Thank you. It shall be fixed.Hi Bill,
You have a problem with the polarity of the diodes in figure 20.
A diode AND gate, redrawn, gone wrong.
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Thank you. It shall be fixed.Hi Bill,
You have a problem with the polarity of the diodes in figure 20.
Do you want to vary the pulse rate (frequency), or the duty cycle? If you just want to adjust the frequency check this thread out, I think it will do what you want...Mr. Marsden
Could you modify your Fig. 8 circuit with the Two LEDs ? I need one that will work on a battery voltage of 6v to 12v and I know the 555 I.C will handle that range. Also I need a variable resistor of 50k, I have lots of them in stock, so that I can have Either One LED or use Two but can adjust the flashing rate to simulate a flame in a wood stove. Can you please include all the parts in a circuit that I can put together ?
P.S. I am the same person who sent you a P.M about this.
Thanks For Time and Help.
Jure George
This site (AAC) is full of good ideas. One I've been working with another gentleman on concerns using 555's (or 556's, or CD40106's) to make 3 different LED throbbers. The idea is if done correctly it can make millions of colors.This going to be quite a read and educational too.
I was at an electronics store these caught my eye.
Not being advanced enough to make up my own circuit maybe you can direct me to something simple and basic to incorporate these components and learn:
"hands on"
Was it okay I left a comment here? I started at your blog, you requested comments to be posted here, right?and if you want to leave a comment or question this is the place to do it. Thank you.
Bill,Chapter 3: The LED / Resistor Only Bargraph
LEDs tend to drop a constant voltage when they are conducting. Its not perfect, but it can be used. Take the following schematic in figure 3.1. Ive included a schematic for a simple variable power supply using transistors and two 9V batteries if you want to experiment with them.
..........................Figure 3.1
You do not need any gating diodes when two lines from a 4017 are taken to a common point on the base of a transistor. Just use a 3k3 from each output to the base, and the transistor will turn on when either line is active "HIGH." The mid-point of the two 3k3 resistors will be "half-rail voltage" but the base will drop this to 0.6v and the base will see sufficient current to turn on.Chapter 11: Making Patterns
Light chasers are cool, but sometimes you want to do more. By steering the current from the 4017 IC sequence other special effects can be created, such as this one.
...................................................................Figure 11.1
This circuit will make the LED light sweep back and forth, a popular Hollywood effect. We have also added transistor drivers that will give the LEDs 20ma without significantly loading U2, which means this particular circuit should last. You may need to add some power supply capacitors, but in general battery circuits are pretty stable without them, as the batteries share some of the same characteristics as capacitors. The voltages from 9V batteries tend to drop fast, down to 7.5 volts, and then stabilize, so be aware. The 555 oscillator will go as low as one cycle every 3 seconds, with the other end being faster than the eye can follow, so it is very open ended for the user.
Another popular design shown in Figure 11.2 is the flasher used in emergency vehicles. This can get you a ticket if you try to use it on a street vehicle, but the basic design is pretty simple.
..
.................................................Figure 11.2.....................................................................................................Figure 11.3
Of course, a design like this practically screams bright lights, so I've shown several options in Figure 11.3. Toys usually use 9V batteries, which can drop as low as 7.5, so this limits what can be done. Some blue LEDs can have a Vf of 3.8V (and 3.8V X 2 = 7.6V). I'd use single transistor in Common Collector configuration shown in Figure 11.3 (also shown in Figure 9.1) to drive individual chains. If the circuit is drawing 100ma for the LEDs, and the transistor has a gain of 50, the current pulled from the CD40XX chips is around 2ma. At 9V and a Vf of 3.8V the LED current is are getting 21ma, if the Vf is 3.5V the LED current is 22ma. At 7.5 and Vf of 3.8 the LED current is 14ma. These calculations show this circuit tries to minimize the current variance for the LEDs. This was covered in the Current Limiting chapter.
If you have a stable 12V then the options are more open. Since you can put more LEDs per chain the total current per LED is reduced a bit. The calculated current for this layout is 21ma. If the Vf is 3.5 then the current would be 24ma. Again, the variation is minimized.
But what if you want a lot of LEDs, say 100 of them (50 chains)? This would be a current of 1 amp. A transistor with a gain of 50 would use 20ma through the base, more than the CMOS IC could provide. This would be a good time to use a Sziklai pair as shown. Q2 would definitely have to be a power transistor, but other than that it is pretty straight forward. This would bring the CMOS requirement to 0.4ma, which solves the problem.
I mentioned earlier that the CD40XX ICs could go above their individual counts. The datasheet shows how to do this, as well as Bill Bowden's Website. Figure 11.4 shows how this is done.
.................................................................Figure 11.4
The number of transistors and resistors used makes the method shown in Figure 9.1 to drive LEDs more appealing, doesn't it? U3 can be repeated for even more counts, if need be.
This was meant to be a quick and dirty power supply thrown together using common parts. This was explained in the first paragraph.This circuit Chapter 3: The LED / Resistor Only Bargraph has two faults. Firstly the darlington transistor is not needed and it doesn't have to be a "power" transistor.
Wasn't meant to, it is designed for the voltage the power was supplied for. The math explained this, if you had read the article.Secondly, the 75R current-limiting resistor does not give you very much tolerance for changes in supply voltage.
Actually, this is incorrect. Many LEDs appear as bright at 10ma as at 20ma. They are very nonlinear devices where brightness is concerned. These are indicators, not illuminators. The whole point of being a bargraph is some will be off while other will be on. Voltage, once Vf is achieved, in not important. They are current devices.A change of 1v will alter the current through the LEDs by 13mA. Most LEDs run on 15-25mA MAX and if the voltage drops by 1v, they will dim considerably.
Again, you prove you didn't read what was written, it was the setup for the following circuit, which was explained in paragraph 3.In addition, the 470R resistors do not serve any purpose AT ALL.
Actually, a 4017 is rated at 6.8ma, which was explained in the transistor driver section. Reading datasheets isn't your thing either.The 4017 will deliver 10mA per output and they can drive LEDs directly, so I do not see why you have said there is any reason why gating diodes are needed.
Depends on your definition of tested, I've built well over 20 LED bargraphs in my career, and I stated such in my later posts. Again, that reading issue. I did not build the power supply, nor did I feel the need to test it and see if it worked.You have also produced a circuit Chapter 3: The LED / Resistor Only Bargraph to which I have taken you to task as you have obvioulsy not tested the circuit at all.
A threat! Of course, you neglected to mention your website, which is OK with me. My reputation is intact, with or without your misinformation. Wonder how much of what you say will be out of context?In both cases you have not tested or tried the circuits and I have put them on my website under: "Spot the Mistake" to show others how faulty circuits are appearing on the web.
by Jake Hertz
by Jake Hertz
by Duane Benson