0
\$\begingroup\$

I have a simple adjustable power supply (1.8V - 16V / <1A) that I put together with an LM350, works great for the various small bench-top things I use it for. Typically on a load being applied, I see the typical small voltage drop as current spikes, all expected.

Last night I hooked it up to a hacked together transistor-as-audio-amp circuit I threw together on a breadboard to test the DAC on a new MCU. Eveything went as expected, sound played fine, but I noticed that my voltage went from 8.4V to 9.75V, around a 16% increase! I'm not completely new to building low-ish power circuits, but I'm trying to get better at the theory, and I'm not seeing why this increase occurs. It is constant, not a spike, and a jump of 1.35V seems like a lot.

Of course using a BC547 as an audio is non-optimal, and the transistor only does an ok job handling the current, but I only turn the circuit on for a few seconds at a time. I was thinking that there is some transistor-related phenomenon that I am unaware of that would account for the voltage, but it does occur in circuit simulators, which leads me to believe that it's a real parts vs ideal parts phenomenon. Can anyone point me to a source for understanding why this happens?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
1
  • 1
    \$\begingroup\$ One cause of this problem involves the fact that LM350 can supply current to a load, but it cannot sink current. An energy-storage component like a capacitor or inductor is also involved. \$\endgroup\$
    – glen_geek
    Sep 14, 2018 at 14:16

3 Answers 3

3
\$\begingroup\$

Probably you don't have a large enough capacitor on the output of your regulator to keep the voltage relatively steady. Try something like 1000uF. The capacitor gives the relatively sluggish regulator time to catch up.

Your peak current could be relatively large- 8.4V/4 ohms is over 2A so the transistor will be limiting.

Note also that the regulator cannot sink current- it can only source it- so without some kind of passive load (called preload in the regulator datasheet) the regulator will tend to err on the high side. A capacitor will smooth it out so that's less of an issue.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Thanks! I know enough to have been able to figure it out, but just putting 100uF on the output near the speaker solves the problem nicely. The transistor is limiting, but I happened to have one sitting on my desk within arm's reach and I was just verifying the DAC output (I could have used a MM, but where's the fun in that?) \$\endgroup\$ Sep 14, 2018 at 15:17
1
\$\begingroup\$

Also check for high frequency noise from the audio transistor. Even with a large bypass capacitor, if your audio circuit is generating a lot of noise, that can also affect the feedback loop of the regulator. You could even consider adding some smaller capacitors (like 0.1u or 0.01u) in parallel with your large electrolytic cap to help with noise suppression.

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

You have inadvertently created a bit of a boost converter! Look at a generic boost converter circuit and recognize that your speaker is the inductor.

enter image description here

The LM350 will not sink current by itself.

\$\endgroup\$
2
  • \$\begingroup\$ The boost would be at the transistor collector rather than the regulator output, no? \$\endgroup\$ Sep 14, 2018 at 15:12
  • 1
    \$\begingroup\$ @SpehroPefhany That is correct. It would depend on where he is measuring this voltage. \$\endgroup\$ Sep 14, 2018 at 15:29

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.