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Min Voltage to turn on curcuit

D

Ducky_Doug

Jan 1, 1970
0
I need a simple if not a single device to turn on a relay at about 10 volts
or so. Idle voltage is about 6 volt but when the voltage rises above 10-11
volts I would like the it conduct and turn the relay on. The 12 volt relay
approx 65 ohms DC. I don't want any idle voltage to the relay.
 
J

Joerg

Jan 1, 1970
0
Hello Ducky,

Look at the Texas TL431. It comes in SOT23, is quite precise and cheap.
Unless it's a really big relay you need two resistors to set the trip
voltage, plus a flyback diode and maybe other stuff to protect it from
the relay.

Regards, Joerg
 
J

Joerg

Jan 1, 1970
0
Hello Ducky,

You might also need some hysteresis to prevent the relay from chattering
when the voltage is near the cut-off. Unless it's a rapid enough rise
and fall.

Regards, Joerg
 
J

John Popelish

Jan 1, 1970
0
Ducky_Doug said:
I need a simple if not a single device to turn on a relay at about 10 volts
or so. Idle voltage is about 6 volt but when the voltage rises above 10-11
volts I would like the it conduct and turn the relay on. The 12 volt relay
approx 65 ohms DC. I don't want any idle voltage to the relay.
The device normally used to make such decisions is called a
comparator. They have two voltage inputs labeled + and -. When the
one labeled - is more positive than the one labeled +, the output
turns on and pulls down to the negative supply rail. However, to make
a comparator based decision like you need, you also need some sort of
voltage reference to compare the supply voltage to, and a switching
device able to boost the few milliampere output current to a current
capable of driving the relay coil
(65 ohms / 10 V =154mA in this case).

Here is the kind of thing I am talking about: (View with font set to
something fixed width per character, like Courier)



+-----+-------+------+--------+---+--- +
| | | / | |
| | 22k | .-. o o | |
| .-. | | | |( - 1N4148
z | | | | | |( ^
A | | | '-' |(_ |
| '-' | |10k +--+
| | | | |
| |\|LM393 | |
+-----|-----|-\ | |<
| | | >-----+-------| 2N4401
| +--+--|+/ | |\
| | .-. |/| | |
| | | | | | |
.-. .-.| | | | |
| | | |'-' | | |
| | | | |220k| | |
'-' '-' +----|------+ |
|10k |10k | |
+-----+-------+----------------+------ -

(created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)

For all voltages lower than 10, the zener diode will drop more voltage
than the 22k resistor, but for voltages higher than 11, the 22k will
drop more. So for voltages higher than 11, the - input of the
comparator will be more positive than the + input and the output of
the comparator (actually, the LM393 contains 3 comparators, so you van
make 2 of these with one chip) will pull down to the - supply rail,
turning the 2N4401 transistor on, and energizing the relay. The diode
across the coil prevents a large voltage spike when the comparator
turns off. The 200k feedback resistor lowers the switching threshold
voltage once the relay is on, to make sure the decision doesn't chatter.
 
J

Jamie

Jan 1, 1970
0
Ducky_Doug said:
I need a simple if not a single device to turn on a relay at about 10 volts
or so. Idle voltage is about 6 volt but when the voltage rises above 10-11
volts I would like the it conduct and turn the relay on. The 12 volt relay
approx 65 ohms DC. I don't want any idle voltage to the relay.
9.1 volt zener from radio shaft used to feed a Resistor of lets say 680
ohms to the base of a common 2n2222 or some larger type transistor.
one end of the R goes to the base, the other end goes to the anode
of the Zener Diode, the cathode of the diode comes from your 6..12 volt
source. the collector will connect to one side of the relay coil, the
emitter will go to common/ground. the other side of the relay coil will
connect to the same location as the Cathode side of the zener to obtain
power for the relay unless you have a constant voltage some where else?
in any case, the Zener will turn off any biasing to the transistor
that will release the coil of the relay and not allow any noticeable
amount of current to flow.
also you need a diode across the relay coil to keep flyback effects
from damaging the transistor and other things in your voltage source.
so what you have is a a circuit with 1 zener, 1 common low power
silicone diode, 1 resistor and 1 transistor..
that should do it.
 
E

ehsjr

Jan 1, 1970
0
Joerg said:
Hello Ducky,

Look at the Texas TL431. It comes in SOT23, is quite precise and cheap.
Unless it's a really big relay you need two resistors to set the trip
voltage, plus a flyback diode and maybe other stuff to protect it from
the relay.

