Login Join Maker Pro

# any comments on this circuit?

A

#### andy

Jan 1, 1970
0
It's meant to send a 1 second pulse through an electromagnet at the end of
every day, to start a plant watering system, as i said in an earlier post.
I've tested it on breadboard, but with only one transistor and a single
coil winding. I'm trying splitting the coil into 3 parts to get more
current through it. It would be good to know what people think of this
circuit before i build the final version - it's only the second circuit
i've built.

http://www.niftybits.ukfsn.org/electronics/daily-water.png
http://www.niftybits.ukfsn.org/electronics/daily-water.gif

andy.

J

#### John Fields

Jan 1, 1970
0
It's meant to send a 1 second pulse through an electromagnet at the end of
every day, to start a plant watering system, as i said in an earlier post.
I've tested it on breadboard, but with only one transistor and a single
coil winding. I'm trying splitting the coil into 3 parts to get more
current through it. It would be good to know what people think of this
circuit before i build the final version - it's only the second circuit
i've built.

A

#### andy

Jan 1, 1970
0
i've had it working with just one output transistor, so i'm not chucking
it out just yet. it would help if you'd say what's wrong with it, or where
i could look to do it right, rather than just trash it with one word.

A

#### andy

Jan 1, 1970
0
It's meant to send a 1 second pulse through an electromagnet at the end of
every day, to start a plant watering system, as i said in an earlier post.
I've tested it on breadboard, but with only one transistor and a single
coil winding. I'm trying splitting the coil into 3 parts to get more
current through it. It would be good to know what people think of this
circuit before i build the final version - it's only the second circuit
i've built.

http://www.niftybits.ukfsn.org/electronics/daily-water.png
http://www.niftybits.ukfsn.org/electronics/daily-water.gif

andy.

I've been through it again, and worked out the component values for the
second half of the circuit again. The new version is at:

http://www.niftybits.ukfsn.org/electronics/daily-water2.png
http://www.niftybits.ukfsn.org/electronics/daily-water2.gif

If there's something wrong with the basic idea of how i've laid it out,
i'd rather be told why than just have it written off.

J

#### John Popelish

Jan 1, 1970
0
andy said:
It's meant to send a 1 second pulse through an electromagnet at the end of
every day, to start a plant watering system, as i said in an earlier post.
I've tested it on breadboard, but with only one transistor and a single
coil winding. I'm trying splitting the coil into 3 parts to get more
current through it. It would be good to know what people think of this
circuit before i build the final version - it's only the second circuit
i've built.

http://www.niftybits.ukfsn.org/electronics/daily-water.png
http://www.niftybits.ukfsn.org/electronics/daily-water.gif

andy.

I've been through it again, and worked out the component values for the
second half of the circuit again. The new version is at:

http://www.niftybits.ukfsn.org/electronics/daily-water2.png
http://www.niftybits.ukfsn.org/electronics/daily-water2.gif

If there's something wrong with the basic idea of how i've laid it out,
i'd rather be told why than just have it written off.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with
HTML:
 or [attachment] in the subject line.[/QUOTE]

Observe the orientations of the 3 output transistors.

The output transistors are NPN types, but a PNP symbol is used.

If you use NPN output transistors, they are operating as emitter
followers, so the emitter voltage will always be at least .6 volts
more negative then the base voltage.

The base current for the output transistors comes entirely from the
base pull up resistor, which provides no current unless there is
voltage across it.  This uses up some more of the 12 volt supply.

The BD139 must sink lots more current through the output drive
resistor than that resistor ever supplies to the output transistors.

There is little reason to expect that what output drive current there
is will divide into three equal parts where it meets 3 bases.

I don't want to ruin all your fun by designing the circuit for you,
but think about these comments and show me what you come up with as a
result.

J

#### John Popelish

Jan 1, 1970
0
John said:
It's meant to send a 1 second pulse through an electromagnet at the end of
every day, to start a plant watering system, as i said in an earlier post.
I've tested it on breadboard, but with only one transistor and a single
coil winding. I'm trying splitting the coil into 3 parts to get more
current through it. It would be good to know what people think of this
circuit before i build the final version - it's only the second circuit
i've built.

http://www.niftybits.ukfsn.org/electronics/daily-water.png
http://www.niftybits.ukfsn.org/electronics/daily-water.gif

andy.

