# Changing the circuit

### Changing the circuit

Let's look at what would happen if we change the values of the resistors or the capacitor.

If we make the capacitor bigger, it will take longer for it to charge and discharge. The rate at which the LED blinks on and off will get slower. If we make the capacitor smaller, the LED flashes on and off more quickly.

The capacitor charges through both resistors, but discharges only through R2. This means the LED will always be on longer than it will be off. If we make R2 very large, and R1 very small, this difference will decrease until it is barely noticeable.

But we can't decrease R1 too much, or the circuit won't work. Remember that when the capacitor reaches 6 volts, pin 7 goes to zero, and the battery drains directly through R1. If R1 is too small, it will overheat, and the battery will not last very long. If the resistance of R1 was zero (like it was replaced with a plain wire), all the current from the battery will flow directly from one terminal to the other, in what we call a short circuit (the current takes a shorter path than it would if it went through the whole circuit). If that happens, the chip will not get any current, and it will stop working. For that reason, we don't use less than 1,000 ohms for R1.

If we keep R1 as low as possible, so that the on and off times are as equal as possible, then the timing is controlled mainly by R2 and the capacitor. Making R2 large means that the capacitor charges and discharges slowly. We can use a large R2 and a small capacitor, or a small R2 and a large capacitor, and get the same timing.

### A new chip -- the 7555

Since the original 555 came out, low power versions have come out. The low power version (the 7555) is better for use in battery powered devices, since it uses less power. The chip itself not only uses less power, but because its trigger and threshold pins are more sensitive, you can use higher value resistors, so less battery current is wasted. Higher resistances allow you to use a smaller (cheaper) capacitor. The 7555 can also reach higher and lower frequencies. In the 7555, the reset pin usually does not need a capacitor, making the circuit simpler and cheaper.

The potential drawbacks to the 7555 are that it is sensitive to static electricity until it is wired into a circuit, so you have to be careful to discharge your hands before handling it (touch a large metal object to remove any static electricity from your fingers). The original 555 chip can supply more current (up to 200 milliamperes) to the LED than the 7555 can.

The amount of current the 7555 can provide depends on the power supply voltage, and on whether the LED (or some other load) is connected to the positive side of the power supply, or is connected to ground. If the chip is delivering current through the LED to ground, we say it is "sourcing" the current. If the LED is connected to the positive side of the power supply, and the current goes through the 7555 to ground, we say the chip is "sinking" the current.

The 7555 can source only about 2 milliamperes at 5 volts, but can sink 8 milliamps. At 12 volts things are a little better -- it can source 10 milliamps and sink 50. At 15 volts, it can source and sink 100 milliamps.

While this limited current capability might be a problem if we were driving a lot of LEDs, or trying to run a motor, for blinking a single LED we can use it to our advantage, and eliminate the current limiting resistor for the LED. If the chip can't deliver more than 20 milliamps to the LED, then the resistor is not needed.

Because both the 555 and the 7555 can source current as well as sink current, we can alternately drive two LEDs, each one off while the other is on.

R1: ohms
R2: ohms

Here we have eliminated the capacitor on pin 5, leaving it unconnected, since the 7555 version of the chip does not need it for this circuit. We also now have two LEDs connected to pin 3 of the chip. One goes to ground, through a 10,000 ohm resistor, and the other goes to the positive power supply through another 10,000 ohm resistor.

Note that we have two LEDs and two resistors making a direct link between power and ground. Without the 7555 chip and the rest of the circuitry, these LEDs would both simply light up. With the 7555 in the circuit, pin 3 alternates between +9 volts and ground. This makes one or the other LED turn off. We need the high resistor values because the chip can't source or sink very much current, so the current going through the LEDs without the chip in the circuit must be less than the current the chip can source or sink. If we didn't have high value resistors, one or both LEDs would simply stay on.

You may have noticed that the LED takes a while to turn on at first. When the capacitor is completely discharged, as it is when we first apply power, the voltage has to climb all the way from zero to 6 volts before the top LED turns off. After that, however, it never gets down to zero volts, since it starts charging again when it gets down to 3 volts. So the first time the LED lights, it stays on longer than on subsequent oscillations.

If we want to keep the on and off times the same, the 7555 allows us to simplify the circuit.

R1: ohms

Instead of using the discharge pin (pin 7), we use the output pin (pin 3) to discharge the capacitor. The 7555 can pull pin 3 quite close to zero volts, acting much like pin 7. So we can leave pin 7 unconnected. We could almost eliminate the current limiting resistor for the LED, since it will not get enough current to be damaged (the 7555 can't sink as much current as the original 555), but the 7555 isn't strong enough to pull pin 3 down to ground unless there is some resistance between the LED and the positive power supply, so we still need a small resistor.

The result is a simpler circuit, where the frequency of the oscillation is controlled by a single resistor and a single capacitor. And because the capacitor drains through the same resistor that it charges from, the on time is the same as the off time.

One last note. The schematic shows a connection from pin 2 to the capacitor crossing over the connection between pin 1 and ground. The little bump in the line shows that these two wires are not electrically connected.

### If you want the math...

The frequency of this last circuit is:

Since the natural log of 2 is a constant (approximately 0.6931471806), and twice that is also a constant (1.386294361), and our values for the resistor and capacitor are only good to a few percent (so we don't need all that much precision in our constant) we can simplify this to:

For the first circuit, with two resistors, the frequency is:

The "on" time will be 0.7 (R1+R2) C.

The "off" time will be 0.7 R2 C.