# 555 confusion

1. Apr 23, 2015

### helofrind

So I built a 555 astable oscillator. I wanted to get 50 KHz on my output, which I have achieved, but the values of my components do not match the formula I have calculated. I have:
R1 = 98 ohms
R2 = 67 ohms
Cap = 100nF
So I have:
1.44/(98 + 2 67) 100nF
Which should be 62KHz. Also can anyone explain why im getting 1volt on output with 5 volts in. Thanks in advance for any help.

2. Apr 23, 2015

### phyzguy

You need to include the circuit diagram so we know which components you are talking about.

3. Apr 23, 2015

### Merlin3189

Reasons could be
for frequency error- component tolerances: 24% is a lot, but maybe possible. How are you measuring f?
for voltage output- output high is typically 1 to 1.5 V below supply
- mark to space ratio of about 2.5:1 could reduce the average output to about 50% of supply, ie. 2.5V
- but I can't get it down to 1V, so: how are you measuring it? what is your power supply? Could there be errors there?

4. Apr 23, 2015

### Svein

I need to see the schematic, but I think one of the problems is that the values of R1 and R2 are too low. I would scale them both up by a factor of 10 and reduce the capacitance by the same amount. See if that helps.

5. Apr 23, 2015

### helofrind

Thank you everyone for responding. Here is a snapshot off of electrodroid, its the same circuit. At the bottom it does say to use resistors above 1K for both R1 and R2, so I will scale up the resistance I'm using a 2072 rigol o-scope to measure, and a bench power supply. I haven't generated a frequency this high, so my first thought was a frequency that high things probly get a little unstable.

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6. Apr 23, 2015

### Eddie Sines

Do you have a pull up resistor on the output?

7. Apr 23, 2015

### helofrind

No I dont have a pull up resistor, but ill try it. I wouldnt know what value to use though

8. Apr 23, 2015

### Eddie Sines

Try a 10K.. its safe... tie it to +5VDC to output pin... then connect scope to output pin... let me know the result. The 10K will not hurt the 555...

9. Apr 23, 2015

### Svein

Try 3.3kΩ.

10. Apr 23, 2015

### Eddie Sines

Sure that will work... even a 1K will do the trick..

11. Apr 23, 2015

### Svein

It just goes to show - a pullup resistor is not a critical component. My 3.3k is a relic from the TTL days.

12. Apr 23, 2015

### Eddie Sines

I have a book on the 555... let me see if I can find it...
the other thing is have you simulated it.. sometimes that is a lot of fun... and it saves time

One last thing is are you sure the parts are connected as like in the schematic? Often you will find that something is not connected... it may look connected, but it not.
I did breadboarding for many years... always cross-check

13. Apr 23, 2015

### Eddie Sines

Helofrind,
The 555 chip will do a lot more that 65 KHz... all you need to do is adjust the RC...I have used this chip since the 70's... It a great chip.

14. Apr 23, 2015

### helofrind

I have reconfigured the circuit. It looks a lot better mathmaticaly. But just for fun im going to try a pull up resistor with a 10K pot from source to output. I have checked the connections multiple times and even reset the scope. Ill have to try these new things a little later when I get home. Ill keep everyone posted. New configuration in attachment.

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15. Apr 23, 2015

### jim hardy

1. You haven't said whether you are using the CMOS or bipolar version.

2. Your schematic shows 98 ohms for R1.
That means when your discharge pin 7 is "low", it must sink 50 milliamps through R1 plus some more from R2.
The IC should do that, but do not think pin 7 will go down to zero volts while sinking that much current.
So - if pin 7 is at a fraction of a volt instead of zero it'll slow down your discharge, which affects timing..

Take look at the spec sheets.
Look here for cmos version http://www.ti.com/lit/ds/symlink/tlc555-q1.pdf
page 5 of 15 , specifications with 5 volt supply
parameter "Discharge-switch on-state voltage" about middle of page
it's allowed half a volt at just 10 milliamps.
Next, page 6 shows that with 15 volt supply it's allowed 1.7 volts at 100 milliamps
see page 8 fig 2, chart of "Discharge switch on state resistance" and observe it's probably around 10 ohms at 50 ma. That's a half volt.

Look here for bipolar version http://www.ti.com/lit/ds/symlink/ne555.pdf
page 5 0f 36, parameter "DISCH switch on state voltage" about middle of page
it's allowed 4/10 volt at just 8 milliamps.
Now look at page 7 figure 1 , the curve of output volts versus current with 5V supply . Assuming the output and discharge transistors are similar, at 50 milliamps you might see more than a whole volt.

I think that's the source of your discrepancy.
Try making your discharge pin currents smaller by using larger R1 and R2.
Then the device will do a better job of discharging C1 so you should find yourself operating closer to the formulas.

Last edited: Apr 23, 2015
16. Apr 23, 2015

### jim hardy

i see svein already suggested that.

Those spec sheets should let you pick values that'll work better.

Let us know how it works out ?

Late entry - i see you came to same conclusion while i was typing.... 10:21 pm..

always glad to see 'hands-on' experimenters. We learn best by doing.

old jim

Last edited: Apr 23, 2015
17. Apr 23, 2015

### Eddie Sines

Helofrind,

See attached... I didn't take a lot of time... but this circuit does simulate. I can send you the simulation model if you provide an email. You can download a Free Spice simulation tool via web go to http://www.linear.com/designtools/software/ down load LTSPICE... now you can simulate before you build... good luck.. -e

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18. Apr 23, 2015

### Eddie Sines

Hi Jim,
New member here.. just stared today... looks like a nice place to spend some time. later -e

19. Apr 24, 2015

### helofrind

Jim Hardy that was a great and helpful explanation, thank you. I am using a CMOS. so I increased the resistance and the components
better match the formula now. also the pull up resistor had no affect. I did however ad a capacitor from source to ground to filter noise,
and that help clean up the square wave (not shown in the picture).
R1 is now 1488
R2 is now 4754
C1 is 2.2 nF
I said before I was getting 1V, but actually its more like 500 mV. can I amplify this output?
(below I attatched the output signal)

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20. Apr 24, 2015

### Svein

It looks to me as if you are using a 10x probe - which means multiply the amplitude by10.