I've been through it again, and worked out the component values for the
second half of the circuit again. The new version is at:

http://www.niftybits.ukfsn.org/electronics/daily-water2.png
http://www.niftybits.ukfsn.org/electronics/daily-water2.gif

If there's something wrong with the basic idea of how i've laid it out,
i'd rather be told why than just have it written off.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with
HTML:
 or [attachment] in the subject line.[/QUOTE]

Observe the orientations of the 3 output transistors.

The output transistors are NPN types, but a PNP symbol is used.

If you use NPN output transistors, they are operating as emitter
followers, so the emitter voltage will always be at least .6 volts
more negative then the base voltage.

The base current for the output transistors comes entirely from the
base pull up resistor, which provides no current unless there is
voltage across it.  This uses up some more of the 12 volt supply.

The BD139 must sink lots more current through the output drive
resistor than that resistor ever supplies to the output transistors.

There is little reason to expect that what output drive current there
is will divide into three equal parts where it meets 3 bases.[/QUOTE]

Whoops.  I forgot that there is a 2.7 ohm coil in each emitter so
there is something that helps force the base drive to split 3 ways.
Sorry.

A

#### andy

Jan 1, 1970
0
andy said:
It's meant to send a 1 second pulse through an electromagnet at the end of
every day, to start a plant watering system, as i said in an earlier post.
I've tested it on breadboard, but with only one transistor and a single
coil winding. I'm trying splitting the coil into 3 parts to get more
current through it. It would be good to know what people think of this
circuit before i build the final version - it's only the second circuit
i've built.

http://www.niftybits.ukfsn.org/electronics/daily-water.png
http://www.niftybits.ukfsn.org/electronics/daily-water.gif

andy.

I've been through it again, and worked out the component values for the
second half of the circuit again. The new version is at:

http://www.niftybits.ukfsn.org/electronics/daily-water2.png
http://www.niftybits.ukfsn.org/electronics/daily-water2.gif

If there's something wrong with the basic idea of how i've laid it out,
i'd rather be told why than just have it written off.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with
HTML:
 or [attachment] in the subject line.[/QUOTE]

Observe the orientations of the 3 output transistors.

The output transistors are NPN types, but a PNP symbol is used.[/QUOTE]

i drew the circuit for PNPs, but bought NPN ones by mistake - hence the
wrong part no. (Maplins had hardly any power transistors, so i had to go
through the catalogue calling out numbers until i found some, and i think
i lost it somewhere along the line.)
[QUOTE]
If you use NPN output transistors, they are operating as emitter
followers, so the emitter voltage will always be at least .6 volts
more negative then the base voltage.

The base current for the output transistors comes entirely from the
base pull up resistor, which provides no current unless there is
voltage across it.  This uses up some more of the 12 volt supply.[/QUOTE]

I think these points are OK with PNP transistors there.
[QUOTE]
The BD139 must sink lots more current through the output drive
resistor than that resistor ever supplies to the output transistors.[/QUOTE]

That's one thing i'm not sure of - how to choose a value for a resistor
like this - as far as i can see all it has to do is pull the bases of
the output transistors back to the rail when the BD139 switches off.
[QUOTE]
There is little reason to expect that what output drive current there
is will divide into three equal parts where it meets 3 bases.[/QUOTE]

I think this was OK like you said in the follow up. i decided to split it
into 3 resistors anyhow, as it seems like it would be more stable.
[QUOTE]
I don't want to ruin all your fun by designing the circuit for you,
but think about these comments and show me what you come up with as a
result.[/QUOTE]

http://www.niftybits.ukfsn.org/electronics/daily-water3.png
http://www.niftybits.ukfsn.org/electronics/daily-water3.gif

I switched to NPNs for the output stage (having ended up buying them),
changed to a different low power transistor, and worked out all the
component values for the second half of the circuit again. The input stage
is much the same - i'm wondering whether to redo this as a
proper differential amplifier/comparator. The reason the resistor values
are all so high is I want the whole thing to run from a 50mA solar cell,
which according to the table i have only gives about 1.75mA average
current in uk winters, so any parts of the circuit that are always on need
to be kept as low power as possible.

one thing i'm not sure of is what happens if you try to drive a transistor
a long way past saturation - does the base current keep going up with the
collector current more or less constant, or is there some maximum base
current for any given collector current?

