1
\$\begingroup\$

schematic diagram

Hi community, what I'm trying to do with this circuit is to get a 5V signal on the output whenever an input AC signal is present. That simple. It works perfectly with 9VAC on the input but when it goes down to around 5VAC, the transistor starts supplying the regulator less (around 4-5Volts) and output voltage goes down to 3Volts.

I am still learning electronics and just found out that I have to limit the current going into the base of the transistor with a resistor. What else am I missing here? Any comment would help. Seemed like an easy application.

\$\endgroup\$
  • \$\begingroup\$ Base resistor yes, and you have it wired as an emitter follower to the regulator. I believe you will not get the voltage gain on the collector from 12v. What are you trying to power with this application? \$\endgroup\$ – Archaeus May 31 '17 at 23:33
  • \$\begingroup\$ A voltage at the base in this config with a load at the emitter will be Vbase-0.7 for the junction drop. \$\endgroup\$ – Archaeus May 31 '17 at 23:43
  • \$\begingroup\$ @Archaeus I'm not powering anything. It is going into Arduino just like a data output. Yes, I never get 12 Volts. what would be the right way? \$\endgroup\$ – Alian4life May 31 '17 at 23:44
  • \$\begingroup\$ Move the output to regulator to the collector side if you wish to still keep experimenting with this circuit. The answer below addresses the other concerns. \$\endgroup\$ – Archaeus Jun 1 '17 at 0:02
  • \$\begingroup\$ This is all wrong for your application. You don't use a 7805 voltage regulator to set the level of a signal - it's for making power supplies. You also don't use an emitter-follower as a switch. \$\endgroup\$ – brhans Jun 1 '17 at 1:39
2
\$\begingroup\$

Your initial difficulty is that the 7805 regulator has a dropout voltage (below which it no longer regulates) of about 2 V. Look at the datasheet.
This means to get regulated 5 V from your 12 V supply the input needs to be greater than 7 V.

Your second difficulty is whether you need a simple level detector or more. Since what you show is a half wave rectifier, you will get approximately 0.8 V less than the positive peak voltage.

If you want an accurate level detection, then you should use a comparator to drive a BJT switch to turn on your 12 V input.

I notice in your comment that you don't need power ...you simply want a signal into an Arduino. Then all you need is a peak detector and comparator, and feed that signal into the Arduino. You wouldn't even need the 12 V supply.

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

This is my 'go to' circuit for detecting AC control inputs to a micro-controller. Vary the components to suit the input voltage, and operating times.

schematic

simulate this circuit – Schematic created using CircuitLab

Pick the zener voltage to suit your processor rail.

Pick C2 and R2 low enough impedance to deliver enough current at your input voltage and frequency to have enough voltage across R1 to turn the zener on, high enough impedance to protect the other components against normal and fault voltages on the input.

Pick C1 big enough to hold up between cycles, small enough for operating speed.

\$\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.