# 555 timer

Discussion in 'General Electronics Chat' started by swappo, Apr 29, 2008.

1. ### swappo Thread Starter Member

Apr 18, 2008
10
0
Hello,

I built the timer circuit using ne555p in the attached file. I am having trouble with my results when I breadboard it. The first pulse is longer than the rest of the pulses. I am assuming this is because the cap is charging from 0V to 2/3V for the first pulse and from 1/3V to 2/3V for all additional pulses. When I measure the second pulse (t1) I get a pulse for 5.2 s and t2(no current out) is 4.4 s. Theoretically, I should get t1=7.6s and t2=6.9s. I tried this circuit with different resistor and capacitor values and have found the same discrepancies between theoretical and experimental results. I thought maybe I fried the chip so I got a new timer and the results are different: t1=8.8s, t2=4.3s. The strange thing is the LED lights up weakly for 2.5s then is brightly lit for 6.4s during the t1 pulses. Also, the current out is 26mA for the first timer and 49.5mA for the second timer. The transient analysis on LTSpice puts the current at 53mA. Anyone have any ideas what the problem is? I believe I have hooked everything up correctly. Its pretty straight forward. BTW, this is a great site. I've been hanging out here for a couple of months and have learned a lot. Thanks.

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2. ### Caveman Active Member

Apr 15, 2008
471
0
First of all the first pulse should be longer for exactly the reasoning you said. I would measure the battery just to see if its voltage is moving a bit when the LED turns on. The battery voltage is defining everything, so if it moves, then everything changes.

3. ### Audioguru New Member

Dec 20, 2007
9,411
896
If the 9V battery voltage drops to 8V when the LED is on, the LED is a 2.0V red one and the output high voltage of the NE555 is at the typical high voltage of only +6.5V when the battery is 8V, then the 150 ohm resistor has 4.5V across it and the current is 30mA.

The output low voltage of the NE555 is about 0.015V in your circuit which is nowhere near high enough to light the LED.

The capacitor is probably an electrolytic type. Then its value tolerance is probably -20% and +50%. So it could be anywhere within 8uF to 15uF.

All electronic circuits need a supply bypass capacitor, especially when powered from a little battery. The 555 causes a supply current spike of up to 400mA each time its output switches and the capacitor can supply it but the battery by itself cannot so the supply voltage drops low for a moment. Use 100uF.

4. ### Caveman Active Member

Apr 15, 2008
471
0
Even with a 100uF capacitor, you will see a voltage drop with such long loading times. You can either ignore it or use a linear regulator.

5. ### swappo Thread Starter Member

Apr 18, 2008
10
0
Thanks for the replies. Sorry but I forgot to mention the supply voltage is from a 9V 200mA walwart. I am not sure if I used the right symbol in the diagram. It is my first time using LTSpice. I measured the voltage from the source while the circuit was running and got a constant 8.85V. The voltage at R3 is 5.35V when the LED is brightly lit and ranges from 105-195mV when the first pulse starts. The voltage seems to ramp up to about 195mV over 2.5s then instantly jumps to 5.35V. The LED was a large type (not sure what kind since it didn't say on the package). I tried a smaller T-1 size and got the same effect. Even though the voltage is 105mV (.7mA), the LED still lights dimmly. I have seen LED's light well below the 20mA optoelectric characteristic though.

6. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
If you are using a DMM (digital voltmeter) you won't see the transients (spikes). DMMs are great for accuracy, but they are quite slow to respond. You would need to use an oscilloscope to see the transients.

LEDs will light up (albiet very dimly) with even small currents, just as long as the Vf (forward voltage) of the LED is slightly exceeded. The amount of light the LED will output is (roughly) porportional to the amount of current flowing through it.

Adding a capacitor across the power supply will most certainly help to minimize the transients, once the capacitor is charged. When a discharged capacitor is connected directly across a voltage source, at the first instant it appears as a dead short. If the capacitor is very large, it could cause the fuse in the power supply to "blow" aka burn up aka become electrically open. If you look at schematics for power supplies, you'll usually see two capacitors on the output; a 10uF and a 0.1uF. The 10uF cap is usually electrolytic; it takes care of low frequency transients, and the 0.1uF is usually tantalum or ceramic, it takes care of the high frequency transients.