1
\$\begingroup\$

I want to install amplifier module based on TDA7297 dual bridge amplifier in my car. The kit is based on the low-cost example shown in datasheet (fig. 3, page 5).

Is there any recommended way to protect the input? I don't want to fry my phone because there is spike in the voltage when starting the car or something like that.

\$\endgroup\$
2
  • \$\begingroup\$ See electronics.stackexchange.com/questions/17108/… . Note that the input circuit shown is capacitively coupled, so if you use suitably rated capacitors (100V?) the input device should be safe even if the amplifier isn't. But you need to protect the amplifier as well, not just wire it directly in to the "12V" rail. \$\endgroup\$
    – pjc50
    Dec 7, 2015 at 12:19
  • 1
    \$\begingroup\$ @FakeMoustache 20V is way too little for automotive spec. Power spikes and load dumps are insane and just a simple inductor can make it ring at those 40 to 100V levels if you unluckily choose wrongly. You need more in cars if you want to be 100% sure to be using it a long time. \$\endgroup\$
    – Asmyldof
    Dec 7, 2015 at 13:18

1 Answer 1

Reset to default
3
\$\begingroup\$

Load dumping... be prepared: -

enter image description here

Maxim say: -

The most aggressive of these surges is the so-called "load-dump pulse" (Figure 1). This transient occurs when the engine is running and the battery lead is disconnected, due to a poor connection or accidentally while the alternator is charging the battery. The magnitude of that transient depends on the alternator's rotational speed and field excitation at the moment of disconnection. The surge can last several hundred milliseconds and reach levels of over 100V, which is potentially lethal to semiconductor circuits.

Do you need to take care of this potential problem? It's up to you. Here's the article.

It offers a few ideas about what to do to protect against load dumps such as the circuit below (obviously using a maxim device): -

enter image description here

So, protect at source and if this can't be done protect as close to the source as possible and if that means the two wires feeding your amp then so be it.

\$\endgroup\$

Your Answer

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

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