Would anyone mind simulating this simple circuit for me? - Electronic Circuits Projects Diagrams Free

William At MyBlueRoom · 6th April 2006, 04:03 PM

I don't have a circuit simulator program and my analog design skills are lax. If anyone would mind running this very simple circuit through a simulator I'm looking for values for R & C so that 20ma flows through the LED when the relay is energised. The circuit is pulled to ground via a ULN2803 darlington array.

The Relay datasheet can be found here.
http://aromat.thomasnet.com/item/pow...=spec&filter=0

Thanks in advance.

Optikon · 6th April 2006, 05:00 PM

I think your circuit is incorrect. You want the resistor to be in series with the LED. You dont need the capacitor. It doesnt even need to be simulated.

Assume the UN2003 saturates at ~ 100mV when "ON", give the relay contact about 50milli-Ohms of resistance.. both of these are unimportant to calculating the LED current. LED might have about a 1.5V drop at 20mA (check datasheet - its based on color) so resistor ~ 175.. so 200 Ohms resistor is fine..

If you are worried about relay contact bounce, I doubt you'll see that in the LED. If you insist, at least return the capacitor to ground and not the 2003 driver.

Fix your circuit though.

evandude · 6th April 2006, 05:58 PM

I think the point of the capacitor is to allow enough of a current kick to actually switch the relay, and then when it charges up after a short time the current is reduced to just enough to keep it on. thus the whole energy-saving idea.

audioguru · 6th April 2006, 06:22 PM

Do it like this:

William At MyBlueRoom · 6th April 2006, 08:28 PM

Actually the circuit is from a small article in Elektor magazine. Although the LED will show when the relay is energized it also acts as part of the circuit. It's a power saving circuit, the cap should act like a short till it takes a charge then the LED and resistor limit the current to 20ma to the relay.

I'll find the article and post it. I do recall he mentioned GREEN LED, 20ma, 10 to 1K ohms and 100 to 400 uf for the Cap @ 6.3v

Claims to reduce the power draw to approx 50% or so without using a PWM.

justDIY · 7th April 2006, 12:46 AM

none of those parts are expensive or hard to find ... why not just simulate it in the real world (build it on a breadboard?)

William At MyBlueRoom · 7th April 2006, 01:24 AM

Quote:

why not just simulate it in the real world

True, I could just build it. But I was hoping to get optimal values for the design.

Russlk · 7th April 2006, 02:26 AM

That is a 9 volt relay, it is not going to work on 5 volts anyway.
If you use 9 volts, anything over 250uF should work.

panic mode · 7th April 2006, 02:45 AM

JW1SN family has 5VDC member as shown in his circuit.
i don't think this circuit will work on only 5V because of considerable
voltage drop on LED and relay is probaly not going to hold reliably.
Circuit is not worth simulating because relay is mechanical device
and factors like spring force, friction and holding current are not known.
if you insist on energy saving don't use relay and if you must, you will
also have to do some experimenting to get the values right.
diferent color LEDs have different voltage drop and if i recall correctly
lowest Vf is for red LEDs. this should work ok even with 5V but LED might need to go or it could be added in parallel branch as shown by audioguru.
to save power, series resistor for LED would have to be increased to
reduce LED current to lowest acceptable level (it affects brightness).

edit:

I'll take that back. Circuit and shown values do make sense but we won't know
for sure without actual test. I checked datasheet and it does show pickup and
drop out voltages (at least the max values). Coil current is some 100mA, and
this is why resistor was in parallel with LED. With coil resistance of 47 Ohm and
that current of 100mA, voltage drop would be 4.7V. LED will require some
1.7-2.2V to operate and can take 20mA (that's about 100 ohm).
With 120 Ohm in parallel and coil in series, whole circuit would draw some
50mA which could be enough to hold relay. Capacitor size is not critical
(doesn't affect current saving or circuit consumption) and can be increased
if required until relay can pickup reliably.

justDIY · 7th April 2006, 03:04 AM

Quote:

Originally Posted by panic mode

JW1SN family has 5VDC member as shown in his
if you insist on energy saving don't use relay and if you must, you will

yea relays are so 19th century

@William

an article I read not too long ago on EDN showed how to take a common pair of N chan mosfets and use them to switch AC currents... i'll see if I can dig that up

audioguru · 7th April 2006, 03:14 AM

Don't worry about the 5V relay not working. It is guaranteed to pull-in at 3.5V and drop-out at only 0.5V.