G

#### Gordon Youd

Jan 1, 1970
0
Would it not be better to use a 555 servo circuit, use the servo to open the
valve, less parts, less current used.

We are here to help not smack you down like some people.

Regards, Gordon.

----------------------------------------------------------------------------
----------------------
andy said:
andy said:
On Thu, 22 Jul 2004 18:41:48 +0100, andy wrote:

It's meant to send a 1 second pulse through an electromagnet at the end of
every day, to start a plant watering system, as i said in an earlier post.
I've tested it on breadboard, but with only one transistor and a single
coil winding. I'm trying splitting the coil into 3 parts to get more
current through it. It would be good to know what people think of this
circuit before i build the final version - it's only the second circuit
i've built.

http://www.niftybits.ukfsn.org/electronics/daily-water.png
http://www.niftybits.ukfsn.org/electronics/daily-water.gif

andy.

I've been through it again, and worked out the component values for the
second half of the circuit again. The new version is at:

http://www.niftybits.ukfsn.org/electronics/daily-water2.png
http://www.niftybits.ukfsn.org/electronics/daily-water2.gif

If there's something wrong with the basic idea of how i've laid it out,
i'd rather be told why than just have it written off.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with
HTML:
 or [attachment] in the subject line.[/QUOTE]

Observe the orientations of the 3 output transistors.

The output transistors are NPN types, but a PNP symbol is used.[/QUOTE]

i drew the circuit for PNPs, but bought NPN ones by mistake - hence the
wrong part no. (Maplins had hardly any power transistors, so i had to go
through the catalogue calling out numbers until i found some, and i think
i lost it somewhere along the line.)
[QUOTE]
If you use NPN output transistors, they are operating as emitter
followers, so the emitter voltage will always be at least .6 volts
more negative then the base voltage.

The base current for the output transistors comes entirely from the
base pull up resistor, which provides no current unless there is
voltage across it.  This uses up some more of the 12 volt supply.[/QUOTE]

I think these points are OK with PNP transistors there.
[QUOTE]
The BD139 must sink lots more current through the output drive
resistor than that resistor ever supplies to the output transistors.[/QUOTE]

That's one thing i'm not sure of - how to choose a value for a resistor
like this - as far as i can see all it has to do is pull the bases of
the output transistors back to the rail when the BD139 switches off.
[QUOTE]
There is little reason to expect that what output drive current there
is will divide into three equal parts where it meets 3 bases.[/QUOTE]

I think this was OK like you said in the follow up. i decided to split it
into 3 resistors anyhow, as it seems like it would be more stable.
[QUOTE]
I don't want to ruin all your fun by designing the circuit for you,
but think about these comments and show me what you come up with as a
result.[/QUOTE]

http://www.niftybits.ukfsn.org/electronics/daily-water3.png
http://www.niftybits.ukfsn.org/electronics/daily-water3.gif

I switched to NPNs for the output stage (having ended up buying them),
changed to a different low power transistor, and worked out all the
component values for the second half of the circuit again. The input stage
is much the same - i'm wondering whether to redo this as a
proper differential amplifier/comparator. The reason the resistor values
are all so high is I want the whole thing to run from a 50mA solar cell,
which according to the table i have only gives about 1.75mA average
current in uk winters, so any parts of the circuit that are always on need
to be kept as low power as possible.

one thing i'm not sure of is what happens if you try to drive a transistor
a long way past saturation - does the base current keep going up with the
collector current more or less constant, or is there some maximum base
current for any given collector current?

