Lowest voltage needed to turn on a FET?

Question

With inspiration from Ladyada's DIY boost calculator, I'm attempting to build a simple, low-current (<50mA) 40V boost circuit to drive an antique voltmeter. The PWM source for the transistor will be an AVR that can operate on as little as 1.8V, so that's the minimum voltage I'm targeting. I've been prototyping with a BC547 (BJT NPN) and a 5V supply, but I'm not able to get more than 36V with the components I have on hand, possibly due to an undersized Schottky diode.

I have a rudimentary understanding of BJTs but I've never worked with FETs, and I'd like to learn more. For this specific application, I believe I'm looking for a power MOSFET with a relatively low on-resistance. Max Vds should be well in excess of the 40V I want to produce.

Should I be looking at the gate-source voltage threshold (Vgs(th)?) to identify a FET that will work at the 1.8V I'm targeting? Filtering on that parameter lead me to this STP60NF06L, but the Output characteristics graph (fig 5, p6/16) shows that the current is extremely limited at that voltage, and may not be within the operating margins of the device.

Answer 1

This does not sound like a good fit for a FET since you only have 1.8V drive available. FETs that switch natively with such a low voltage aren't likely to withstand 40V drain to source.

A bipolar only needs one diode drop to put enough current thru the base to turn it on well. Getting one to withstand 40V with good gain is easy. You didn't say how much current you need at 40V and what the inductor saturation current is, so let's say the switch needs to handle 500 mA. You can easily find a NPN that can do 40V with a minimum guaranteed gain of 50, so that means you need 10 mA base current. Let's say the B-E drop is 700mV, so the drop on the resistor is 1.1V. 1.1V / 10mA = 110Ω As long as the processor can reliably provide 10mA out, that's all you need.

What power voltage are you boosting from? If its the 1.8V, then the turn on voltage of the NPN will be a significant fraction of that. Another possibility is to make a small analog booster to tun the micro off of 5V or so, then have it drive a FET directly. FETs that turn on well with 5V on the gate and can withstand 40V will be a little less scarce. That FET can still switch the 1.8V to make the 40V, but will get switched from the intermediate 5V supply.

Answer 2

The table on page 4 says \$V_{GS(th)}\$ > 1V, but that's when the FET just starts to conduct, and indeed the conditions say \$I_D\$ = 250 \$\mu\$A.
That's one thing. You mention figure 5, but that's not so clear; everything interesting is in the lower 2 mm. By the way, if you look at the vertical scale it doesn't look too bad. How many amperes do you want. No, then figure 6 tells us a lot more: drain current doesn't start to flow until the gate voltage is well above 2V.
What's typical about (MOS)FETs is that there can be a big tolerance on \$I_D\$ vs \$V_{GS}\$. While one FET has enough with 1V another one may need 1.3V. The values are not as close together as with BJTs.
Note that for the \$V_{GS(th)}\$ a minimum value is given, no maximum, while that would be much more interesting. Now we know that below 1V it won't do anything, but we have no idea how much is needed to have it open guaranteed.