0
$\begingroup$

I want to run an AC sweep from 1 Hz - 10 MHz and calculate the output of the circuit (the voltage on RL), however spice shows that my voltage is 0 in a lot of parts that it shouldn't. I believe the circuit to be correct since it is copied from a book. I must have misunderstood something because I run into similar problems using the sine source (transient sweep not ac).

enter image description here

$\endgroup$
5
  • $\begingroup$ Where is it showing 0V, where you are expecting an alternate value. Does the values match your analysis? $\endgroup$ Commented May 31, 2020 at 22:56
  • $\begingroup$ The voltage on RL is the main problem, I have to run an ac sweep and find how it's amplitude changes but it is always 0 like my ac source is not working. $\endgroup$ Commented Jun 1, 2020 at 9:22
  • $\begingroup$ Do you know if the transistors switching on? What voltage are you expecting at $V_{RL}$ $\endgroup$ Commented Jun 1, 2020 at 16:44
  • $\begingroup$ I have no idea why the wouldn't switch on, how do I check? I also try to run an transient simulation with an sine source (50mv 1kHz) but that doesn't work either. The voltage on RL is always 0 ( apart from a few nanoseconds in the start of the simulation where it's unstable, but I don't think that counts) $\endgroup$ Commented Jun 1, 2020 at 18:40
  • $\begingroup$ Did you delete the previous post? Take a look at the datasheet for the transistor to figure cutoff. active and saturation regions for the transistor. Here is the datasheet for BC238 Also take note the BC238 transistor is tag obsolete. $\endgroup$ Commented Jun 1, 2020 at 21:16

1 Answer 1

0
$\begingroup$

It seems to be a missing ... Wire from top R5 and collector of BC238. However, gain seems low ... -35 dB from 10 Hz to 200 kHz. Adapter for high voltage piezo or same.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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