Converting Ladder to cmos logic

Allen Bong · Oct 27, 2008

Hi group,

+12v
|
NC |
| |, .-. |
IN -+--------,-´-------( X )-+
| ´| | '-' |
| |
| | |NO .-----. |
+---| |--+------| |--+
| | | | '-----' |
| | RELAY |
| T | |
| --- | |
+---o o--+ |
|
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

I have the above ladder diagram consisting a relay, a buzzer and a
push button

switch.

I would like to sound the buzzer "x" when the input is low and when
the button is

pressed the relay would be switched on and holds itself through the
"NO" contact and

the buzzer is switched off due to the "NC" contact is opened. Both
the NO and NC

contacts belong to the same relay. When the input goes high, the
relay releases and

the circuit goes back to normal waiting for the next alarm.

I would like to convert this into a circuit using cmos chips. I use
an AND and

inverters and a 4013 FF and came out with the circuit below.

VCC
VCC +
.-. |
ALARM=L | |10K .-----.
NORM=H | | | |BUZZER
'-' '-----'
IN|\ | _4081 |
-| >O---+-------| \ __4081 |
|/ | | | )------| \ ___ |/
| +--|__/ | )--|___|--| VCC
| +-----+ +---|__/ 10K |> |+
| | | | | == | o | | | | /-\
| PB |=| | | === |
| o | | | GND |
| | | | +-------------------+
| | | | .-----. .-.
| === | +--|Q S D|-VCC | |
| GND | | | /| | |10K
| | |_ <|----O< |---+ '-'
| | |Q R | \| | |
| | '-----' | |
| | | 4013 | == | | | | GND
+----------|---------|--------------+
| | H -> L
| |\ |
+---| >O--+
|/
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

The C and R at the set input of the 4013 is a power-up set to make the
Q output high

during power on. The PB silences the buzzer until a new alarm is
detected.

The circuit seems too complicated and I was cracking my head how to
simply it. I

would also like the 4013 to be toggled by both H to L as well as L to
H transitions.

How should I do it?

Is there any simpler way to do it? Any help very much appreciated.

Allen

Allen

ehsjr · Oct 28, 2008

Allen Bong wrote:

Hi group,

+12v
|
NC |
| |, .-. |
IN -+--------,-´-------( X )-+
| ´| | '-' |
| |
| | |NO .-----. |
+---| |--+------| |--+
| | | | '-----' |
| | RELAY |
| T | |
| --- | |
+---o o--+ |
|
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

I have the above ladder diagram consisting a relay, a buzzer and a
push button

switch.

I would like to sound the buzzer "x" when the input is low and when
the button is

pressed the relay would be switched on and holds itself through the
"NO" contact and

the buzzer is switched off due to the "NC" contact is opened. Both
the NO and NC

contacts belong to the same relay. When the input goes high, the
relay releases and

the circuit goes back to normal waiting for the next alarm.

I would like to convert this into a circuit using cmos chips. I use
an AND and

inverters and a 4013 FF and came out with the circuit below.

VCC
VCC +
.-. |
ALARM=L | |10K .-----.
NORM=H | | | |BUZZER
'-' '-----'
IN|\ | _4081 |
-| >O---+-------| \ __4081 |
|/ | | | )------| \ ___ |/
| +--|__/ | )--|___|--| VCC
| +-----+ +---|__/ 10K |> |+
| | | | | ===
| o | | | | /-\
| PB |=| | | === |
| o | | | GND |
| | | | +-------------------+
| | | | .-----. .-.
| === | +--|Q S D|-VCC | |
| GND | | | /| | |10K
| | |_ <|----O< |---+ '-'
| | |Q R | \| | |
| | '-----' | |
| | | 4013 | ===
| | | | GND
+----------|---------|--------------+
| | H -> L
| |\ |
+---| >O--+
|/
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

The C and R at the set input of the 4013 is a power-up set to make the
Q output high

during power on. The PB silences the buzzer until a new alarm is
detected.

The circuit seems too complicated and I was cracking my head how to
simply it. I

would also like the 4013 to be toggled by both H to L as well as L to
H transitions.

How should I do it?

Is there any simpler way to do it? Any help very much appreciated.

