2
\$\begingroup\$

I've been working on a boost converter based on an LM2733-Y integrated circuit, as follows:

enter image description here

My question is, how to shut it down completely?

If the SHDN pin is in low state, I get the input voltage on the output.

I want to add a switch between a battery and a PIN 5. What should I use?

The converter must be controlled by a microcontroler unit (0 - 3.3V). I mustn't use any mechanical element e.g. relay. I'm thinking of an SSR, but it requires a lot of power to control it and it is relatively expensive.

A PNP transistor is not an option as well. Could you recommend me any solution?

What do you think of using this solution? enter image description here

\$\endgroup\$
  • \$\begingroup\$ A P-FET driven by an N-FET? \$\endgroup\$ – Wesley Lee Dec 17 '18 at 11:11
  • \$\begingroup\$ @WesleyLee any easier solution? What do you think of the above one? \$\endgroup\$ – Y. Markov Dec 17 '18 at 12:07
  • \$\begingroup\$ Why don't you put it before the whole circuit? I thought the point was to have a microcontroller turn the boost converter off? That placement between the diode and the feedback resistors doesn't make much sense to me. \$\endgroup\$ – Wesley Lee Dec 17 '18 at 12:25
  • \$\begingroup\$ We need to know what is the input voltage VBAT. \$\endgroup\$ – peufeu Dec 17 '18 at 13:48
  • \$\begingroup\$ The input voltage is up to 8 V \$\endgroup\$ – Y. Markov Dec 17 '18 at 14:47
1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Place a uC controlled P-mosfet between Vin and the inductor. Pull down the shutdown pin with 10K to ground. Connect shutdown to mosfet drain (between mosfet and inductor). When the mosfet is off, the shutdown will be pulled low, but the chip will still be powered from Vin. When mosfet is on, shutdown is high.

added

I have modified the schematic. If Vin is higher than uC voltage, the NPN is unavoidable. Also, the other answer says that the 10uF needs to be after the mosfet. This is prudent, but not essential. Doing so keeps the surge current of the switch away from the mosfet. In this case, an additional 0.1uF may be required to keep Vin steady at the chip. The chip Vin draws 3mA max from datasheet http://www.ti.com/lit/ds/symlink/lm2733.pdf. Current limiting in this chip is done between switch and ground pins, so putting the mosfet between Vin and the inductor is not a problem.

There may potentially be a few cycles of lag, since the 10uF cap will keep the shutdown pin high before voltage falls below shutdown threshold 0.5V. If this is not desirable, the 10uF may be moved back, but one has to ensure that the mosfet will survive the peak current [The AO3401 will survive].

\$\endgroup\$
  • \$\begingroup\$ My mistake I haven't said what the input voltage is. I have 7.4 - 8V in the input. I'm not sure if it won't damage the microcontroller. \$\endgroup\$ – Y. Markov Dec 17 '18 at 14:49
  • \$\begingroup\$ @Y.Markov Any way you look at it, you need a small transistor to isolate the 8V from the uC, whether it's the shutdown pin or any other pin. So at a minimum you will need a NPN to drive the P-mos. ...... unless you want to lift the ground of the switching converter and put a N-mos there.. in which case the shutdown pin is not actually used. \$\endgroup\$ – Indraneel Dec 17 '18 at 17:58
  • \$\begingroup\$ What is the purpose of the L1 coil? Noise limiting? Or is it a part of a boost converter circuit? \$\endgroup\$ – Y. Markov Dec 18 '18 at 13:52
  • \$\begingroup\$ @Y.Markov The 10uH inductor and 10uF capacitor are from your schematic of the boost converter. I think this should work. You may have to adjust the gate and base resistors for optimal switching speed, but it is not critical if you shutdown rarely. Make sure you check the number of cycles latency in having the 10uF cap near the shutdown pin. \$\endgroup\$ – Indraneel Dec 18 '18 at 16:44
  • \$\begingroup\$ @Y.Markov Do remember to accept my answer if it solves your problem. \$\endgroup\$ – Indraneel Dec 19 '18 at 23:15
1
\$\begingroup\$

The best option these days for a high-side load switch is either an integrated load switch, or a discrete job with a PMOS.

At the same RdsON, PMOS will be slower than NMOS (higher Qg) and use more silicon area, which is why integrated switches often use a NMOS driven by an integrated charge pump.

Since switching speed is not important, using a PMOS is not really a performance disadvantage.

I won't repeat the schematic for a discrete PMOS high side switch as it can be found everywhere, for example here.

I will consider maximum switch current for LM2733 as 1.5A from datasheet, so if we put the switch on the input, we need an 1.5A switch at least. If we put it on the output, its current rating can be lower. This depends on your load current.

Note that if the switch is on the input, it has to be BEFORE the input cap of the boost converter. The input cap C6 on your schematic should be very close to the boost chip, do not put the switch between them.

Full VBAT range is 3.7 - 8V

OK, this is important. If we put a PMOS switch on the input, we want it to be fully ON with a Vgs corresponding to the lowest input voltage. So we should select a 2.5V drive PMOS. If we put a PMOS which is ON with Vgs=5V but not with Vgs=3.7V then it will not work...

Let's go integrated first. Pop up the search engine: TPS22810 a nice little chip with integrated charge pump, about 80mOhm NMOS plus accessories in a neat package, takes logic input, very simple to use. This would work well on the input side. Good thing with a NMOS and charge pump chip is that the RdsON does not depend on input voltage, which is a plus.

If you want to put it on the output side, you can use an automotive load switch like this one. Since this is the voltage range used in automotive electronics, you'll get lots of options.

If you want to go discrete, on the input side the most important thing is to select a PMOS which will actually turn on with the lowest battery voltage, so you need tyo go there and select "drive voltage" <= 2.7V or something like that.

Here's a PMOS that would work. Note the datasheet specs RdsON < 64mOhm for Vgs 2.5V. So it is guaranteed to turn on at 2.5V. If it is not specified, then it is not guaranteed...

Now you can also put the switch on the output of course, and you need to put it AFTER the output capacitors C8/C9. Again, do not put the switch between the boost chip and its caps. In this case select a PMOS for 30V or more and suitable RdsON. You will need to ensure its maximum Vgs voltage rating is not exceeded though, so you will need to add a Zener to the high-side switch schematic above.

We could even do it like so:

schematic

simulate this circuit – Schematic created using CircuitLab

In this one, when the boost delivers enough volts on the output, something like the Zener voltage plus whatever is needed to turn on the FET, it will turn on. With only 8 volts, it will turn off due to the zener. I'm only mentioning this one as a quick hack since it will use a lot more idle current than a proper load switch, and also the FET can enter linear mode and burn if the output is overloaded and the voltage sags. So much for the elegant hack.

Also, some boost converters have a fully-OFF feature. They include an extra FET to block the current path from input to output when turned off.

\$\endgroup\$

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.