# Lowest voltage needed to turn on a FET?

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.

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.

• On the same line as your last paragraph, he could make a 12V boost from which to drive a FET through a BJT. Jul 12, 2011 at 21:00
• The voltmeter has a resistance of 100kΩ, so 0.04mA @ 40V. I want to light some LEDs, too, so less than 30mA total. The inductor I'm prototyping with (RLB1014-563KL) is rated for 18mA (not sure if that's the saturation current), so I need to find a different one. I do want to run the whole system (µC and boost) off 1.8V. Can you give an example of the "small analog booster"? A similar boost circuit with a 555 for a PWM source? Jul 12, 2011 at 21:48
• @blalor: What voltmeter? What does this voltmeter have to do with anything? I see you added that you need 50mA from the 40V supply. That's 2W. If the booster is 80% efficient (will be tricky to achieve without fanciness from such a low input voltage) then it will need 2.5W in, which means 1.4A from the 1.8V supply. That's average, so you need a inductor to have at least 2A saturation current, and then you have to keep it in continuous mode without low ripple, which means switching fast. What you're asking for is not trivial or a "starter" switcher. Jul 12, 2011 at 21:56
• Doh! I apparently mixed up trying to say "without much ripple" and "with low ripple", but what my fingers typed came out completely backwards. Sorry for the confusion. Jul 12, 2011 at 23:58

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.

• So it looks like I'm on the right path, but I need to scrutinize the datasheets closely… Jul 12, 2011 at 20:12