Allen

You may consider this simpler than the relay or the CMOS
approach:

+12 ---+------------+----+-------+
| | k| |
| | [D1] [Buzzer]
| | | |
[1K] [1K] +-------+
| | |
| | c/
| +----------| NPN
| | e\
| --- |
| \ / SCR [D2]
| --- |k
| _ / | |
+---o o---+ +------------+
|
IN-----------------+

The SCR does not conduct until the putton is pressed.
Because the SCR is not conducting, the base of the NPN
is +12 and the NPN can conduct when IN goes low. When
that happens, the buzzer energizes through the NPN.
When the button is pressed, the SCR conducts, and the
base of the NPN goes low, turning the NPN off and
de-energizing the buzzer. The SCR will continue to
conduct, even after the button is released, until IN
goes high.

D2 creates a voltage drop to compensate for the voltage
drop across the SCR so that the emitter and the base
are at about the same level. D1 absorbs inductive kick
from the buzzer, if needed.

Ed

Allen Bong · Oct 29, 2008

On Oct 28, 10:58 am, ehsjr <eh...@NOSPAMverizon.net> wrote:

Allen Bong wrote:
Hi group,

+12v
|
NC |
| |, .-. |
IN -+--------,-´-------( X )-+
| ´| | '-' |
| |
| | |NO .-----. |
+---| |--+------| |--+
| | | | '-----' |
| | RELAY |
| T | |
| --- | |
+---o o--+ |
|
|
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

I have the above ladder diagram consisting a relay, a buzzer and a
push button

switch.

I would like to sound the buzzer "x" when the input is low and when
the button is

pressed the relay would be switched on and holds itself through the
"NO" contact and

the buzzer is switched off due to the "NC" contact is opened. Both
the NO and NC

contacts belong to the same relay. When the input goes high, the
relay releases and

the circuit goes back to normal waiting for the next alarm.

I would like to convert this into a circuit using cmos chips. I use
an AND and

inverters and a 4013 FF and came out with the circuit below.

VCC
VCC +
.-. |
ALARM=L | |10K .-----.
NORM=H | | | |BUZZER
'-' '-----'
IN|\ | _4081 |
-| >O---+-------| \ __4081 |
|/ | | | )------| \ ___ |/
| +--|__/ | )--|___|--| VCC
| +-----+ +---|__/ 10K |> |+
| | | | | ==> > | o | | | | /-\
| PB |=| | | === |
| o | | | GND |
| | | | +-------------------+
| | | | .-----. .-.
| === | +--|Q S D|-VCC | |
| GND | | | /| | |10K
| | |_ <|----O< |---+ '-'
| | |Q R | \| | |
| | '-----' | |
| | | 4013 | ==> > | | | | GND
+----------|---------|--------------+
| | H -> L
| |\ |
+---| >O--+
|/
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

The C and R at the set input of the 4013 is a power-up set to make the
Q output high

during power on. The PB silences the buzzer until a new alarm is
detected.

The circuit seems too complicated and I was cracking my head how to
simply it. I

would also like the 4013 to be toggled by both H to L as well as L to
H transitions.

How should I do it?

Is there any simpler way to do it? Any help very much appreciated.

Allen

You may consider this simpler than the relay or the CMOS
approach:

+12 ---+------------+----+-------+
| | k| |
| | [D1] [Buzzer]
| | | |
[1K] [1K] +-------+
| | |
| | c/
| +----------| NPN
| | e\
| --- |
| \ / SCR [D2]
| --- |k
| _ / | |
+---o o---+ +------------+
|
IN-----------------+

The SCR does not conduct until the putton is pressed.
Because the SCR is not conducting, the base of the NPN
is +12 and the NPN can conduct when IN goes low. When
that happens, the buzzer energizes through the NPN.
When the button is pressed, the SCR conducts, and the
base of the NPN goes low, turning the NPN off and
de-energizing the buzzer. The SCR will continue to
conduct, even after the button is released, until IN
goes high.

D2 creates a voltage drop to compensate for the voltage
drop across the SCR so that the emitter and the base
are at about the same level. D1 absorbs inductive kick
from the buzzer, if needed.

Ed- Hide quoted text -

- Show quoted text -

Wow, I never thought that the solution can be this simple. I'll try
it out tonight and let you know about it tomorrow. Have to go out and
buy a SCR.