Regards, Joerg

Exactly. You need a relay driver after the TL431.
Here's a diagram:

+ Vin ------------+----------+-------------+
| | |
R1 33K R3 220 e
| | /
| +----R4-----b| 2N4403
| | 3.3K \
| ----- c
+---------/ \ TL431 |
| --- +------+
| | | |
| | ----- R
| | / \ L
R2 4.7K | --- Y
| | | |
Gnd -------------+----------+-------------+------+


This will turn the transistor on at ~ 10 V. Whether
the relay will energize is another issue, as we don't
know its pull in current. Darn thing draws ~ 200 mA
at 12 volts.
Ed
 
F

Fred Bloggs

Jan 1, 1970
0
ehsjr said:
Exactly. You need a relay driver after the TL431.
Here's a diagram:

+ Vin ------------+----------+-------------+
| | |
R1 33K R3 220 e
| | /
| +----R4-----b| 2N4403
| | 3.3K \
| ----- c
+---------/ \ TL431 |
| --- +------+
| | | |
| | ----- R
| | / \ L
R2 4.7K | --- Y
| | | |
Gnd -------------+----------+-------------+------+


This will turn the transistor on at ~ 10 V. Whether
the relay will energize is another issue, as we don't
know its pull in current. Darn thing draws ~ 200 mA
at 12 volts.
Ed

Looks like you have the values of R3/R4 interchanged.
 
J

Joerg

Jan 1, 1970
0
Hello Ed,
Here's a diagram:

+ Vin ------------+----------+-------------+
| | |
R1 33K R3 220 e
| | /
| +----R4-----b| 2N4403
| | 3.3K \
| ----- c
+---------/ \ TL431 |
| --- +------+
| | | |
| | ----- R
| | / \ L
R2 4.7K | --- Y
| | | |
Gnd -------------+----------+-------------+------+


This will turn the transistor on at ~ 10 V. Whether
the relay will energize is another issue, as we don't
know its pull in current. Darn thing draws ~ 200 mA
at 12 volts.

Watch out for excess base current as you don't limit it anywhere. The
TL431 can draw quite a bit.

Also, you'd need a little hysteresis in there.

Regards, Joerg
 
F

Fred Bloggs

Jan 1, 1970
0
I need a simple if not a single device

I bet it has to be simple...
to turn on a relay at about 10 volts
or so. Idle voltage is about 6 volt but when the voltage rises above 10-11
volts I would like the it conduct and turn the relay on. The 12 volt relay
approx 65 ohms DC. I don't want any idle voltage to the relay.

Lessee- could it be that the 6V "idle voltage" means you have an
identical relay in series? Whatever- be sure you bypass all relays with
supressor diodes. One way to do it with garden variety discretes:
View in a fixed-width font such as Courier.
..
..
.. input >--+------+-------------+
.. voltage | | |
.. | | |
.. [1k] | 2n4403 |
.. | | e
.. | | |/
.. +------|---[560]---|
.. | | |\
.. | | c
.. | | |
.. | [560] +---------+--------------------+
.. | | | | |
.. | | | | |
.. | 10v - | | |
.. | zener^ [22k] - 9v |
.. | | | ^ zener + |
.. c | | | +-----+
.. \| | | | 1n4001 V |relay|
.. |----+---+--[1k]---+---|<|--+ R | |
.. /| | | | +-----+
.. e | | | - |
.. | 2n4401 [1k] [560] | |
.. | | | | |
.. +----------+---------+--------+-----------+
.. |
.. ---
.. ///
..

It would be better to get your relay logic act together...
 
F

Fred Bloggs

Jan 1, 1970
0
Exactly. You need a relay driver after the TL431.
Here's a diagram:

+ Vin ------------+----------+-------------+
| | |
R1 33K R3 220 e
| | /
| +----R4-----b| 2N4403
| | 3.3K \
| ----- c
+---------/ \ TL431 |
| --- +------+
| | | |
| | ----- R
| | / \ L
R2 4.7K | --- Y
| | | |
Gnd -------------+----------+-------------+------+


This will turn the transistor on at ~ 10 V. Whether
the relay will energize is another issue, as we don't
know its pull in current.

The Vref for the TL431 is 2.5V, and not 1.24V, so that you have
R3=R2*(Vtrip/Vref-1), neglecting leakage thru relay due to feedback
resistor, for Vtrip=10V, and R2=3.3k, R3= 3.3*(10/2.5-1)=10k. Then if
you want this condition to sustain itself to 9.5V or so, you would want
R3=3.3*(9.5/2.5-1)=9.25k so that an R5 feedback such that R5||R2=9.25k
would be R5~100k. Also, at 12V, you want Ic/Ib=10 for low loss and clean
switching. This makes R4=(12-0.8-2.5)/20m=470 ohms. The circuit should
look more like this:

View in a fixed-width font such as Courier.
 