In your circuit, with the 2.2V LED in series with the low resistance (5V, 47 ohms) relay coil then the LED and 220 ohm resistor's current is 60mA. The 220 ohm resistor will conduct 10mA leaving 50mA through the LED, but its absolute max current is only 30mA.

If your green LED's forward voltage is exactly 2.2V then the resistor could be 56 ohms and the LED's current will be a safe 21mA.
You could use a 3.3V blue LED and increase the resistor's value to 200 ohms to reduce the total current even more to 36mA.

William At MyBlueRoom · 7th April 2006, 03:36 AM

Thank audioguru, Blue LEDs are too expensive to use in something you would probably bury near the furnace.

How does the cap affect the circuit?

audioguru · 7th April 2006, 03:48 AM

Blue LEDs are fairly inexpensive today.
In the first moment after the circuit switches on, the capacitor has 0V across it so the full 5V is across the relay's coil, activating it. After a short time when the capacitor is charged then it doesn't draw any current and the resistor in series with the relay coil reduces its current.

William At MyBlueRoom · 7th April 2006, 04:19 AM

audioguru? are you in Toronto?

k7elp60 · 7th April 2006, 04:27 AM

This relay circuit is a neat idea.
A number of years ago I did something similar, but the purpose was completly different. The reason was to dissipate less power in the relay coil than the on resistance of a Power Mosfet. A number of years ago the Rds of the power mosfet's was no where it is today. Even today the power dissipated by the poser mosfet is = Id squared by the Rds. Relays have contact resistance and this on new relays is very low. So if one is to calculate the power dissipated in relay contact resistance and add the power consumed by the energized coil the two can exceed the power dissipated in turned on power mosfet. If once the relay is energized the coil voltage is reduced to less than than 1/2 nominal coil voltage the power consummed by the coil is 1/4 that at normal coil voltage.
In simple terms a number of years ago a power relay with half the coil voltage dissipated less power than a power mosfet. To confirm this I actually measured the contact resistance of a DC relay at full coil voltage and then at 1/2 coil voltage. One would assume the with less coil voltage the pull from the magnetism on the contacts would be somewhat less. On a new relay I could not measure any appreciable difference. I was measuring contact resistance in the range of .002 ohms

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6th April 2006, 05:00 PM	(permalink)
Optikon	I think your circuit is incorrect. You want the resistor to be in series with the LED. You dont need the capacitor. It doesnt even need to be simulated. Assume the UN2003 saturates at ~ 100mV when "ON", give the relay contact about 50milli-Ohms of resistance.. both of these are unimportant to calculating the LED current. LED might have about a 1.5V drop at 20mA (check datasheet - its based on color) so resistor ~ 175.. so 200 Ohms resistor is fine.. If you are worried about relay contact bounce, I doubt you'll see that in the LED. If you insist, at least return the capacitor to ground and not the 2003 driver. Fix your circuit though.

6th April 2006, 05:58 PM	(permalink)
evandude	I think the point of the capacitor is to allow enough of a current kick to actually switch the relay, and then when it charges up after a short time the current is reduced to just enough to keep it on. thus the whole energy-saving idea. __________________ EEgeek.net

6th April 2006, 08:28 PM	(permalink)
William At MyBlueRoom	Actually the circuit is from a small article in Elektor magazine. Although the LED will show when the relay is energized it also acts as part of the circuit. It's a power saving circuit, the cap should act like a short till it takes a charge then the LED and resistor limit the current to 20ma to the relay. I'll find the article and post it. I do recall he mentioned GREEN LED, 20ma, 10 to 1K ohms and 100 to 400 uf for the Cap @ 6.3v Claims to reduce the power draw to approx 50% or so without using a PWM. __________________ Bill Home of the Firefly PIC Tutor Inchworm ICD2 http://www.blueroomelectronics.com

7th April 2006, 12:46 AM	(permalink)
justDIY	none of those parts are expensive or hard to find ... why not just simulate it in the real world (build it on a breadboard?)

