Anyone care to help out a circuit design n00b?

Corn-Picker · 22nd December 2003, 09:49 PM

Ok, here's the background. I have three gauges for my car I am installing. These gauges have warning lights, for example, if gas temperatures at the exhaust manifold exceed 1600°F then a warning LED lights. A LED is OK, but I want a loud buzzer to go off when the LED lights. I have designed a circuit for this but I'm not sure if it will work

Any suggestions or comments are greatly appreciated.

Corn-Picker · 22nd December 2003, 10:36 PM

A few things I forgot to mention, the MOSFET I got from Radio Shack requires 2-4 Volts (positive at the gate) to open the drain-source path. This should not be a problem, as I think the relay has a higher resistance than the MOSFET, and if not I can always throw a resistor in there after the relay. The only thing I need for sure is 9.6 Volts minimum across the relay to get the relay to open up.

The reason I stuck the diodes in there is because I am afraid of too much current being demanded by other gauges. Only one of the LEDs may be on at any one time. For instance, my oil temperature may be too high but my oil pressure and exhaust gas temperature may be fine, so only one LED out of the three would be on.

The values of 12 and 10.4 Volts were measured with a multimeter, so there's a 1.6 Volt drop across the LED.

Any help is appreciated, even condescending remarks like "you can't use a diode that way dumbass" would be appreciated.

Any just because I like you guys so much you get the dancing banana smiley

sreten · 23rd December 2003, 12:03 AM

Looks too complicated to me !

What are the current requirements of the buzzer ?
(or whats its impedance)

What are the current requirements of the relay ?
(or whats its impedance and minimum voltage)

What current are the LEd's run at when lit ?
(or what is the load resistance to ground ?)

sreten.

sam9 · 23rd December 2003, 02:56 AM

Think you need to start with a schematic of the existing, standard circuitry if you possibly can. With luch you can loose the MOSFET and the diodes and do it all with some relays and resistors.

Corn-Picker · 23rd December 2003, 02:56 PM

Here's a picture of the original circuit with proposed additions. I'm using a MOSFET because I'm afraid anything else might draw too much current from the LED's circuits... I've also added a diode across the relay's coil to prevent voltage spikes from destroying anything.

Edit: Almost forgot. LED current is 8.3 mA. MOSFET impedance is 0.6 ohms. I'm uncertain of relay coil impedance at the moment, but I do have that number at home... Piezo buzzer current requirement is about 80 mA. A npn transistor would probably do the job fine, but component cost isn't much of an object (do I spend $0.79 on one BPT from Radio Shack or $1.95 on one MOSFET from Radio Shack, not much of an issue for a one time thing).

li_gangyi · 23rd December 2003, 03:53 PM

do you really have to use a relay?? seems like the relay will take more current than just the piezo buzer...and since u are going for a mosfet...might as well drive the buzzer straight...

sreten · 23rd December 2003, 04:23 PM

What we can't see is whether the LED are turned on
by completing the ground connection or the 12V.

I can't see the point of the relay.

If they are turned by completing ground then :

Connect the emitter of a PNP transistor to 12V via a 5R resistor.

Connect the collector via the buzzer to earth.

Connect the base to the points marked 10.4 volts via 3 diodes.

If any LED lights up, the buzzer will sound.

(LED voltage = 1.6 V, diode drop = 0.6V, base-emitter
drop=0.6V, leaving 0.4V across 5R, I = V/R = 80mA.)

If they are turned on by being connected to 12V then
a NPN emitter follower can replace the FET and relay
in your circuit.

sreten.

li_gangyi · 23rd December 2003, 04:57 PM

yeah...I dun see a need for a relay till u're tryin to isolate 2 different circuits...but need a way to trigger another...

Corn-Picker · 23rd December 2003, 05:55 PM

Quote:

Originally posted by li_gangyi
do you really have to use a relay?? seems like the relay will take more current than just the piezo buzer...and since u are going for a mosfet...might as well drive the buzzer straight...

You're right. I had the relay in there and never questioned myself as to whether or not it was a necessary piece

I'll just drive the buzzer with the MOSFET.

I have to check the LED leads when the LED is off, all of the above voltages were measured when the LEDs were on.

