control the electrical appliances using PC

Hi,

I used the Circuit Diagram mentioned in the below link to "control the electrical appliances using PC". But it does not work. Did any one worked recently on this, if yes does it works?

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I used the following components

  1. 6 VDC Relay
  2. C2235 NPN transistor
  3. 1N4007 diode
  4. 4.7 K resisitor
  5. LED

Only LED glows when parallel port passes +5v. As it is metioned in the link relays switches when parallel port passes +5v which is not happening in my case. Please help me in fixing issue. Relay switch is happening in before parallel port passes +5v.

Reply to
ryadav
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what is the wrong in the circuit? why it does not work?

| Vcc for Relay

| |

/ -------+--------| |---------|

P data 4.7K B / c | 1N4002 |_| |-------

-----------/\\/\\/\\/\\---------+-----| NPN _____ | relay |home device | \\ e /_\\ |-| |------- LED X \\ +--------| |---------|

P Ground | | |

----------------------------+-------+--------+

Relay Ground |

---------------------------------------------+

Reply to
ryadav

Relay will be always glows since diagram shows relay VCC is passed to cathod end of diode and relay ground given to anode end of diode. should it be like that?

Reply to
ryadav

The circuit is wrong, use the one from

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Reply to
cbm5

That circuit is defective.

The base-emitter junction forward voltage should never exceed about 0.7V. That's not enough to light the LED which is therefore spurious. You say the LED lights which suggests that you miswired the transistor.

Graham

Reply to
Pooh Bear

The diode is there to 'catch' the flyback pulse when the relay switches off ( inductive parts have stored energy ). It's shown connected correctly.

Graham

Reply to
Pooh Bear

cbm5 is correct.

The circuit you posted is wrong. The relay should be in series with the transistor not in parallel. Use the link he supplied.

Graham

Reply to
Pooh Bear

it needs extra power for the relay. pin 1 of the joystick socket has 5V which might be anough, otherwise use a 6V plugpack.

Bye. Jasen

Reply to
Jasen Betts

that circuit in that document is bullshit, try the coffee howto instead

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It gives this circuit which is very conventional. you ca probably make it work with the parts you have.

Vcc | +------+ | __|__ Relay /^\\ Diode 1N4002 Coil /---\\ | | +------+ | | / 4.7K B |/ C parallel port >-\\/\\/\\/\\/---| NPN Transistor: BC547A or 2N2222A data pi |\\ E | \\ V parallel port >--------------+ ground pin |

I'm submitting a bug report on the Home-Electrical-Control document.

Bye. Jasen

Reply to
Jasen Betts

Thank you all, finally it worked with new circuit provided in the URL posted.

:-)

Reply to
ryadav

In addition to what others have posted, note that NT-based versions of Windows (NT, XP, 2000) don't allow direct access to the ports, so you will probably need a special ring 0 driver like GIVEIO for them. That won't be needed for Win9x and earlier.

Best regards,

Bob Masta dqatechATdaqartaDOTcom D A Q A R T A Data AcQuisition And Real-Time Analysis

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Home of DaqGen, the FREEWARE signal generator

Reply to
Bob Masta

On 11 Dec 2005 10:26:51 -0800, via , "ryadav" spake thusly:

Quite a bit seems to be wrong. The thing that catches my attention is that, as drawn, the relay will always be "on," regardless of the state of the applied data. You've got "Vcc" and "Relay Ground" hard-wired across the relay.

Reply to
Big Mouth Billy Bass

if the power supply for the relay is short-circuit protected and rather wimpy and the NPN has enough gain. it could work by the NPN shorting out the supply and turning the relay off.

if not the best you can hope of is for it to fail in an unspectacular way.

Bye. Jasen

Reply to
Jasen Betts

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