do the job?

All the Zener voltage limiters I see have current limiting resistor in between the input and load before the zener. I cannot have that topology as limiting resistor between input and load drops the voltage to load.

What should be the resistor value and zener wattage?

Thanks in advance. bhav

Reply to
bhavanireddy
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ong as it is less than 12V and remain at 12V when the input voltage raises = above 12V. =A0Will the above circuit do the job?

Nope. The 12 V output should come from the top of the zener. But the the 275 ohm coil load is too much for a zener reference. (At least the ones I know of.)

How about a three terminal voltage regulator (LM7812.. or cousin.) (Hmmm, maybe not so good at the low voltage end.) George H.

tween the input and load before the zener. I cannot have that topology as l= imiting resistor between input and load drops the voltage to load.

Reply to
George Herold

as it is less than 12V and remain at 12V when the input voltage raises above

12V. Will the above circuit do the job?

the input and load before the zener. I cannot have that topology as limiting resistor between input and load drops the voltage to load.

How many of these are you gonna build? Stated another way...is this something you can tweak once and be done? Or is it a production device that has to work over a wide range of relay vendors and temperatures?

You can't reliably run a 12V relay on 8V. That's why they call it a

12V relay. Yes, you can often make ONE work.

Use a 5V relay and drive it with a current source. LM317 and a resistor?

Reply to
mike

ong as it is less than 12V and remain at 12V when the input voltage raises = above 12V. Will the above circuit do the job?=20

tween the input and load before the zener. I cannot have that topology as l= imiting resistor between input and load drops the voltage to load.=20

You would need something like this, not pretty: Please view in a fixed-width font such as Courier.

. . . . . 8V-18V ---------+----------+---------. . | | | . | | | . | + | | + . | | ) 12V . ---/ [2.7K] ) 275R . 1N4742A // \ 12V | ) relay . --- zener | | coil . | 1W '---------+ - . | - | . | | . | 1N4002 | . | .--|>|----+ . | | | . | | | . | | |< . +----------+-------| TIP42C . | |\ . | | . | | . [390] | . | --- . | /// . | . | . | . D . ON/OFF ---- G NUD3124 . S . | . | . | . --- . /// . . .

I am assuming this is automotive application and the 8V is due to cranking = cycle. In that case, the 8V will hold the relay in provided it was pulled i= n while BATT was at 12V. There are integrated constant current relay driver= s, mostly working on the PWM step down principle. See

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/ds/symlink/uc3702.pdf and=20
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maybe others- you look them up.

Reply to
bloggs.fredbloggs.fred

Using a depletion mode FET should be better, almost no voltage drop until it reaches 12V, then starts to hold. Just as an idea to get you started:

8-18V --------o-------' | | | | \ | / ' \ | / |-+ |
Reply to
Joerg

it is less than 12V and remain at 12V when the input voltage raises above 12V. Will the above circuit do the job?

--
No.

If Vcc drops to 8V there is no mechanism in the circuit which can
 Click to see the full signature
Reply to
John Fields

as it is less than 12V and remain at 12V when the input voltage raises above

12V. Will the above circuit do the job?

between the input and load before the zener. I cannot have that topology as limiting resistor between input and load drops the voltage to load.

Not necessarily. One might consider a "kicker circuit" that produces a spike out of an inductor and then drops back to 8V. For most 12V relays

8V is enough hold voltage. [...]
--
Regards, Joerg

http://www.analogconsultants.com/
Reply to
Joerg

drop

Maybe not an unqualified success over variation in VGS(TH) and all relays a= nd environmental conditions. The problem: may not drop out. This hysteria a= bout not getting enough voltage across the relay is unfounded. Most manufac= turers comply with the SAE minimums so relay will pull-in at low cranking v= oltages and across temperature. Here is a detailed characterization: http:=//pewa.panasonic.com/assets/pcsd/catalog/jj-m-catalog.pdf All the OP needs to worry about is definitely down regulating anything over= 12V in such a way that there is not an outrageous overhead drop at lesser = voltages. This does not require anything extreme, like going to a depletion= mode or using a 6V relay.

Reply to
bloggs.fredbloggs.fred

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You are right, it may not drop off if the relay tends to still "stick" at 2V or so.

I tested some relays a couple weeks ago for a project and they would not reliably pull in unless I went to about 65% of nominal coil voltage.

Or just use a regulated PWM driver from iC-Haus or similar.

--
Regards, Joerg

http://www.analogconsultants.com/
Reply to
Joerg

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environmental conditions. The problem: may not drop out. This hysteria about not getting enough voltage across the relay is unfounded. Most manufacturers comply with the SAE minimums so relay will pull-in at low cranking voltages and across temperature. Here is a detailed characterization:

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12V in such a way that there is not an outrageous overhead drop at lesser voltages. This does not require anything extreme, like going to a depletion mode or using a 6V relay.

That's the second time today, in this thread, that I've seen the word "most" as justification for a circuit topology. If you'd said that in a design review, I'd have sent you to your room without supper.

The only justification for using a 12V relay in this design is if the relay is in some subassembly over which you have no control. In that case, a topology that relies on a parameter of an uncontrolled component is ludicrous. If you have control, design a circuit that works by design, not by accident. Don't confuse clever design with stupid design. The difference is that clever designs work.

