- Published: Friday, 30 December 2011
- Written by Jon Chandler
- Hits: 9454
In Power Protection Circuits, I outlined some methods of circuit protection. In this article, I'll take a closer look at the practical aspects of using a series diode for reverse polarity protection.
A series diode is about the simplest form of protection again a power supply being connected incorrectly. For some applications it will work well but with others, the voltage drop across the diode in normal operation will create problems.
A series diode will conduct when the power supply is correctly connected but will block the current flow if the power supply is connected with reverse polarity, protecting the circuit from damage. The downside of this simple circuit is the voltage drop across the diode. Depending on the power supply and circuit arrangement, this may or may not be a problem.
Let's start by examining the voltage drop. For this experiment, I used a common 1N4001 silicon diode (which is available at RadioShack). The diode will handle up to 1 amp and 50 volts PIV (peak inverse voltage). I used my bench power supply set to 5.0 volts as the power supply for this test and my adjustable dummy load to vary the output current. A Fluke 45 was used to measure the output voltage across the "circuit." The plot below shows the results.
Just inserting the resistor drops the 5.0 volts of the supply to about 4.6 volts at no load. There's a 0.4 volt drop at no load! The voltage quickly drops t0 4.3 volts with a load of 10-20 mA and continues to fall as the load increases. At the maximum load of 1 amp, the voltage drops almost to 4 volts. The voltage drop is also a function of temperature and can be as large as 1.6 volts according to the data sheet.
What's the practical affect of this voltage drop? It depends on the circuit and power supply arrangement.
5 Volt Regulated Wall Wart
If our circuit relies on a 5 volt regulated wall wart, such as the cell phone chargers I like to use, the voltage supplied to the circuit depends on the load and could drop to as little as 4 volts (3.4 volts minimum according to the data sheet). If any of the components in the circuit require tightly-regulated 5 volts, this low voltage will be a problem.
A quick look at the PIC18F2520 shows it is only rated for full speed operation down to 4.2 volts. The series diode may create problems depending on load and temperature.
Let's consider a case that is even worse. Suppose we're using the PIC's ADC with a VDD reference to measure a voltage and illuminate a bargraph according to voltage level. As more LED segments are illuminated, the current draw of the circuit increases, lowering the reference voltage. As the reference voltage falls, even more segments will be illuminated. The result is an inaccurate and unstable circuit!
We have to conclude that a series diode and a regulated power supply can have some series limitations.
Circuit With On-board 5 Volt Regulator
Another common arrangement is a non-regulated power supply (or a regulated supply at a higher voltage) with a 5 volt regulator on the circuit board. In this case, the series diode may work ok. As long as the power supply voltage is greater than the diode drop plus the regulator dropout voltage, the circuit will have the correct supply voltage. An LM7805 has a dropout voltage of 2 volts. As long as the power supply puts out 1.6 volts +
2.0 volts + 5 volts = 8.6 volts, the 5 volt regulator will function properly.
The diode voltage drop means the power supply must supply almost 2 volts more than would be needed by the regulator alone. Let's consider the power dissipation if the circuit draws 1 amp (far more than typical circuits draw). The power dissipated in the diode is equal to the current x the voltage drop = 1 amp x 1.6 volts = 1.6 watts. The power dissipated by the voltage regulator is equal to the current x the voltage drop across the regulator. If we have a 9 volt supply, we drop at most 1.6 voltages across the diode, leaving a drop of 2.4 volts across the regulator. The regulator power dissipation = 1 amp x 2.4 volts = 2.4 watts. A total of 4 watts is wasted in this arrangement. A small night-light bulb draws 4 watts, so this is a significant waste.
If we're using a series diode for reversal polarity protection, a 5 volt regulated wll wart won't provide a good solution and may result in an unstable circuit. In order to account for the diode voltage drop, we've forced to provide a power supply of higher voltage and a regulator on the circuit board. The simple solution of a series diode for reverse polarity protection results in a more complex, energy-wasting circuit. This method may work for circuits with a small current draw but in general, it's not a good solution.