# Tag Info

39

I'm not a 555 fan and only used them a couple of times over several decades but they do have their uses. Why is the 555 so prevalent, or ubiquitous; what about it so useful? It was probably the first general purpose timer to market. (Timing is everything!) It was featured in all the hobby electronics magazines of the 1970s, etc., so it became very well ...

36

I know that a square can be constructed from sine waves. However, the 555 does not generate sine waves or multiple frequencies, it generates a single square pulse. So where have these harmonic frequencies come from? Congratulations on your explanation of what you are seeing and your experimentation. The key issue is that not only CAN a square wave be ...

29

A square wave can be viewed as a sum of the odd harmonics of a single frequency. A square wave can be generated by summing a bunch of sine waves. A square wave can also be generated by simply toggling the power on and off at the primary frequency of the square wave. In either case, the spectrum will look the same. You cannot tell how a square wave was ...

24

In general, when a circuit works with a device from one maker and not from another it is telling you is that your design is incorrectly using some feature of the device on the verge of the design characteristics. In this case how you are driving the reset pin. That is a bad thing. You say it works with a 555N, but I am willing to bet if you tested enough ...

21

Your measured short-term stability is about +/-0.01%, which isn't bad for an uncompensated RC timer. You can improve it by using low temperature coefficient resistors and capacitors in the timing circuit, maybe by bypassing pin 5 to ground, by isolating the circuit thermally and electrically, in the extreme controlling the temperature in an oven, powering ...

20

The granularity of the light sequence suggests a 555 oscillator period of 0.1 seconds (10 Hz). From there, it is a matter of counting and decoding: simulate this circuit – Schematic created using CircuitLab

19

Many ages ago not long after the first caveman whittled a piece of silicon into a transistor, they learned how to make a bunch of transistors on the same hunk of silicon. This led to all kinds of things that we consider as always having been around, like the 741 opamp, 7805 regulator, and 2N2222 transistor. Yes, these things actually had beginnings. They ...

19

Systemic problems and upsides of the NE555 Let's refer to a common datasheet, the TI NE555 datasheet. Power usage The NE555 is really power-hungry. As in, it realistically requires a supply voltage >= 5V, and uses a typical current of 10 mA at no load, no switching; that's at least 50 mW for doing nothing. Modern microcontrollers work at lower voltages, ...

18

The original 555 with 5K resistors: And here is a CMOS version with 40K resistors: The choice of resistors for R7, R8, R9 (bipolar version) would be influenced by two things- 1) The desire to minimize power consumption (as high value as possible without using up too much chip area) 2) The desire to minimize temperature variations due to beta changes of ...

18

Your design is flakey. You are using the conductance of water to trip the reset line and this is not a great idea given that the reset pin current (the leakage current from the reset pin) is in the realm of 0.1 mA to 1.5 mA depending on the voltage state of that pin. Read the DS. So, if it is producing 0.1 mA (ignoring the water sensor effect), this ...

17

The obvious answer here is to not even try to do this with evil 666 555 timers. You want three signals that need to be kept in phase, and one of them is a double pulse. While this could eventually be accomplished with a mess of 555 timers, it is actually quite simple to do in firmware. All you need is a micro with 3 outputs. Even the tiny PIC 10F200 can ...

17

The IC was designed in 1971 by Hans R. Camenzind under contract to Signetics and introduced in 1972. The 555 is still in widespread use due to its low price, ease of use, and stability. It is now made by many companies in the original bipolar and in low-power CMOS. As of 2003, it was estimated that 1 billion units were manufactured every year. The 555 is the ...

17

What I expected to see was a single peak at 2.5kHz. I don't know why. You need to reset your expectations. Think of this this way: If you just had a single peak, then the input would by definition be a sine wave. But you're feeding it a square wave, so how do you account for the difference? I know that a square can be constructed from sine waves. ...

16

Low low low cost is the aim. A small volume maker with no expertise in anything other than manual soldering (if that) can make these. They can be even made at home by workers if desired. Single side phenolic board. Cheap. Not only manual assembly but component size and hole spacings not well matched and leads are hand bent and nobody cares. A design ...

15

When you are holding a hammer, the world looks like a nail. Roughly speaking, a spectrum analyser captures a time record and represents the resulting capture as a unique linear combination of sinusoids. It does not mean that whatever generated the signal generated separate sinusoids, only that the resulting signal can be represented in this (very useful) ...