Allen

Allen Bong · Oct 30, 2008

On Oct 28, 10:58 am, ehsjr <eh...@NOSPAMverizon.net> wrote:

Allen Bong wrote:
Hi group,

+12v
|
NC |
| |, .-. |
IN -+--------,-´-------( X )-+
| ´| | '-' |
| |
| | |NO .-----. |
+---| |--+------| |--+
| | | | '-----' |
| | RELAY |
| T | |
| --- | |
+---o o--+ |
|
|
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

I have the above ladder diagram consisting a relay, a buzzer and a
push button

switch.

I would like to sound the buzzer "x" when the input is low and when
the button is

pressed the relay would be switched on and holds itself through the
"NO" contact and

the buzzer is switched off due to the "NC" contact is opened. Both
the NO and NC

contacts belong to the same relay. When the input goes high, the
relay releases and

the circuit goes back to normal waiting for the next alarm.

I would like to convert this into a circuit using cmos chips. I use
an AND and

inverters and a 4013 FF and came out with the circuit below.

VCC
VCC +
.-. |
ALARM=L | |10K .-----.
NORM=H | | | |BUZZER
'-' '-----'
IN|\ | _4081 |
-| >O---+-------| \ __4081 |
|/ | | | )------| \ ___ |/
| +--|__/ | )--|___|--| VCC
| +-----+ +---|__/ 10K |> |+
| | | | | ==> > | o | | | | /-\
| PB |=| | | === |
| o | | | GND |
| | | | +-------------------+
| | | | .-----. .-.
| === | +--|Q S D|-VCC | |
| GND | | | /| | |10K
| | |_ <|----O< |---+ '-'
| | |Q R | \| | |
| | '-----' | |
| | | 4013 | ==> > | | | | GND
+----------|---------|--------------+
| | H -> L
| |\ |
+---| >O--+
|/
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

The C and R at the set input of the 4013 is a power-up set to make the
Q output high

during power on. The PB silences the buzzer until a new alarm is
detected.

The circuit seems too complicated and I was cracking my head how to
simply it. I

would also like the 4013 to be toggled by both H to L as well as L to
H transitions.

How should I do it?

Is there any simpler way to do it? Any help very much appreciated.

Allen

You may consider this simpler than the relay or the CMOS
approach:

+12 ---+------------+----+-------+
| | k| |
| | [D1] [Buzzer]
| | | |
[1K] [1K] +-------+
| | |
| | c/
| +----------| NPN
| | e\
| --- |
| \ / SCR [D2]
| --- |k
| _ / | |
+---o o---+ +------------+
|
IN-----------------+

The SCR does not conduct until the putton is pressed.
Because the SCR is not conducting, the base of the NPN
is +12 and the NPN can conduct when IN goes low. When
that happens, the buzzer energizes through the NPN.
When the button is pressed, the SCR conducts, and the
base of the NPN goes low, turning the NPN off and
de-energizing the buzzer. The SCR will continue to
conduct, even after the button is released, until IN
goes high.

D2 creates a voltage drop to compensate for the voltage
drop across the SCR so that the emitter and the base
are at about the same level. D1 absorbs inductive kick
from the buzzer, if needed.

Ed- Hide quoted text -

- Show quoted text -

You may consider this simpler than the relay or the CMOS
approach:

+12 ---+------------+----+-------+
| | k| |
| | [D1] [Buzzer]
| | | |
[1K] [1K] +-------+
| | |
| | c/
| +----------| NPN
| | e\
| --- |
| \ / SCR [D2]
| --- |k
| _ / | |
+---o o---+ +------------+
|
IN-----------------+

I tried your circuit and it works. If I change the NPN transistor to
a Darlington like 2sd768, do I need to add one more diode in series
with D2? The SCR I used is FOR3G.

Thank you.

Allen

ehsjr · Oct 31, 2008

Allen Bong wrote:

On Oct 28, 10:58 am, ehsjr <eh...@NOSPAMverizon.net> wrote:

Allen Bong wrote:

Hi group,

+12v
|
NC |
| |, .-. |
IN -+--------,-´-------( X )-+
| ´| | '-' |
| |
| | |NO .-----. |
+---| |--+------| |--+
| | | | '-----' |
| | RELAY |
| T | |
| --- | |
+---o o--+ |
|
|
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

I have the above ladder diagram consisting a relay, a buzzer and a
push button

switch.