Thanks a lot for the advice guys. I look foward to working on this tonight and getting my car's gauges to work the way I want them to

Corn-Picker · 24th December 2003, 11:24 PM

Ok, I did some more multimetering. If the LED is on (as in the pictures above), the positive side is 12 V and the negative side is 10.4 V. If the LED is off, the positive side is still 12 V but the negative side is 10.8 V. Is there any way to tell from these numbers whether the LED is turned on by completeing the ground or 12 V connection?

If not, how can I tell whether the LED is turned on by completing the 12 V side or completing the ground side?

Thanks again, I think I'm close to getting this to work

23rd December 2003, 03:53 PM	#6
li_gangyi diyAudio Member Join Date: Jun 2003 Location: Woodlands Circle	do you really have to use a relay?? seems like the relay will take more current than just the piezo buzer...and since u are going for a mosfet...might as well drive the buzzer straight... __________________ Kids in the back seat cause accidents...Accidents in the back seat cause kids...

23rd December 2003, 04:57 PM	#8
li_gangyi diyAudio Member Join Date: Jun 2003 Location: Woodlands Circle	yeah...I dun see a need for a relay till u're tryin to isolate 2 different circuits...but need a way to trigger another... __________________ Kids in the back seat cause accidents...Accidents in the back seat cause kids...

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22nd December 2003, 10:36 PM	#2
Corn-Picker diyAudio Member Join Date: Oct 2002 Location: Pennsylvania	A few things I forgot to mention, the MOSFET I got from Radio Shack requires 2-4 Volts (positive at the gate) to open the drain-source path. This should not be a problem, as I think the relay has a higher resistance than the MOSFET, and if not I can always throw a resistor in there after the relay. The only thing I need for sure is 9.6 Volts minimum across the relay to get the relay to open up. The reason I stuck the diodes in there is because I am afraid of too much current being demanded by other gauges. Only one of the LEDs may be on at any one time. For instance, my oil temperature may be too high but my oil pressure and exhaust gas temperature may be fine, so only one LED out of the three would be on. The values of 12 and 10.4 Volts were measured with a multimeter, so there's a 1.6 Volt drop across the LED. Any help is appreciated, even condescending remarks like "you can't use a diode that way dumbass" would be appreciated. Any just because I like you guys so much you get the dancing banana smiley

23rd December 2003, 12:03 AM	#3
sreten diyAudio Member Join Date: Nov 2003 Location: Brighton UK	Looks too complicated to me ! What are the current requirements of the buzzer ? (or whats its impedance) What are the current requirements of the relay ? (or whats its impedance and minimum voltage) What current are the LEd's run at when lit ? (or what is the load resistance to ground ?) sreten.

23rd December 2003, 02:56 AM	#4
sam9 diyAudio Member Join Date: Jun 2002 Location: Left Coast	Think you need to start with a schematic of the existing, standard circuitry if you possibly can. With luch you can loose the MOSFET and the diodes and do it all with some relays and resistors.

23rd December 2003, 04:23 PM	#7
sreten diyAudio Member Join Date: Nov 2003 Location: Brighton UK	What we can't see is whether the LED are turned on by completing the ground connection or the 12V. I can't see the point of the relay. If they are turned by completing ground then : Connect the emitter of a PNP transistor to 12V via a 5R resistor. Connect the collector via the buzzer to earth. Connect the base to the points marked 10.4 volts via 3 diodes. If any LED lights up, the buzzer will sound. (LED voltage = 1.6 V, diode drop = 0.6V, base-emitter drop=0.6V, leaving 0.4V across 5R, I = V/R = 80mA.) If they are turned on by being connected to 12V then a NPN emitter follower can replace the FET and relay in your circuit. sreten.

24th December 2003, 11:24 PM	#10
Corn-Picker diyAudio Member Join Date: Oct 2002 Location: Pennsylvania	Ok, I did some more multimetering. If the LED is on (as in the pictures above), the positive side is 12 V and the negative side is 10.4 V. If the LED is off, the positive side is still 12 V but the negative side is 10.8 V. Is there any way to tell from these numbers whether the LED is turned on by completeing the ground or 12 V connection? If not, how can I tell whether the LED is turned on by completing the 12 V side or completing the ground side? Thanks again, I think I'm close to getting this to work

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