LED Circuit with Reversable Polarity

[Elektrotechnik]

In the attached circuit, the LED circuit can be illuminated, regardless of the polarity.

Can someone walk me through how the diodes are working in this circuit to allow for this to happen? I'm familiar with diodes, but only the basics. I'm hoping that someone here can walk me through the steps of this circuit.

Also, should the 4 LEDs connect back to the (-) on the voltage source? I'm not sure why the LEDs don't return to the power source.

Thanks!

 

[gar]

130731-2351 EDT

Only the lower circuit will illuminate the one single diode, D11.

The upper circuit has D17 reversed biased and no current flows from the battery.

If you want a circuit that will supply a unidirectional output current from either polarity at the input, then you need a bridge rectifier.

Ideal diodes conduct in only one direction. So trace the paths that will allow current flow to analyze your two circuits.

There would need to be a separate current limiting resistor in series with each LED. I have no idea where the other LEDs are to be connected, or what the purpose of the circuit is, and what function the LEDs perform.

.

 

[eHunter]

In the attached circuit, the LED circuit can be illuminated, regardless of the polarity.

Can someone walk me through how the diodes are working in this circuit to allow for this to happen? I'm familiar with diodes, but only the basics. I'm hoping that someone here can walk me through the steps of this circuit.

Also, should the 4 LEDs connect back to the (-) on the voltage source? I'm not sure why the LEDs don't return to the power source.

Thanks!

The upper circuit will not illuminate.
In the top diagram using electron current flow, from the battery negative(-) terminal, through D15, through the cathode(-) of D4, through current limiting R2, through D16, back to the negative(-) battery lead, this is an incomplete circuit, no current will flow and will not illuminate the D4 LED.
Diode D17 is blocking battery positive (+)

In the lower diagram using electron current flow, from the battery negative(-) terminal, through forwqard biased D20, through the forward biased cathode(-) of D11, through current limiting R3, through forward biased D19, back to the positive(+) battery terminal, this is a complete circuit and will illuminate the D11 LED.

Yes, to operate properly and illuminate, all 4 LEDs will require both the cathode(-) connected to the battery negative and the anode(+) connected to the battery positive through a proprly sized current limiting resistor.

The easiest method is to use a bridge rectifier or construct the equivalent of a bridge rectifier using discrete rectifier diodes, then the polarity of the power source is never incorrect. With the addition of a capacitor most DC circuits would likely accept either an AC or DC power source.


OOPS, late to the party, sorry gar.

 

[suemarkp]

What is the purpose of this circuit? Is it a reverse polarity indicator?

There are LEDs that work in both directions. They are tri color LEDs -- red in one direction, green in the other, and yellow if you reverse fast enough (e.g. 60 Hz) that the colors blend in your eye

 

[Elektrotechnik]

Can I still use a bridge rectifier if I am powering the circuit using a 12 V DC source?

Also, what are the F1 and S2 components in your schematic, eHunter? And what are the squigly lines which you have inside of your diode bridge?

 

[eHunter]

Can I still use a bridge rectifier if I am powering the circuit using a 12 V DC source?

Also, what are the F1 and S2 components in your schematic, eHunter? And what are the squigly lines which you have inside of your diode bridge?

Yes, select size current, voltage and temp appropriately for your application.
F1 is a fuse.
S1 is a switch.
Squigly lines on BR1 indicates AC.
A bridge rectifier will accept AC or any polarity DC voltage input and output DC.

 

[Elektrotechnik]

Thanks, eHunter!


This is where I start to get confused. The circuit design. But I want the practice!


I have a 14 V Power Source, forward current of 30mA for each LED, ~ 3 V per LED forward voltage. What should I look for when choosing a diode based on these parameters? Will I need a current driver if my power supply is voltage driven?


Thanks for your help!

 

[gar]

130801-0929 EDT

Elektrotechnik :

You need to start by understanding the basic characteristics of the components that you use.

Find some V-I curves for a diode, LEDs are also diodes. Zener diodes are units with a special controlled reverse voltage characteristic. See http://en.wikipedia.org/wiki/Diode .

See http://www.oup.com/us/pdf/microcircuits/students/diode/1n4148-general.pdf for some curves. The forward characteristic is plotted with a log scale for current, and linear for voltage.

Since a diode has a low voltage drop in the forward direction and this voltage changes only a small amount with current, and you want to supply energy from a higher voltage source a current limiting method must be employed. Roughly speaking the diode forward voltage can be considered a constant voltage relative to a much higher source voltage for the energy source. Whenever a constant voltage load is supplied with energy there must be some method to limit current. Such devices are neon bulbs, diodes, base-emitter junction of a transistor (basically a diode characteristic), fluorescent bulb, an arc discharge, glow discharge, a battery, etc.

LEDs produce light output when forward biased.

The circuit you have shown is far too complex for you to study basic characteristics. A single battery, diode, current limiting resistor, and LED is where you should start. A question you need to ask is how to size the current limiting resistor, and why do you need this resistor? What is the limiting factor on how low the resistance can be? You need to assemble this simple circuit, choose different resistance values and measure current and the various voltages.

.

