LED indicates power source
A zener diode lets detects the presence or absence of a power source.
Brian Conley, PE, Circuitsville Engineering LLC; Edited by Martin Rowe and Fran Granville -- EDN, August 26, 2010
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LED circuits with current-limiting
resistors find extensive use as
power indicators and for debugging circuits
(Reference 1). In some cases, however,
your design may require a different
approach. Bipolar transistors have a little-
discussed behavior: reverse active region.
For low voltages and small currents,
an NPN transistor can operate in
reverse with a significantly lower gain, which can be undesirable. Some linear
regulators also operate in this way.The circuit in Figure 1 gets its inputvoltage
power primarily from a wallwart
dc-power supply that can provide
7 to 12V. It may also get 5V from a USB
(Universal Serial Bus) port. This design
requires a circuit that indicates whether
the board is receiving voltage from the
wall wart or from the USB port.
The circuit uses Q1, a 2N7002 FET, and zener diode D1 to solve the problem. The FET is in series with LED1 and current-limiting resistor R1. Diode D1 is a Vishay AZ23C4V3-V, which has a typical reverse voltage of 4.3V within a range of 4 to 4.6V. When Q1’s gate-to-source voltage exceeds its threshold-voltage range of 1 to 2.5V, the LED turns on. The voltage coming from the USB port is insufficient to turn on LED1 because of the voltage drop across D1. Thus, the LED illuminates when the board receives voltage from the wall wart, but not from the USB port.
Under testing, the LED illuminates when the input voltage is at least 7.1V. When it is below that voltage, the LED is off, indicating that the USB port is powering the circuit.
Resistor R3 comprises two 1-kΩ resistors in parallel. This setup is necessary because the input voltage is 12V and the zener diode’s minimum voltage is 4V. A voltage of 8V appears across R3, producing 0.128W—too much power for one resistor in a 0805 package.
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Talkback
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My humble opinion is that this is one of those design ideas that should never have passed the publication threshold. The circuitry is neither novel or "elegant". It uses components in a very inefficient way and some components are stressed, as other readers already have noted.
Christian de Godzinsky - 2010-9-9 00:29:39 PDT -
A way to avoid the objections to this circuit would be to first, get rid of the capacitor, it really is not serving any important purpose, then put the current limiting resistor below the FET, between the source and ground, and then increase the value of the gate pull-down resistor. That resistor really only needs to keep the zener pre-turn on current from turning on the FET. So it may take some actual zener diode characteristic data to discover the correct value for R3
William Ketel - 2010-7-9 16:41:06 PDT -
I need to point out that applying C1 in series with a zener diode could distroy the zener diode due to unlimited transient during power up.
Vladimir Doubovis - 2010-6-9 14:29:44 PDT -
I have three issues with this circuit, the first is that the resistor R1 is more likely to burn out than R3 because when Vin is 12V and the LED drops 2V then R1 gets 10V * 20mA = 200mW which means that this resistor must be built from several standard SMT resistors or a power SMT must be used.
Secondly the user might not be so happy with the fact that the brightness of the LED varies with the voltage, since the input varies from 7 to 12V this causes the LED current to vary between approx 10 and 20mA due to R1 as limiting resistor getting between 5 and 10V.
Thirdly, this circuit is quite inefficient seeing that the input current at 12V is 16mA drained via R3 while the output is 20mA through the LED and R1 so about 40% of the energy it uses is wasted. May not be a big deal until you run on battery power. All these issues can easily be avoided using the same part count:
Take an ordinary NPN transistor with high HFE, Collector to Vin and a voltage divider connected to its Base: 5k6 to Vin and 4k7 to ground. Also from Base to ground a zener diode of 3V3. From Emitter to ground the LED and a 39 Ohm resistor.
At 5V the voltage divider delivers 2.3V to the Base, so Emitter voltage is about 1.8V and since the green LED has 2V voltage drop, it will not light up. At 7V the Base receives 3.2V so the Emitter is at 2.6V and the LED receives approx 16mA.
At Vin voltages over 7V the zener will limit the Base voltage, so that the LED brightness will not vary with input voltage. Current in the voltage divider is a meager 8.8V/5k6 = 1.6mA so more than 90% of the current drawn is actually flowing through the LED.
Cor van de Water - 2010-28-8 10:32:19 PDT
