Tell me more ×
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free, no registration required.
up vote 3 down vote favorite

I have a 555 timer setup in astable mode, with a frequency of about 1Hz which I measured and verified without load. Out of curiosity, I connected an 8 ohm speaker, expecting not to hear anything. However, I heard a tone, and when I looked at the signal on a scope, the frequency was about 1.4KHz. How can the frequency change like this when I have a load?

schematic enter image description here

Update:

@s3c: I added a 47 ohm series resistor at the output of the 555, and re-measured the signals. I no longer see the strange 1.4KHz output. The 555 output looks normal (yellow scope trace). Should the speaker input (blue trace) be an AC coupled, reduced voltage version of the 555 output?

share|improve this question
Can you confirm the scope output with and without a load? Also can you post a complete schematic? Did you build this on a breadboard? – s3c Jun 19 '12 at 1:44
I attached the schematic just now. I'll have access to a scope in a couple of days, and I'll post waveforms. Yes, this was built on a breadboard. – Ravi Jun 19 '12 at 3:07
You don't need to reduce the voltage, only the current. Depends on what you are trying to do but I don't see any real need to AC couple the speaker in this application. – s3c Jul 16 '12 at 19:51
The series current limiting resistor is reducing the voltage seen at the speaker. And I thought AC coupling would prevent the voice coil from being biased in one direction. – Ravi Jul 17 '12 at 3:35

1 Answer

up vote 3 down vote accepted

The 555 is only rated to about 200mA input/output, I'm guessing the output current through your 8 Ohm speaker is pulling the voltage down low enough to screw with it's operation. Try a smaller load and see what you get (20mA or so, so try with a 300 Ohm resistor).

If you need more juice use the 555 to drive a transistor.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.