I would like to sound the buzzer "x" when the input is low and when
the button is

pressed the relay would be switched on and holds itself through the
"NO" contact and

the buzzer is switched off due to the "NC" contact is opened. Both
the NO and NC

contacts belong to the same relay. When the input goes high, the
relay releases and

the circuit goes back to normal waiting for the next alarm.

I would like to convert this into a circuit using cmos chips. I use
an AND and

inverters and a 4013 FF and came out with the circuit below.

VCC
VCC +
.-. |
ALARM=L | |10K .-----.
NORM=H | | | |BUZZER
'-' '-----'
IN|\ | _4081 |
-| >O---+-------| \ __4081 |
|/ | | | )------| \ ___ |/
| +--|__/ | )--|___|--| VCC
| +-----+ +---|__/ 10K |> |+
| | | | | ===
| o | | | | /-\
| PB |=| | | === |
| o | | | GND |
| | | | +-------------------+
| | | | .-----. .-.
| === | +--|Q S D|-VCC | |
| GND | | | /| | |10K
| | |_ <|----O< |---+ '-'
| | |Q R | \| | |
| | '-----' | |
| | | 4013 | ===
| | | | GND
+----------|---------|--------------+
| | H -> L
| |\ |
+---| >O--+
|/
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

The C and R at the set input of the 4013 is a power-up set to make the
Q output high

during power on. The PB silences the buzzer until a new alarm is
detected.

The circuit seems too complicated and I was cracking my head how to
simply it. I

would also like the 4013 to be toggled by both H to L as well as L to
H transitions.

How should I do it?

Is there any simpler way to do it? Any help very much appreciated.

Allen

You may consider this simpler than the relay or the CMOS
approach:

+12 ---+------------+----+-------+
| | k| |
| | [D1] [Buzzer]
| | | |
[1K] [1K] +-------+
| | |
| | c/
| +----------| NPN
| | e\
| --- |
| \ / SCR [D2]
| --- |k
| _ / | |
+---o o---+ +------------+
|
IN-----------------+

The SCR does not conduct until the putton is pressed.
Because the SCR is not conducting, the base of the NPN
is +12 and the NPN can conduct when IN goes low. When
that happens, the buzzer energizes through the NPN.
When the button is pressed, the SCR conducts, and the
base of the NPN goes low, turning the NPN off and
de-energizing the buzzer. The SCR will continue to
conduct, even after the button is released, until IN
goes high.

D2 creates a voltage drop to compensate for the voltage
drop across the SCR so that the emitter and the base
are at about the same level. D1 absorbs inductive kick
from the buzzer, if needed.

Ed- Hide quoted text -

- Show quoted text -

You may consider this simpler than the relay or the CMOS
approach:

+12 ---+------------+----+-------+
| | k| |
| | [D1] [Buzzer]
| | | |
[1K] [1K] +-------+
| | |
| | c/
| +----------| NPN
| | e\
| --- |
| \ / SCR [D2]
| --- |k
| _ / | |
+---o o---+ +------------+
|
IN-----------------+

I tried your circuit and it works. If I change the NPN transistor to
a Darlington like 2sd768, do I need to add one more diode in series
with D2? The SCR I used is FOR3G.

Thank you.

Allen

Why do you want to use a darlington?

Setting my question aside, a second diode should
not be necessary if you change to a darlington -
you could even remove D2 when you use it.

With the SCR you are using, you should add a 1/4 watt,
1K resistor from the gate to the cathode, according to
the datasheet. You may need to reduce the value of the
1K that is connected between +12 and the pushbutton to
330 ohms.

Ed

Allen Bong · Oct 31, 2008

On Oct 31, 9:43 am, ehsjr <eh...@NOSPAMverizon.net> wrote:

Allen Bong wrote:
On Oct 28, 10:58 am, ehsjr <eh...@NOSPAMverizon.net> wrote:

Allen Bong wrote:

Hi group,

+12v
|
NC |
| |, .-. |
IN -+--------,-´-------( X )-+
| ´| | '-' |
| |
| | |NO .-----. |
+---| |--+------| |--+
| | | | '-----' |
| | RELAY |
| T | |
| --- | |
+---o o--+ |
|
|
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

I have the above ladder diagram consisting a relay, a buzzer and a
push button

switch.