 

[eHunter]

Thanks, eHunter!


This is where I start to get confused. The circuit design. But I want the practice!


I have a 14 V Power Source, forward current of 30mA for each LED, ~ 3 V per LED forward voltage. What should I look for when choosing a diode based on these parameters? Will I need a current driver if my power supply is voltage driven?


Thanks for your help!

I agree with gar, understand the mechanics of what you are attempting to accomplish.
One such website than can help is:

http://www.allaboutcircuits.com/vol_3/chpt_3/1.html

More specifically about your questions, use ohm's law and solve for the resistance and then you can determine the power dissipation of the resistor.
You apparently have already selected the LED(diode) and know the operating parameters.
For a bridge rectifier, choose a device that has at least 2x the voltage of your supply maximum and 2x the maximum current load specs and do the same for any power dissipation spec. For reliable operation do not exceed the max temp spec for any component.
Return showing your process and work product and I am sure we will evaluate your results.

 

[ronaldrc]

I think this is supposed to be a circuit to keep you from putting the wrong
polarity on the LEDs.

The circuit as shown is short one power diode.

I added Diode #18 to correct the circuit.

Ronald

 

[steve66]

I think this is supposed to be a circuit to keep you from putting the wrong
polarity on the LEDs.

The circuit as shown is short one power diode.

I added Diode #18 to correct the circuit.

Ronald

That looks much better.

The resistor shown is 820 ohms. If we assume each diode drops about 0.7 volts, then the 12 volt supply minus 3 times 0.7 volts = 9.9 volts left to drop across the resistor.

That would supply the LED with about 9.9V /820 ohms = 12 mA. That should be about right for a typical LED.

For a 14 volt supply, you might want a slightly higher resistor - maybe 1K.

 

[ronaldrc]

That looks much better.

The resistor shown is 820 ohms. If we assume each diode drops about 0.7 volts, then the 12 volt supply minus 3 times 0.7 volts = 9.9 volts left to drop across the resistor.

That would supply the LED with about 9.9V /820 ohms = 12 mA. That should be about right for a typical LED.

For a 14 volt supply, you might want a slightly higher resistor - maybe 1K.

Steve thanks for the reply

I changed my graphic to make it a little clearer.

Ronald

 

[suemarkp]

That looks much better.

The resistor shown is 820 ohms. If we assume each diode drops about 0.7 volts, then the 12 volt supply minus 3 times 0.7 volts = 9.9 volts left to drop across the resistor.

That would supply the LED with about 9.9V /820 ohms = 12 mA. That should be about right for a typical LED.

For a 14 volt supply, you might want a slightly higher resistor - maybe 1K.

I thought the "Classic" forward voltage drop across a LED was 1.7V, not .7 like a rectifier diode. Some of the newer different color ones may be different. These LEDs are in parallel so they supply voltage needs to be dropped to what a single LED needs, not a series of them. Then what matters is how much current you want each LED to consume. The classic value used to be 10mA, but again with high power ones it can be all over the map. So you need to drop 12V - 1.7V from the supply, or 10.3V. Divide by the current (say 4 * 10mA) = 40ma or .04A. The resistor would be 10.3V / .04 = 258 ohms.

Then there's those other diodes in the circuit. You'll probably lose .7V through each of those too, so that should be factored into the voltage equation. Perhaps that's where the 3 * .7V came from and not all the LEDs. Too many different diodes and schematics to keep from getting confused...

 

[GoldDigger]

When you put LEDs in parallel with only one current limiting resistor, you are depending on the diodes being 100%a identical so that their I eternal resistance causes a perfect current sharing. In real life you may find that one LED is brighter than the rest.

 

[ronaldrc]

Thanks Mark

I agree I just used what was there, I'm more worried about the power rectifier polarity than anything.

I have been accused of being Bipolar :happysad:

Ronald

 

[ronaldrc]

I agree each one should have its own resistor, because if one fails and opens
it would cause the others to also fail.

 

[Elektrotechnik]

If I wanted ESD Protection, where would a zener be placed?

 

[steve66]

I thought the "Classic" forward voltage drop across a LED was 1.7V, not .7 like a rectifier diode.

Yes, its been a while since I've worked with diodes, and I forgot LED's have a higher forward voltage.

If I wanted ESD Protection, where would a zener be placed?

Probaby after the diodes shown, in parallel across the input voltage., Like from the junction of D15 and D17 to D16 and D18. If I remember right, zeners are reverse biased, so the arrow would point up toward D15 and D17.

I think you also need a resistor in series with the input voltage to limit the current through the diode. You need to calculate the resistor value so the power and current ratings of the zener aren't exceeded, but you also have the keep the resistor value small enough so that it doesn't significantly affect your input voltage under normal conditions.

 

[gar]

130805-2149 EDT

At the very input of your circuit put a 100 ohm 1/4 W resistor in series with each battery lead. On the circuit side of the two resistors connect a Transorb.

Now you do some research on how these work, and how to select an appropriate unit. Note: on an excessive energy transient, like plugging the input into a 120 V outlet, the resistors will burn out without damage to the Transorb. Transients of lesser energy won't damage anything.

.