Looking for a calc that will solve for R1, R2, C in astable 50% DC with entering desired freq (1KHZ)and DC(50%). All the calcs I find require entering the component values to solve for freq and DC. Pain. Thanks JR
JR North wrote in news:hMKdnVrC7bBFZtXZnZ2dnUVZ snipped-for-privacy@seanet.com:
You'll still have to enter at least one component value as well as frequency and duty cycle.
Lostgallifreyan wrote in news:Xns97ACAE1B1E2F3lostgallifreyangmail@140.99.99.130:
Google, "NE555 calculator", second result:
That does what you're wanting. It assumes a practical value for one of the components and calculates the other values from the freq and DC input.
Did you notice Ra for 50%? Zero. The chip will try to discharge the power supply. ;-)
Cheers! Rich
Rich Grise wrote in news: snipped-for-privacy@example.net:
It had crossed my mind as odd. :) I didn't think that through though...
Try
You can get a 50% DC with a cap from the trigger and threshold pins to ground and a single resistor to the output. You need the CMOS version of the 555 for an exact 50% DC. The regular version output doesn't swing all the way between the rails, so there will be a small error.
Frequency = 1.44 / 2RC
-Bill
I guess it's just from all of these billions and billions of years of experience. ;-) Or maybe because I had tried it once, so I already knew the answer, and was just tweaking you. ;-)
I was seriously curious how the program would deal with it, and I guess it's another example of why computers will never replace people. ;-) ;-)
One time, I think (it's kinda lost in the mists of time) I actually used two resistors with a diode in series with one of them in place of Rb, and got pretty durned close to 50%, but I like the one-resistor, schmitt- triggeroid version more, I think. :-)
Cheers! Rich
Cheers! Rich
...and another example of why bad programmers (who don't do limits-checking) won't displace good programmers.
impossible with the normal 2-resistor circuit, to get 50% duty cycle the easiest way is the 1 resistor circuit,
----+--- vcc | +-[R1]--------|--------+ | | | | +--------+ | | | | | | | . . . .|. . . . | | | . VCC(8) . | | | . . | | +--RES(4) OUT(3)--+--> out1 (totem pole) | . 555 . +-------TH(6) DIS(7)-----> out2 (open collector) | . . +-------TR(2) CV(5)-- C1 | . . ===== . GND(1) . | . . . .|. . . . | | +-------------+ | ---+-- gnd
the frequecy is approximately 0.7/R1*C1
The output must move all the way to +Vcc for a 50% DC. Otherwise, the cap charge time is greater toward 2/3 Vcc and less on the way down, and therefore not equal. From the 555 spec sheet, I see the output moves to
+13.3 with a 15 volt supply and 100mA load. No data for lighter loads. But looking at the schematic, the high level is controlled by a emitter follower, and therefore the high level output will always be less than the supply voltage regardless of load. So, it doesn't look like the DC can ever be exactly 50% using the bipolar 555 with one resistor and cap.-Bill
The standard 555 astable circuit will only give you duty cycles above
50%. But you CAN get a 50% duty cycle with a simple mod to the circuit. Put a diode in PARALLEL with the resistor that connects pin 6 and pin
There is a "standard" circuit for achieving 50% duty cycle with a 555, and it does include a diode. Used it once before I decided 555's were crap.
1KHz is 1millisecond making the charge and discharge times 0.72/F or about 720us. Smaller C is better and R in megohms is about the upper limit for the 555, so C somewhere around 720us/megohms gives somewhere around 720pf. 720pf is almost a nf which is 0.001uf a nice standard value. So make C=0.001uf. Then R has to be like 720K. If R1 is the Vcc-connected resistor and R2 is the other one, maybe making R1=39K and R2=680K will do the trick. The charge time is 0.69*(680+39)K*0.001u=496us, and the discharge time is 0.69*680K*0.001u=469us. The frequency is then 1/(496u+469u)=1.036KHz and the duty is 496/(496+469)=0.513. Both of these numbers are nearly one half stock tolerance of R's,C's, and the 555 itself. If you're not worried about a little current, then make R1=10K and R2=720K. This makes charge time 0.69*730K*0.001u=504us and discharge 0.69*720K*0.001u=497us for a frequency of 1.001KHz and duty of 0.503. Close enough yet?
Rich Grise wrote in news: snipped-for-privacy@example.net:
I haven't used 555's much actually. :) I'd have to go back to the data sheet and some experimentation just to contribute specifically to the thread. Like Budgie, I don't like them much, they seem very loose for a standard, not an approach to an ideal the way the average op-amp is. I did use a 556 for a servo tester once though, it was helpful for that, but even then some simple means of direct voltage control would have been nice.
About programs, I agree. While good programming can solve things like this value predictor, there are limits to models. I never use SPICE modelling, I think electronics is as much art as science, and if I want that much detail I explore the real component's's oddities to see what helps and what doesn't. I actually do non-recommended things with op-amps at times when they work in the context I want, and I suspect a model wouldn't do this for me even if I could afford that kind of expensive software.
Chain curves have caused problems for mechanics computer design, and there's recently been work on imaging real ones to speed up the computer modelling. Likewise, modelling a bass guitar is harder than learning to play bass...
I suppose the more intelligent way to use that calculator is to enter duty cycles that are arbitarily close to and above but not exactly equal
50%. Of course, you wouldn't know to do this if all you've ever done is use calculators. Entering 50.3% , 1KHz, and C=0.001u, values obtained with some fairly rough insight into 555 operation, returns RA=8.64K and RB=716K- not std values according to my other formula derived mysteriously from additional applications insight, almost as good as, but a lot less fun than, the manual method.Have something to add? Share your thoughts — no account required.
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