Shutdown circuit for LM3478 - Page 1
Forum section: electronics / projects

author: shadewind
author_rank: Frequent Contributor
timestamp: November 23, 2012, 11:07:15 pm
content: I'm building a boost converter for Nixie tubes around the LM3478 (http://www.ti.com/lit/ds/symlink/lm3478.pdf) and it's been working great and the output has been stable. I wanted to add the ability to shut it down using a control signal since it actually does draw some power even the tubes are not lit. I want it to be a single control signal that can be controlled with a pin on an MCU.The LM3478 shuts down when the FA/SD (the pin doubles as frequency adjust) is pulled high and my initial design of the shutdown feature used a combination of a NMOS and a PMOS to pull the pin to VIN (which is 12 V) when a high signal is given on the control input. For reasons that are currently beyond me, I never caught the fact that the maximum voltage allowed on this pin is 7 V and indeed, it did break with 100% certainty when I tried it.The data sheet lists an example circuit on page 13 but this requires some voltage source between 1.3 V and 7 V to be present.The idea I currently have simply involves using a diode with the cathode connected to the FA/SD pin. When a high signal is applied to the anode of the diode, it becomes forward biased and the LM3478 shuts down. When theres a low signal applied to the anode, the diode becomes reverse biased and acts as a short circuit. Would this work? Any better ideas?

author: c4757p
author_rank: Super Contributor
timestamp: November 24, 2012, 12:59:57 am
content: I'd go with the example circuit as it's simple. Surely you can find a voltage source between 1.3V and 7V? How about your MCU's VDD? Your diode idea should work, except that they want the shutdown signal to be fed in through a resistor. Just add the resistor in series with the diode. Make sure, however, to choose the value correctly - you'll end up forming a voltage divider between that resistor and the frequency set resistor, and you need to make sure the final voltage, both attenuated by the voltage divider and dropped by the diode's forward voltage, exceeds the 1.3V threshold.

author: Bored@Work
author_rank: Super Contributor
timestamp: November 24, 2012, 10:59:04 am
content: Sounds like you could drive it directly with an MCU pin if your MCU has tri-state outputs:LM3478 enable: Set the MCU output pin to high-Z.LM3478 disable: Set the MCU output pin to high.Make sure the frequency adjust resistor to ground is high enough to not overload the MCU output in high state.

author: shadewind
author_rank: Frequent Contributor
timestamp: November 24, 2012, 05:25:00 pm
content: Quote from: c4757p on November 24, 2012, 12:59:57 amThe thing is, I want this module to the self contained and not depend on any particular MCU. Also, I don't want to run another wire for 3.3V.Quote from: c4757p on November 24, 2012, 12:59:57 amThat seems to only be the case for that particular circuit. The resistor makes it possible to pull the signal down at all.Quote from: Bored@Work on November 24, 2012, 10:59:04 amYeah that should probably work. The MCU I'm using is an LPC1758 which has all sorts of GPIO features. However, I want to be able to control it using an AVR (for example, using an Arduino) as well and the AVR series has no explicit support for tri-stating. However, the same effect should be possible by simply setting the pin to input, shouldn't it?Quote from: Bored@Work on November 24, 2012, 10:59:04 am165k should be plenty enough.Would the diode idea work otherwise? It feels like a more set and forget solution that doesn't depend on the control signal being of any particular kind. Any drawbacks to this?

author: c4757p
author_rank: Super Contributor
timestamp: November 24, 2012, 09:10:52 pm
content: Yeah, the diode idea should work.Of course, if you did use their example circuit, you could also use a voltage divider between VIN and ground to get the lower shutdown voltage.

author: shadewind
author_rank: Frequent Contributor
timestamp: November 24, 2012, 11:35:57 pm
content: Quote from: c4757p on November 24, 2012, 09:10:52 pmAny advantages to the example solution? The diode solution uses fewer components but maybe there's some disadvantage.

author: c4757p
author_rank: Super Contributor
timestamp: November 25, 2012, 12:55:42 am
content: Safety - if you use an on-board pulldown resistor on the gate, then the default mode is "off". With the diode you must feed in a signal to shut it off. And as far as fewer components, resistors are cheap as hell. (I'd be putting one in series with the gate as well, if the signal is coming from off board.)Speaking of safety, you should probably make sure that the voltage rating on whatever device you're using (reverse voltage for the diode, VDS- and from that VDG- for the FET) can withstand the output voltage of the power supply. I presume the MCU board is connected to buttons and such, controlled by a human. This way, if something goes horribly wrong on the power supply, the MCU board and anybody touching it are insulated from the voltage.