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What: I'm designing a DC-DC Boost converter for a solar panel maxmimum-power-point-tracker (MPPT).

Setup: I'm using a TPS2819 MOSFET driver connected to a FQD16N25C N-MOSFET. The TPS2819 is driven by a 5V PWM source at ~500Hz. With the duty cycle set to roughly 50%.

Problem: The FQD16N25C gets very hot with a 48ohm resistive load on the output.

Shown below is the schematic of the boost converter stage:

enter image description here

A DC power supply was connected to input (Vsolar and GND). A 48ohm resistor was placed across the output (Vbatt and GND)

Shown here is the Vds wave form of the mosfet. With an input voltage of 2.64V, the current draw is 1.687A

Vgs waveform

From what I've read about N enhancement MOSFETs, what may be happening is that the MOSFET is operating in the linear triode region as opposed to full saturation. However, I'm not exactly sure how to solve this

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  • \$\begingroup\$ The circuit you've shown - where's the MOSFET you refer to and any chance you can compact it to make it more visible? \$\endgroup\$ – Andy aka Jul 18 '14 at 18:53
  • \$\begingroup\$ what is the size of your inductor L1??? For driving the boost converter at ~500Hz your inductor would have to be pretty sizeable. \$\endgroup\$ – Kvegaoro Jul 18 '14 at 20:43
  • \$\begingroup\$ Sorry about that, the inductor is 50uH rated for 16A and the NMOS has been made more clear. \$\endgroup\$ – Dub Jul 28 '14 at 23:19
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You appear to be driving the MOSFET with a 5V p-p signal - this is usually nowhere near enough to adequately turn these devices on - that is why it gets so warm. I'll also add that you haven't chosen a great FET for this application - it's a 250 volt rated FET with an on-resistance of typically 0.22 ohms. Clearly, the maximum voltage you are generating is going to be about 50V (due to the 63V rated cap on the output) - you should be looking for a MOSFET with about one-tenth the on-resistance and driving it at least 10V p-p.

In short, I'd say this is a bad choice of MOSFET.

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  • \$\begingroup\$ In particular, I'd highlight that all else being equal, a MOSFET with a lower maximum source-drain voltage will have a lower on resistance. Of course different manufacturing techniques can compensate, but these usually come with associated cost differences too. \$\endgroup\$ – Phil Frost Jul 18 '14 at 20:44
  • \$\begingroup\$ Thanks for the feedback Andy and Phil. One area I don't understand is the relation between gate charge and Vgs saturation. My understanding was that the mosfet driver allows more current to be delivered to the gate allowing the mosfet to be switched faster. \$\endgroup\$ – Dub Jul 18 '14 at 20:44
  • \$\begingroup\$ @DsaR If you want the simple answer to what gate voltage you should apply, just see the datasheet. It will specify an absolute maximum (which results in an almost instant, violent destruction of the device if exceeded) and a value a little below that which the manufacturer uses when testing all the other values. Use that. Usually it's somewhere between 8 and 15 volts, for power MOSFETs. \$\endgroup\$ – Phil Frost Jul 18 '14 at 20:47
  • \$\begingroup\$ I'm still confused about the Vgs. According to page 2 of the NMOS data sheet here: fairchildsemi.com/ds/FQ/FQD16N25C.pdf Under "On Characteristics", it says Vgs min = 2V and max = 4V I realize now that the Vds is way higher than it needs to be resulting in a higher than necessary Rds(on), but shouldn't the NMOS still fully switch on and off? \$\endgroup\$ – Dub Jul 28 '14 at 23:05
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    \$\begingroup\$ @Dub - this is Vgs(threshold) i.e the point where it's just starting to conduct a tiny bit - not a useful area to use as a switch. \$\endgroup\$ – Andy aka Jul 29 '14 at 7:40
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The min = 2V and max = 4V you refer to are the Vgs(th), or threshold Vgs voltage range. The threshold voltage can vary due to manufacturing process variations but the datasheet guarantees that it will be between 2 and 4V.

Vgs(th) is the voltage at which the mosfet turns on, but is not fully on. The mosfet does not operate in a digital fashion. The on-resistance, Rds(on), is dependent on Vgs. Generally, the higher the Vgs, the better the on-resistance. However, Vgs is limited by the thickness of the gate oxide (a dielectric). Too high a Vgs and the dielectric breaks down.

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  • \$\begingroup\$ I see...So this is where the Vds>Vgs-Vth and Vgs>Vth relationships govern full saturation of the FET. Two questions: 1. How do I select a NMOS knowing that I will be controlling it via a 5v P-P PWM-> Am I limited to Logical Level NMOS? What happens if I want to drive this particular Power MOSFET? 2. Does the MOSFET driver amplify voltage or does it only increase the gate-source current at a particular logic level (to increase the switching speed of the FET)? \$\endgroup\$ – Dub Jul 29 '14 at 2:10
  • \$\begingroup\$ Yes. 1. A gate driver IC is needed to translate the 5V (or TTL) PWM signal generated by a microcontroller to a higher voltage one that drives a power MOSFET. Indeed, TPS2819 is such a gate driver 2. Yes, you can use this same driver but would need to supply at least 10V to VCC/VDD. Note that TPS2819 can generate its own VCC using the internal regulator (see p14-15 of the datasheet). The typical VCC generated in this mode is 11.5V, sufficient to drive your MOSFET. \$\endgroup\$ – 2over0 Jul 29 '14 at 3:20
  • \$\begingroup\$ So I've tried connecting the mosfet drive directly to the mosfet as shown in this schematic dropbox.com/s/44hgis5kojnftxt/8-1-2014%206-10-55%20PM.png The problem I have is that the circuit is drawing a tremendous amount of power (~4.7V at .4A) regardless of whether or not the resistor is connect to the source of the NMOS. If I disable the PWM and hold the input high or low, large amounts of current is not drawn and the voltage can go as high as the max of the PSU (36V). Any ideas? \$\endgroup\$ – Dub Aug 1 '14 at 23:15
  • \$\begingroup\$ Unfortunately, the regulator does not work with 4.7V input. Can you probe the gate-to-source voltage of the MOSFET? You are still driving it at a low voltage. See Figure 16 of the TPS2819 datasheet. Will 4.5V be the typical value for Vsolar? If so, you can improve your circuit by picking a better MOSFET. The existing MOSFET is a 250V rated device and is thus unsuited for this application. Such large MOSFETs normally require 10-15V gate drive signals. What is your target Vsupply? Consider using a 30V (or smaller) MOSFET, which can be driven with 4.5V gate drive. Eg. IR's low voltage mosfets \$\endgroup\$ – 2over0 Aug 2 '14 at 18:22
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What the other responders have said is correct but in addition you need to change the frequency. 500Hz is much too low.

The current in the inductor will be increasing at ~40Amps/ms with 2V input. Since 500Hz is 1ms on time the current through the FET is trying to be 40A. You need to reduce the on time to 50kHz or so. The current will then only increase to about 400mA with 50kHz drive.

kevin

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