0
\$\begingroup\$

In the below circuit, the DC component (6V) of a bridged car head unit speaker wire turns on a LM53601 buck converter. The LM53601 accepts voltages up to 42V. The +12V input into Q2 is protected to withstand the pulses described in ISO 7637. I have used this circuit in my car for some months now, and it works fine.

But now I want to extend the functionality so that the LM53601 is turned on either with the +6V speaker signal or when the ignition is turned on. The EN signal of the LM53601 accepts up to 42V, but I believe that I will have to protect this input for the pulses of ISO 7637 as well. My proposed solution is shown in blue where I suggest to use TVS-diodes and a reverse polarity diode. This seems like an overkill solution to me, but I have not found any other way to protect the input.

Proposed circuit

My question is regarding Q2. Should I put a diode (STPS2200U) or a resistor (or both) in the point where I have drawn a red arrow? The 3906 PNP transistor specifications are listed below.

3906 specifications

Update: The transient protection circuitry was taken from a TI reference design using a similar buck converter that has the same voltage input range (see below). This passed ISO 7637 testing so that is why I thought it was smart to use the same design. The design uses a smart diode controller (LM74610) for reverse polarity protection, but I have used a 200V Schottky instead which I believe should be ok.

enter image description here

\$\endgroup\$
  • \$\begingroup\$ all switches must define some load. What is yours? Then we can see if it is overkill. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 22 '17 at 17:59
  • \$\begingroup\$ @Tony Do you mean the characteristics of the enable pin of the LM53603? EN to GND: Min -0.3V, Max 42V EN pin input current at 13,5V: 2,7uA \$\endgroup\$ – JohnDonut Sep 22 '17 at 18:18
  • \$\begingroup\$ Doesn't your radio have a switched accessory output? \$\endgroup\$ – Passerby Sep 22 '17 at 18:43
  • \$\begingroup\$ @passerby Unfortunately not \$\endgroup\$ – JohnDonut Sep 23 '17 at 10:25
0
\$\begingroup\$

Q2 is effectively a diode already so no additional diode is required.

I'm not too convinced about the clamping circuit though and I don't see a pull down on that enable line so it's effectively floating when off which is a no-no according to the data sheet.

I think I'd build it something like this instead. D1 ensures the enable can't wander above the power rail and R1 limits the current through that wire if it does. R2 pulls the enable low when everything else is off.

Not sure why you need C1 though. It will cause a significant delay on disable, especially at 10uF.

schematic

simulate this circuit – Schematic created using CircuitLab

ADDITION: For reverse battery protection too, this is probably better.

schematic

simulate this circuit

\$\endgroup\$
  • \$\begingroup\$ Thanks! Definitely need the pull-down you suggested. I've updated the question with reference to where I found the transient protection circuit. \$\endgroup\$ – JohnDonut Sep 22 '17 at 18:25
  • \$\begingroup\$ @JohnDonut ya that's a bidirectional protection circuit though. Limits to +X Volts, -Y Volts. You only need a unipolar one. \$\endgroup\$ – Trevor_G Sep 22 '17 at 18:30
  • \$\begingroup\$ OK. Forgive me for asking, but I'm a novice when it comes to these things: Why would TI include a bidirectional circuit when they also have reverse polarity protection? They say the following about the TVS diodes in the design specs: The positive clamping device should clamp above double- battery (jump-start) and clamped Load Dump voltages, but lower than the maximum operating voltage of the downstream devices (40V). The reverse clamping device should clamp all negative voltages greater than the battery voltage so that it does not short out during a reverse-battery condition. \$\endgroup\$ – JohnDonut Sep 22 '17 at 18:42
  • \$\begingroup\$ @JohnDonut hmm. Well.. I suppose that is a real issue with autos right enough. But that is not really transient protection though... IN that case I'd move R1 all the way left in the circuit in my answer. \$\endgroup\$ – Trevor_G Sep 22 '17 at 18:56
  • \$\begingroup\$ @JohnDonut mind you, hooking up the battery backwards will cause serious headaches all over. \$\endgroup\$ – Trevor_G Sep 22 '17 at 18:58

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.