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with [html] or [attachment] in the subject line.
[/QUOTE]

J

#### John Fields

Jan 1, 1970
0
I've been through it again, and worked out the component values for the
second half of the circuit again. The new version is at:

http://www.niftybits.ukfsn.org/electronics/daily-water2.png
http://www.niftybits.ukfsn.org/electronics/daily-water2.gif

If there's something wrong with the basic idea of how i've laid it out,
i'd rather be told why than just have it written off.

---
OK, let's start with your output stage which, even after two
iterations and comments from John Popelish, you don't have drawn
correctly .

Instead of three TIP3055's beasts being driven as emitter followers,
why not use a single MOSFET which can handle a 13.3 amp load, (which
is what your three windings will look like in parallel) and make it
common source?

+12>-------+
|
S
VIN>-----G IRF4905
D
|
+----+----+------+
| | | |
[L1] [L2] [L3] [1N4001]
| | | |
GND>-------+----+----+------+

The only current you need to drive it with is what's needed to charge
and discharge the gate capacitance, and with a channel resistance of
20 milliohms and a 13.3 amp load, the IRF4905 will be dissipating
about 3.5 watts for the 1 second it'll be on, which is nothing.

Next, let's look at what you're using to drive the 2N3055's:

+12>------------------------+
|
[50R]
|
+----+------+
| | |
C | +--[10R]--->TO 2N3055'S
VIN>---[340K]--B Q1 | |
E | |
| C |
+---B Q2 |
| E |
[2K2] | |
| | C
+----+-+---B Q3
| E
[1K0] |
| |
GND>-------------------+----+

Notice that in order to keep the 2N3055's turned off, Q3 has to be
turned on, which will be doing your 1.2AH battery no favors since
it'll have to supply current through the 50 ohm resistor ***ALL THE
TIME***. Assuming Q1, Q2, and Q3 are collector-to-emitter shorts, 12V
through 50 ohms is 240mA, which means that the battery will be drained
in about 5 hours after the circuit is connected. Period. End of
story.

But, if you change your driver and output stage to look like this:

+12>-----------+----+
| |
[1K] |
| S
+---G IRF4905
| D
C |
VIN>--[10K]---B Q1 +----+----+------+
E | | | |K
| [L1] [L2] [L3] [1N4001]
| | | | |
GND>-----------+----+----+----+------+

You could use one of your 2N5551's for Q1, and then the only time
current will be drawn from the battery is during the time when the one
second pulse will drive the base of Q1 high.

Assuming no self-discharge and a fully charged battery, with a
capacity of 1.2AH and a load of 13.3A for one second every 24 hours,
the battery would last for 0.09 hours, which is 324 seconds. That
translates to 324 days... How big is your solar panel?

which brings us to the next thing: Generating the 1 second pulse
properly.

But first, let's see your comments on the foregoing and whether you
now understand _why_ your circuit is junk.

J

#### John Fields

Jan 1, 1970
0
Would it not be better to use a 555 servo circuit, use the servo to open the
valve, less parts, less current used.

We are here to help not smack you down like some people.

A

#### andy

Jan 1, 1970
0

i think you didn't see the version i posted after john popelish's comments:

http://www.niftybits.ukfsn.org/electronics/daily-water3.png
http://www.niftybits.ukfsn.org/electronics/daily-water3.gif
Instead of three TIP3055's beasts being driven as emitter followers,
why not use a single MOSFET which can handle a 13.3 amp load, (which
is what your three windings will look like in parallel) and make it
common source?

like i said in my reply to john popelish, they aren't meant to be NPN
emitter followers but PNP switches - I just bought the wrong part by
mistake, and then copied the part number into the diagram without noticing
it was the NPN version. my mistake, but that wasn't the mistake.
+12>-------+
|
S
VIN>-----G IRF4905
D
|
+----+----+------+
| | | |
[L1] [L2] [L3] [1N4001]
| | | |
GND>-------+----+----+------+

paralleling the coils and using a bigger transistor is fair enough -
that's just me not thinking. I used bipolar ones just to stick with what i
know for this project - it's only the second i've built.
The only current you need to drive it with is what's needed to charge
and discharge the gate capacitance, and with a channel resistance of
20 milliohms and a 13.3 amp load, the IRF4905 will be dissipating
about 3.5 watts for the 1 second it'll be on, which is nothing.

