# Tristate Polarity Indicator

#### lkgan

Joined Dec 18, 2009
117
Hi everyone,

I have found the attached circuit in book and couldn't understand something:

Here, a tristate LED is used to indicate the direction and type of current flow. If V is a positive dc voltage, the device emits red light. However, if V is a negative dc voltage, the device emits green light. If a high-frequency ac voltage is applied, it appears as if the device emits a yellow light. The diode which is connected in parallel with the resistor R2 is used to provide reversed-voltage protection when the applied voltage exceeds the maximum reversed voltage. My question is, there's only one diode to protect the reverse bias of gree LED, how about the red LED? No protection for it?

#### Attachments

• 18.6 KB Views: 92

#### bertus

Joined Apr 5, 2008
22,305
Hello,

The diode Vd is used to compensate the brightness difference between the red and the green.
The green is less bright, by shorting a resistor with the diode teh green led gets more current.

Greetings,
Bertus

#### lkgan

Joined Dec 18, 2009
117
Hi Bertus,

But from the book I read, it mentioned that the diode Vd is used for reverse breakdown protection for the LED. From the schematic, it seems that it only protect one way. If I wish to protect 2 LEDs which are green and red, should I just connect another diode in parallel with the existing diode?

#### Wendy

Joined Mar 24, 2008
23,476
The LEDs protect each other, by being back to back. Like Bertus said, the diode compensates for the differences between the two diodes, making the current vary depending on direction of flow.

#### lkgan

Joined Dec 18, 2009
117
So it means that the green LED is brighter than the Red LED because it bypass the R2 resistor right? Isn't it a bad practice to use LED protect back to back as the breakdown voltage of LED is very low compared to diode.

#### peranders

Joined May 21, 2007
88
Isn't it a bad practice to use LED protect back to back as the breakdown voltage of LED is very low compared to diode.
Not in this case. The LED's will have only 1.6-1.8 as reverse voltage and this is OK. Less than 5 V is OK for red LED's.

#### peranders

Joined May 21, 2007
88
So it means that the green LED is brighter than the Red LED because it bypass the R2 resistor right?
This is overkill if you have LED's with similar brightness. The difference in forward voltage is 200-300 mV so this extra resistor and diode are of minor importance.