F

Fred Bloggs

Jan 1, 1970
0
Fred said:
The Vref for the TL431 is 2.5V, and not 1.24V, so that you have
R3=R2*(Vtrip/Vref-1), neglecting leakage thru relay due to feedback
resistor, for Vtrip=10V, and R2=3.3k, R3= 3.3*(10/2.5-1)=10k. Then if
you want this condition to sustain itself to 9.5V or so, you would want
R3=3.3*(9.5/2.5-1)=9.25k so that an R5 feedback such that R5||R2=9.25k
would be R5~100k. Also, at 12V, you want Ic/Ib=10 for low loss and clean
switching. This makes R4=(12-0.8-2.5)/20m=470 ohms. The circuit should
look more like this:

View in a fixed-width font such as Courier.

.
.
. + Vin ------------+----------+-------------+
. | | |
. R1 10k R3 3.3k e
. | | /
. | +----R4-----b| 2N4403
. | | 470 \
. | | c
. | | |
. | | |
. +----------|----R5-------+
. | | 100k |
. | | |
. | | |
. | ----- +------+
. +---------/ \ TL431 | |
. | --- ----- R
. | | / \ L
. R2 3.3k | --- Y
. | | | |
. Gnd -------------+----------+-------------+------+
.
.

Obviously the R3 in the calcs refers to R1. Make the real R3 anything,
3.3k makes it the same as R2.
 
F

Fred Bloggs

Jan 1, 1970
0
This one works well enough with fewer components. Multiplied Vbe of T2
is used for cutoff threshold:

View in a fixed-width font such as Courier.


..
..
..
.. input >-+-----+------+-------------+
.. voltage | | | |
.. | | | |
.. | [1k] [560] 2n4403 |
.. | | | e
.. | | | |/
.. | +------|---[560]---| T1
.. | | | |\
.. | | | c
.. | | | |
.. | | | +---------+----------+
.. | | | | | |
.. === | | | | |
.. 0.47u | 10v - | | + |
.. | | zener^ [15k] | +-----+
.. | | | | - V |relay|
.. | c | | 1n4001 ^ R | |
.. | \| | | | +-----+
.. | T2 |----+---+ | - |
.. | /| | | ---
.. | e | | ///
.. | | 2n4401 [1k] |
.. | | | |
.. +-----+----------+---------+
.. |
.. ---
.. ///
..
 
J

Joerg

Jan 1, 1970
0
Hello Ed,

Sorry, I missed R4 so you do have base current limiting. However, 3.3K
might be a bit much, the beta of the transistor may not be sufficient to
drive a large relay.

Regards, Joerg
 
E

ehsjr

Jan 1, 1970
0
Joerg said:
Hello Ed,



Watch out for excess base current as you don't limit it anywhere. The
TL431 can draw quite a bit.

Also, you'd need a little hysteresis in there.

Regards, Joerg

Hysteresis is an excellent point - I didn't think
of it at all. But current is limited - cathode-anode
current is limited to ~ 54ma by the 220 R3. The 2N4403
base has a 3.3k limiter about 3 mA. And the input pin
on the TL431 is limited to 300 uA

But see Fred's post below yours. He spotted a major
error - I computed with Vref of ~1.25, but it is 2.5.
That changes the values of the R1/R2 divider - 10K
would work for R1 with 3.3K for R2.

Ed
 
E

ehsjr

Jan 1, 1970
0
Fred said:
The Vref for the TL431 is 2.5V, and not 1.24V, so that you have
R3=R2*(Vtrip/Vref-1), neglecting leakage thru relay due to feedback
resistor, for Vtrip=10V, and R2=3.3k, R3= 3.3*(10/2.5-1)=10k.

Thanks! Yup - I was thinking LM317 Vref 1.25. It is 2.5
for the TL431, as you said. So R1 should be 10K, R2 3.3K.

And the hysteresis was something I completely overlooked.

Regarding driving the transistor into saturation, you
are right once again. It will work without that, but
it is better with your design. Bottom line, I *like*
your much better design. Thanks again.

Ed



Then if
 
E

ehsjr

Jan 1, 1970
0
Joerg said:
Hello Ed,

Sorry, I missed R4 so you do have base current limiting. However, 3.3K
might be a bit much, the beta of the transistor may not be sufficient to
drive a large relay.

Regards, Joerg

Hi Joerg,

That's a good point. Fred has a better solution that
drives the transistor into saturation using a much
smaller base R. I computed the base drive 10/3300
at 3 mA, and the hfe for the transistor is 100 min
at 150 mA IC, so it will work, but not as crisply
as a harder drive, and it wastes more power (Ic*Vce)
than when it is saturated with a lower vce. Fred's
version includes the hysteresis you mentioned, too.

Ed
 
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