That does sound like a better idea given the number of bipolars i've had
to use to amplify the signal enough to drive the output stage.
Next, let's look at what you're using to drive the 2N3055's:

+12>------------------------+
|
[50R]
|
+----+------+
| | |
C | +--[10R]--->TO 2N3055'S
VIN>---[340K]--B Q1 | |
E | |
| C |
+---B Q2 |
| E |
[2K2] | |
| | C
+----+-+---B Q3
| E
[1K0] |
| |
GND>-------------------+----+

Notice that in order to keep the 2N3055's turned off, Q3 has to be
turned on, which will be doing your 1.2AH battery no favors since
it'll have to supply current through the 50 ohm resistor ***ALL THE
TIME***.

That's still the same mistake/misunderstanding over PNP/NPN transistors.
yes i did draw it wrong, but the version i put up in reply to john
popelish is right i think. in this version, that triple darlington is
meant to be to switch /on/ a PNP switch, not switch /off/ an NPN follower.
Assuming Q1, Q2, and Q3 are collector-to-emitter shorts, 12V
through 50 ohms is 240mA, which means that the battery will be drained
in about 5 hours after the circuit is connected. Period. End of
story.

Like i say, i did think about this, and apart from the mistake with the
part number (should have been tip2955 for the PNP design i drew), i think
the original circuit was OK for this. What i was trying to do was keep the
current through the two input stage transistors as low as possible (hence
the high resistor values), but not worry so much about the output stage,
because most of the current in this would be used by the coil anyhow.
But, if you change your driver and output stage to look like this:

+12>-----------+----+
| |
[1K] |
| S
+---G IRF4905
| D
C |
VIN>--[10K]---B Q1 +----+----+------+
E | | | |K
| [L1] [L2] [L3] [1N4001]
| | | | |
GND>-----------+----+----+----+------+

You could use one of your 2N5551's for Q1, and then the only time
current will be drawn from the battery is during the time when the one
second pulse will drive the base of Q1 high.

Assuming no self-discharge and a fully charged battery, with a
capacity of 1.2AH and a load of 13.3A for one second every 24 hours,
the battery would last for 0.09 hours, which is 324 seconds. That
translates to 324 days... How big is your solar panel?

Thinking of using a 50mA panel, which should give a long term average
current of about 1.8 mA minimum in uk winters, according to the table i
have.
which brings us to the next thing: Generating the 1 second pulse
properly.

But first, let's see your comments on the foregoing and whether you
now understand _why_ your circuit is junk.

that is what i wanted to know.

J

#### John Fields

Jan 1, 1970
0
i think you didn't see the version i posted after john popelish's comments:

http://www.niftybits.ukfsn.org/electronics/daily-water3.png
http://www.niftybits.ukfsn.org/electronics/daily-water3.gif
Instead of three TIP3055's beasts being driven as emitter followers,
why not use a single MOSFET which can handle a 13.3 amp load, (which
is what your three windings will look like in parallel) and make it
common source?

like i said in my reply to john popelish, they aren't meant to be NPN
emitter followers but PNP switches - I just bought the wrong part by
mistake, and then copied the part number into the diagram without noticing
it was the NPN version. my mistake, but that wasn't the mistake.
+12>-------+
|
S
VIN>-----G IRF4905
D
|
+----+----+------+
| | | |
[L1] [L2] [L3] [1N4001]
| | | |
GND>-------+----+----+------+

paralleling the coils and using a bigger transistor is fair enough -
that's just me not thinking. I used bipolar ones just to stick with what i
know for this project - it's only the second i've built.
The only current you need to drive it with is what's needed to charge
and discharge the gate capacitance, and with a channel resistance of
20 milliohms and a 13.3 amp load, the IRF4905 will be dissipating
about 3.5 watts for the 1 second it'll be on, which is nothing.