If the op wants to build just one and is satisfied that it works most of the time, I'd say, "go for it!"

Reply to
mike

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elays and environmental conditions. The problem: may not drop out. This hys= teria about not getting enough voltage across the relay is unfounded. Most = manufacturers comply with the SAE minimums so relay will pull-in at low cra= nking voltages and across temperature. Here is a detailed characterization:=

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ng over 12V in such a way that there is not an outrageous overhead drop at = lesser voltages. This does not require anything extreme, like going to a de= pletion mode or using a 6V relay.

word &quot;most&quot;

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Can't imagine anyone would even consider not obtaining a detailed datasheet= of the relay line. Maybe practice reading comprehension.

Reply to
bloggs.fredbloggs.fred

&quot;stick&quot;

You can throw it in full reverse to unstick it:

Please view in a fixed-width font such as Courier.

. . . . 8V-18V ---------+----------+---------. . | | | . | | | . | | | + . |+ | ) 12V . ---/ [2.7K] ) 275R . 1N4742A // \ 12V | ) relay . --- zener | | coil . |- 1W '---------+ - . |- | . --- | . 1N4742A \ / 12V | . ---/zener | . / |+ 1W |< . +------------------| TIP42C . | |\ . [390] | . | --- . D /// . ON/OFF ---- G NUD3124 . S . | . | . | . --- . /// .

Reply to
bloggs.fredbloggs.fred

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.

"scribble, scribble. Joerg, SED 7/12"

That goes into the back of my note book. Thanks,

George H.

Reply to
George Herold

as it is less than 12V and remain at 12V when the input voltage raises above

12V. Will the above circuit do the job?

the input and load before the zener. I cannot have that topology as limiting resistor between input and load drops the voltage to load.

if your 12V relay is specified to work with 45% over voltage put a resistor 1/20th the coil resistance in series with the relay coil and stop there.

trying to use a zener or active pass element to limit the voltage is likely to hurt your low voltage performance unneccessarily.

if you need better low voltage erformance use a 5V relay and a 7805 regulator instead,

--
?? 100% natural

--- Posted via news://freenews.netfront.net/ - Complaints to news@netfront.net
Reply to
Jasen Betts

as it is less than 12V and remain at 12V when the input voltage raises above

12V. Will the above circuit do the job?

between the input and load before the zener. I cannot have that topology as limiting resistor between input and load drops the voltage to load.

Joerg, Why don't you educate us on that technique ?:-) Last I heard it takes _current_ to pull in a relay. ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

as it is less than 12V and remain at 12V when the input voltage raises above

12V. Will the above circuit do the job?

the input and load before the zener. I cannot have that topology as limiting resistor between input and load drops the voltage to load.

Ignore my previous suggestion since Fred found that it may not turn off relays that aren't guaranteed to disengage around a couple of volts. This should work better:

8-18V --------o-------o-------------------------. | | | | | | \ | | / | | \ |Depletion mode N-channel | / |-+ - |
Reply to
Joerg

Not quite yet, Fred found a wee bug in it (coil voltage won't drop below about a couple of volts). Paste it over with this instead, same number of parts:

8-18V --------o-------o-------------------------. | | | | | | \ | | / | | \ |Depletion mode N-channel | / |-+ - |
Reply to
Joerg

long as it is less than 12V and remain at 12V when the input voltage raises above 12V. Will the above circuit do the job?

between the input and load before the zener. I cannot have that topology as limiting resistor between input and load drops the voltage to load.

No, it takes _energy_ to pull in a relay :-)

Picture what a boost converter does. It charges an inductor by closing a switch, lets go when a prescribed level has been reached, then the inductor dumps its stored energy into a cap. Now replace this cap with a relay coil. The inductor tries to maintain the current flow through it at all cost. So it'll dump that energy into the relay coil -> relay coil gets a whooping, should be 45mA in this case -> inductance of the "gooser circuit" is big enough to keep this up for a few tens of milliseconds until it's all pulled in -> bingo.

BTW, did you take a look at your smart meter data, whether it uses RF?

--
Regards, Joerg

http://www.analogconsultants.com/
Reply to
Joerg

long as it is less than 12V and remain at 12V when the input voltage raises above 12V. Will the above circuit do the job?

between the input and load before the zener. I cannot have that topology as limiting resistor between input and load drops the voltage to load.

Big-ass extra inductance, and relay has no significant inductance ?:-)

I think you might do better with a resonant charging approach.

It uses RF. Spehro found a flysheet on my type of meter. ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
 Click to see the full signature
Reply to
Jim Thompson

long as it is less than 12V and remain at 12V when the input voltage raises above 12V. Will the above circuit do the job?

between the input and load before the zener. I cannot have that topology as limiting resistor between input and load drops the voltage to load.

The relay coil inductors is not important here. It's usually not high enough to exceed uncomfy voltage levels where you'd run into a limiter and waste energy.

Can you elucidate?

See? Told ya so :-)

--
Regards, Joerg

http://www.analogconsultants.com/
Reply to
Joerg

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