14

The LTC1799 can be programmed by a resistor and a link to produce a vast array of frequencies. Is there a small (6-pin) DIP I can use to generate multi-MHz squarewave clocks with an external crystal? It's certainly small It's 5 pins not 6 It's not DIP but you want "small" and DIPs are not small in comparison to SM devices. It doesn't use a crystal

14

I don't think you will ever get the accuracy and stability you want from a 555 timer. The pulse width is determined by the values of resistors and a capacitor, and the values of these elements will change with temperature and over time. For a precise pulse duration you should be looking at a crystal oscillator with a digital counter to generate the desired ...

13

A CMOS 555 like the TI LMC555 has a supply current of around 300 µA at 9v (the datasheets don't have values specific for 9v, but list a maximum supply current of 400 µA for 12v, so I'm interpolating). Holding the device in reset adds a negligible amount of current (10 pA) which can be ignored. A typical 9v battery has a capacity of 550 mAH. Assuming there ...

13

As awesome as the 555 is, its drift may be a bit too high for periods this long. I recommend you look at using a MCU instead, such as the ATtiny4. It comes in SOT23-6 packaging and has an internal RC oscillator which is reasonably accurate. If you move up to a ATtiny25/45/85 (SOIC-8/DIL-8/QFN-20) then you can program it using the Arduino IDE once you ...

13

It takes almost 2 mA just to charge and discharge the gate of your MOSFET. You're also wasting about 5 mA in R1, since it is grounded through pin 7 about half the time. Your voltage feedback divider is drawing about 1 mA from the high-voltage rail, which translates to more than 20 mA at the input. There's a problem with using a 555 to drive a large MOSFET: ...

12

Yes, you can make a Pierce oscillator with one chip and a resistor and the two load capacitors, plus maybe a series resistor if the drive power is too great for your particular crystal. The inverter must be an unbuffered type or it will likely do all kinds of undesirable things such as oscillating at two frequencies at once. Here is TI application note on ...

11

You'll need to couple the square wave with a capacitor to the transformer because standing DC voltages will just cause heat but there is no problem feeding a transformer with a square wave in principle. In practice, if you fed a 60Hz "power" square wave to a transformer, the higher order harmonics in the square wave would mean that a regular AC power ...

11

The reason you are getting something other than the sum of the two independent signals is because the two circuits are interfering with each other when you connect their outputs. One solution is to put enough resistance in series with each output so that the signal generators can't interfere with each others' operation. 100 kΩ resistors in series ...

11

Design Requirements restated from the question: - Vin = 5V nominal (I assume 4.75V minimum) - Vout = 12V nominal, +/- 0.5V ripple - Iout = 15mA peak Since I'm an applications engineer at Maxim Integrated, I did a quick check for parts we make that can meet those requirements. Regardless of whether you buy from Maxim or not, it's worth looking at the ...

10

Well a larger capacitor needs more energy stored to trigger the threshold pin, this energy is just dumped when the output is low. A lower valued capacitor needs larger resistors for the same timing but waste less power. I'd say make the capacitor as small as possible while keeping the circuit functional to save power. Eventually the resistor needed will be ...

10

The 555 output can Source or Sink 200 mA so unless you want to drive the mosfet in high frequencies it will be more than enough. A single transistor is probably going to drive the mosfet worse than the 555 itself If you are need to use a driver between the mosfet and 555 then you are better off with a push-pull (totem pole) driver configuration that can ...

10

Yes, a transformer will worth the opposite when connected backwards. You have a roughly 16:1 transformer - for every 16 volts you put in you get 1 volt out. In a perfect world, for every 1 volt you then put in the secondary you'd get 16 volts out of the primary. So 5V p-p would ideally give 80v p-p out. However, it's not quite that simple. Transformers ...

10

The actual 555 timer frequency depends on the RC time constant, which is strongly affected by variations in the actual resistance and capacitance. The resistor has initial accuracy spec (like 1% or 0.1%) and also varies with temperature. The capacitor also has initial accuracy spec (often +20%/-80% or worse) and varies with applied voltage. A better way is ...

9

The PNP transistor is part of a constant current source for the capacitor. The current is determined by the voltage across $R_E$ (ignoring the relatively small base current). The voltage across $R_E$ is $V_Z - V_{BE} + V_D$ where $V_D$ is the forward voltage across the diode, so we can say that $I_C \approx$ $(V_Z - V_{BE} + V_D)\over R_E$ ...

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