That does sound like a better idea given the number of bipolars i've had
to use to amplify the signal enough to drive the output stage.
Next, let's look at what you're using to drive the 2N3055's:

+12>------------------------+
|
[50R]
|
+----+------+
| | |
C | +--[10R]--->TO 2N3055'S
VIN>---[340K]--B Q1 | |
E | |
| C |
+---B Q2 |
| E |
[2K2] | |
| | C
+----+-+---B Q3
| E
[1K0] |
| |
GND>-------------------+----+

Notice that in order to keep the 2N3055's turned off, Q3 has to be
turned on, which will be doing your 1.2AH battery no favors since
it'll have to supply current through the 50 ohm resistor ***ALL THE
TIME***.

That's still the same mistake/misunderstanding over PNP/NPN transistors.
yes i did draw it wrong, but the version i put up in reply to john
popelish is right i think. in this version, that triple darlington is
meant to be to switch /on/ a PNP switch, not switch /off/ an NPN follower.
Assuming Q1, Q2, and Q3 are collector-to-emitter shorts, 12V
through 50 ohms is 240mA, which means that the battery will be drained
in about 5 hours after the circuit is connected. Period. End of
story.

Like i say, i did think about this, and apart from the mistake with the
part number (should have been tip2955 for the PNP design i drew), i think
the original circuit was OK for this. What i was trying to do was keep the
current through the two input stage transistors as low as possible (hence
the high resistor values), but not worry so much about the output stage,
because most of the current in this would be used by the coil anyhow.
But, if you change your driver and output stage to look like this:

+12>-----------+----+
| |
[1K] |
| S
+---G IRF4905
| D
C |
VIN>--[10K]---B Q1 +----+----+------+
E | | | |K
| [L1] [L2] [L3] [1N4001]
| | | | |
GND>-----------+----+----+----+------+

You could use one of your 2N5551's for Q1, and then the only time
current will be drawn from the battery is during the time when the one
second pulse will drive the base of Q1 high.

Assuming no self-discharge and a fully charged battery, with a
capacity of 1.2AH and a load of 13.3A for one second every 24 hours,
the battery would last for 0.09 hours, which is 324 seconds. That
translates to 324 days... How big is your solar panel?

Thinking of using a 50mA panel, which should give a long term average
current of about 1.8 mA minimum in uk winters, according to the table i
have.

---
If you take 13.3 amps out of the battery for 1 second it should take
7389 seconds to fill it back up at 1.8mA, but it'll take longer
because there is no free lunch. Figure about 50% more, so that'll be
about 11083 seconds, which is about 3.08 hours. Not bad, but that's
not counting the quiescent current of the circuitry which, if it's
1.8mA will keep the battery from charging in winter...
---
that is what i wanted to know.

---
You're welcome?

Oh, well, it's a slow day, so here are some thoughts about the front
end...

You've got basically four approaches to choose from when deciding how
to generate the 1s pulse:

1. 7555/7556
2. 4538/HC123
3. LM393 or some other comparator array.
4. Discretes/glue logic

The 7555/7556 route looks attractive because of low quiescent current
and a totem-pole output which would eliminate the wasted current drawn
by a pullup resistor-open collector feeding the driver.
Unfortunately, the slow-rising DC voltage from the LDR would have to
be differentiated and AC coupled into the timer, which would require
additional circuitry to do the job.

The 4538 looks very attractive because of a totem pole output and DC
coupled Schmitt trigger inputs, which could help to prevent
retriggering because of noise on the output of the LDR circuitry.
Unfortunately, the amount of hysteresis isn't specified. That is, I
couldn't find it in the spec's. Still in all, there are two timers in
each package, so the second one might be able to be used to hold off
the first for long enough until it gets dark enough outside that there
won't be a chance for a retrigger. That would also eliminate the need
for hysteresis and you could use Fairchild's 4538BC part which has no
hysteresis, but some other nice stuff.

Using LM393's looks attractive because of the flexibility allowed, but
because of no totem-pole output there'll always be a drain on the
battery which can be avoided by using one-shots like 4536's. Also,
LM393's have higher quiescent current requirements than CMOS does.

Discretes... Unless you're doing this because you want to learn how to
do it with discretes, forget it.