I would like to sound the buzzer "x" when the input is low and when
the button is

pressed the relay would be switched on and holds itself through the
"NO" contact and

the buzzer is switched off due to the "NC" contact is opened. Both
the NO and NC

contacts belong to the same relay. When the input goes high, the
relay releases and

the circuit goes back to normal waiting for the next alarm.

I would like to convert this into a circuit using cmos chips. I use
an AND and

inverters and a 4013 FF and came out with the circuit below.

VCC
VCC +
.-. |
ALARM=L | |10K .-----.
NORM=H | | | |BUZZER
'-' '-----'
IN|\ | _4081 |
-| >O---+-------| \ __4081 |
|/ | | | )------| \ ___ |/
| +--|__/ | )--|___|--| VCC
| +-----+ +---|__/ 10K |> |+
| | | | | ==> >>> | o | | | | /-\
| PB |=| | | === |
| o | | | GND |
| | | | +-------------------+
| | | | .-----. .-.
| === | +--|Q S D|-VCC | |
| GND | | | /| | |10K
| | |_ <|----O< |---+ '-'
| | |Q R | \| | |
| | '-----' | |
| | | 4013 | ==> >>> | | | | GND
+----------|---------|--------------+
| | H -> L
| |\ |
+---| >O--+
|/
(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)

The C and R at the set input of the 4013 is a power-up set to make the
Q output high

during power on. The PB silences the buzzer until a new alarm is
detected.

The circuit seems too complicated and I was cracking my head how to
simply it. I

would also like the 4013 to be toggled by both H to L as well as L to
H transitions.

How should I do it?

Is there any simpler way to do it? Any help very much appreciated.

Allen

You may consider this simpler than the relay or the CMOS
approach:

+12 ---+------------+----+-------+
| | k| |
| | [D1] [Buzzer]
| | | |
[1K] [1K] +-------+
| | |
| | c/
| +----------| NPN
| | e\
| --- |
| \ / SCR [D2]
| --- |k
| _ / | |
+---o o---+ +------------+
|
IN-----------------+

The SCR does not conduct until the putton is pressed.
Because the SCR is not conducting, the base of the NPN
is +12 and the NPN can conduct when IN goes low. When
that happens, the buzzer energizes through the NPN.
When the button is pressed, the SCR conducts, and the
base of the NPN goes low, turning the NPN off and
de-energizing the buzzer. The SCR will continue to
conduct, even after the button is released, until IN
goes high.

D2 creates a voltage drop to compensate for the voltage
drop across the SCR so that the emitter and the base
are at about the same level. D1 absorbs inductive kick
from the buzzer, if needed.

Ed- Hide quoted text -

- Show quoted text -

You may consider this simpler than the relay or the CMOS
approach:

+12 ---+------------+----+-------+
| | k| |
| | [D1] [Buzzer]
| | | |
[1K] [1K] +-------+
| | |
| | c/
| +----------| NPN
| | e\
| --- |
| \ / SCR [D2]
| --- |k
| _ / | |
+---o o---+ +------------+
|
IN-----------------+

I tried your circuit and it works. If I change the NPN transistor to
a Darlington like 2sd768, do I need to add one more diode in series
with D2? The SCR I used is FOR3G.

Thank you.

Allen

Why do you want to use a darlington?

Setting my question aside, a second diode should
not be necessary if you change to a darlington -
you could even remove D2 when you use it.

With the SCR you are using, you should add a 1/4 watt,
1K resistor from the gate to the cathode, according to
the datasheet. You may need to reduce the value of the
1K that is connected between +12 and the pushbutton to
330 ohms.

Ed- Hide quoted text -

- Show quoted text -

Why do you want to use a darlington?

Setting my question aside, a second diode should
not be necessary if you change to a darlington -
you could even remove D2 when you use it.

With the SCR you are using, you should add a 1/4 watt,
1K resistor from the gate to the cathode, according to
the datasheet. You may need to reduce the value of the
1K that is connected between +12 and the pushbutton to
330 ohms.

Ed

The reason is I have about 30 pieces of 2sd768 sitting there idling.
The buzzer that I am using has a built-in circuit which flashes some
LED as well as a loud buzzer. I am afraid the current may be too high
and I don't have any TO220 transistors handy unless I go to purchase
them from the nearby town. And I have to make about 25 of these
circuits on a 1" x 1" pcb.

Thanks for your tips on using the SCR that I bought.

Regards

Allen

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