1
\$\begingroup\$

I am trying to create a simple and cheap frequency circuit with a 555 timer to drive a piezoelectric transducer. The transducer I am using calls for a 48v input with 1.66Mhz frequency timing. I've used a CMOS 555 timer to create the frequency (using 2 trim pots to do some testing). I don't have an oscilloscope but my multimeter shows I'm right on track.

Now, I am trying to use a 5v USB cable and I obviously need to boost the voltage to 48v. I'm not sure how or where I should be putting this boost converter, or even if I'm doing this right?

Here's the circuit I'm using for the 555:

schematic

simulate this circuit – Schematic created using CircuitLab

Do I just turn the 5v input into 48v input? or do I have to boost it after the output from the 555?

When I measure the voltage with the multimeter (in place of the crystal) I am getting 1.4v. Obviously not enough to drive the piezo element.

\$\endgroup\$
1
\$\begingroup\$

The simplest solution would be to use a 1:10 transformer to transform the 5V swing of the 555 output to about 50V at the transducer. Put a capacitor (about 0.1 µF) between the 555 and the transformer to eliminate the DC bias.

However, you should note that 1.66 MHz is rather high for a 555 in the first place, and it might not be able to drive much power at that frequency.

Also, you'll probably get better results if you can get the duty cycle of the waveform closer to 50%.

\$\endgroup\$
  • \$\begingroup\$ I'm taking a look for different transformers and researching them now. The 555 that I got is rated up to 3MHz typical, It can drive the frequency, but no, I don't think it can drive that power. \$\endgroup\$ – ntgCleaner Jun 22 '14 at 18:51
0
\$\begingroup\$

You obviously need a boost converter for this. You could then apply the output of the 555 to the base of a transistor (or darlington pair). The transistor would then actually drive the piezo transducer. Then refer to the data sheet of the piezo to find out which capacitor and resistors would be adequate. I have done something similar in the past but with 9v buzzer and the output of an MCU (instead of 555) and it worked fine.

Another solution, but I do not know how stable it would be, would be to use one of those boosters with a shutdown pin, and let the 555 control the shutdown pin, but this would probably mess up the frequency if it even work.

\$\endgroup\$
  • \$\begingroup\$ Please let me know if I'm understanding properly. The output of the 555 would go to a transistor that had the required 48v to the collector then the transistor would switch with the frequency of the 555 and when it switched on, it would allow the 48v to flow through? (Still new to all of this) \$\endgroup\$ – ntgCleaner Jun 21 '14 at 8:12
  • \$\begingroup\$ yes, correct. You will have to check if a single transistor will do or if you need a darlington pair. If you can breadboard your circuit the easiest way is to just try. \$\endgroup\$ – Bitgamma Jun 21 '14 at 8:28
  • \$\begingroup\$ Yes, I have plenty of breadboard space. Now I just need to figure out how to make a boost converter with the parts I have or can get. Any suggestions? I've never made a boost convert and I need to step-up from 5 -> 48v \$\endgroup\$ – ntgCleaner Jun 21 '14 at 8:39
  • \$\begingroup\$ I suggest to buy a boost converter if you want good results. Linear Technologies and Texas Instruments are your best bet. \$\endgroup\$ – Bitgamma Jun 21 '14 at 10:57
0
\$\begingroup\$

I have seen cheap door alarms use a very small transformer to convert two button cells worth of voltage into a nasty voltage which can drive a low power piezo transducer.

Cheap voltage boost transformers

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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