So, my vote is for the 4538, and I'll post a schematic of how to do
the whole thing to alt.binaries.schematics.electronic under "Water
timer" in a few minutes.

A

#### andy

Jan 1, 1970
0
Would it not be better to use a 555 servo circuit, use the servo to open the
valve, less parts, less current used.

The design i'm going for is like this:

| :#: iron cored electro- |
| :#: magnet. |
| m |
========|X------O ballcock valve with |
| magnet (m) on top. |
| |
| |
| |
| |
| |
| |
| |
| |
| |
| |
| |
| =============
| | slow drain
'--------------------------------' into drip hose.

The tank is normally empty, with the magnet holding the ballcock shut
against the electromagnet's core. Then a short pulse through the
electromagnet should make the ball drop down. The tank fills quickly,
the valve latches shut again, and then the tank drains slowly through the
drip hose.

the idea is:

a) a short pulse to cancel the magnet's field should take less current
than having to open a valve against friction. and much less than having to
hold a solenoid valve open for the whole watering cycle.

b) the system always comes back to a stable off (no watering) state by a
purely mechanical process once the initial pulse has been sent. i.e. if
the power goes, it can't get stuck open.

c) it's easy to build out of common or garden parts - don't have to buy an
expensive electrically operated valve.

A

#### andy

Jan 1, 1970
0

why? where is the energy lost?
Figure about 50% more, so that'll be
about 11083 seconds, which is about 3.08 hours. Not bad, but that's
not counting the quiescent current of the circuitry which, if it's
1.8mA will keep the battery from charging in winter...

that's why i was trying to keep that as low as possible. The final design
i posted, this is about 0.3mA, I think.
---
You're welcome?

Oh, well, it's a slow day, so here are some thoughts about the front
end...

You've got basically four approaches to choose from when deciding how
to generate the 1s pulse:

1. 7555/7556
2. 4538/HC123
3. LM393 or some other comparator array.
4. Discretes/glue logic

The 7555/7556 route looks attractive because of low quiescent current
and a totem-pole output which would eliminate the wasted current drawn
by a pullup resistor-open collector feeding the driver.
Unfortunately, the slow-rising DC voltage from the LDR would have to
be differentiated and AC coupled into the timer, which would require
additional circuitry to do the job.

The 4538 looks very attractive because of a totem pole output and DC
coupled Schmitt trigger inputs, which could help to prevent
retriggering because of noise on the output of the LDR circuitry.
Unfortunately, the amount of hysteresis isn't specified. That is, I
couldn't find it in the spec's. Still in all, there are two timers in
each package, so the second one might be able to be used to hold off
the first for long enough until it gets dark enough outside that there
won't be a chance for a retrigger. That would also eliminate the need
for hysteresis and you could use Fairchild's 4538BC part which has no
hysteresis, but some other nice stuff.

Doing it that way is probably better - i designed some hysteresis into my
circuit, but a hold off period is more likely to work right i think.
Using LM393's looks attractive because of the flexibility allowed, but
because of no totem-pole output there'll always be a drain on the
battery which can be avoided by using one-shots like 4536's. Also,
LM393's have higher quiescent current requirements than CMOS does.

Discretes... Unless you're doing this because you want to learn how to
do it with discretes, forget it.

i was - is there any reason why the configuration i've drawn isn't a good
way of doing it? i.e. the bit where the emitter of the transistor is
connected to the wiper of the pot? I thought of doing it using a circuit
from horowitz and hill for a differential amplifier, but the way i thought
of seemed neater for something where the actual level it switches at isn't
so important.
So, my vote is for the 4538, and I'll post a schematic of how to do
the whole thing to alt.binaries.schematics.electronic under "Water
timer" in a few minutes.

the 4538 does look like it would be good for this - i was also thinking of
adding a switch to select whether to water in morning, evening or both, so
the rising and falling edge inputs could be useful. i'll have a go at
working out a circuit with it, then compare to the one you've posted.

Replies
14
Views
1K
K
Replies
8
Views
990
Richard Henry
R
G
Replies
16
Views
1K
E
Replies
7
Views
2K
E
Replies
1
Views
440