7th April 2006, 02:26 AM	(permalink)
Russlk	That is a 9 volt relay, it is not going to work on 5 volts anyway. If you use 9 volts, anything over 250uF should work. __________________ see my website: www.geocities.com/russlk

7th April 2006, 02:45 AM	(permalink)
panic mode	JW1SN family has 5VDC member as shown in his circuit. i don't think this circuit will work on only 5V because of considerable voltage drop on LED and relay is probaly not going to hold reliably. Circuit is not worth simulating because relay is mechanical device and factors like spring force, friction and holding current are not known. if you insist on energy saving don't use relay and if you must, you will also have to do some experimenting to get the values right. diferent color LEDs have different voltage drop and if i recall correctly lowest Vf is for red LEDs. this should work ok even with 5V but LED might need to go or it could be added in parallel branch as shown by audioguru. to save power, series resistor for LED would have to be increased to reduce LED current to lowest acceptable level (it affects brightness). edit: I'll take that back. Circuit and shown values do make sense but we won't know for sure without actual test. I checked datasheet and it does show pickup and drop out voltages (at least the max values). Coil current is some 100mA, and this is why resistor was in parallel with LED. With coil resistance of 47 Ohm and that current of 100mA, voltage drop would be 4.7V. LED will require some 1.7-2.2V to operate and can take 20mA (that's about 100 ohm). With 120 Ohm in parallel and coil in series, whole circuit would draw some 50mA which could be enough to hold relay. Capacitor size is not critical (doesn't affect current saving or circuit consumption) and can be increased if required until relay can pickup reliably.

7th April 2006, 03:36 AM	(permalink)
William At MyBlueRoom	Thank audioguru, Blue LEDs are too expensive to use in something you would probably bury near the furnace. How does the cap affect the circuit? __________________ Bill Home of the Firefly PIC Tutor Inchworm ICD2 http://www.blueroomelectronics.com

7th April 2006, 03:48 AM	(permalink)
audioguru	Blue LEDs are fairly inexpensive today. In the first moment after the circuit switches on, the capacitor has 0V across it so the full 5V is across the relay's coil, activating it. After a short time when the capacitor is charged then it doesn't draw any current and the resistor in series with the relay coil reduces its current. __________________ Uncle $crooge

7th April 2006, 04:19 AM	(permalink)
William At MyBlueRoom	audioguru? are you in Toronto? __________________ Bill Home of the Firefly PIC Tutor Inchworm ICD2 http://www.blueroomelectronics.com

7th April 2006, 04:27 AM	(permalink)
k7elp60	This relay circuit is a neat idea. A number of years ago I did something similar, but the purpose was completly different. The reason was to dissipate less power in the relay coil than the on resistance of a Power Mosfet. A number of years ago the Rds of the power mosfet's was no where it is today. Even today the power dissipated by the poser mosfet is = Id squared by the Rds. Relays have contact resistance and this on new relays is very low. So if one is to calculate the power dissipated in relay contact resistance and add the power consumed by the energized coil the two can exceed the power dissipated in turned on power mosfet. If once the relay is energized the coil voltage is reduced to less than than 1/2 nominal coil voltage the power consummed by the coil is 1/4 that at normal coil voltage. In simple terms a number of years ago a power relay with half the coil voltage dissipated less power than a power mosfet. To confirm this I actually measured the contact resistance of a DC relay at full coil voltage and then at 1/2 coil voltage. One would assume the with less coil voltage the pull from the magnetism on the contacts would be somewhat less. On a new relay I could not measure any appreciable difference. I was measuring contact resistance in the range of .002 ohms __________________ The great thing about electronics is unlimited ways to do the job. The only limit is one\'s imagination. I generally think my way is best. Show me a different way. I have an open mind.