Switching with MOSFET in series - Page 1
Forum section: electronics / beginners

author: acolomitchi
author_rank: Contributor
timestamp: April 02, 2017, 10:53:10 am
content: Before perfboarding this (and potentially releasing the holy smoke without understanding why), I thought I'd rather ask first: could the attached design work for switching under a voltage higher than the Vds(max) of a NMOS by using two (N in the general case) in series?If it may work only under special conditions, what are other conditions which can cause it to fail?The expected behaviour (yes, I know the diodes will introduce a voltage drop, considering them ideal in the followings should not modify the behaviour)a. start conditions - input LO. Q1 and Q2 non-conductive, the mid-point to a voltage half of Vcc. As a result, Q2's gate at the voltage equal with its source (D2 directly polarized).b. the LOHI switch - Q1 goes conductive over tens/hundreds of nanosec. While mid-point voltage (=Vs for Q2) goes down, the voltage of Q2's gate lags behind with 15V - Zener D2 clamps it. So, after td(on) delay, Q1 starts conducting, after 2*td(on) Q2 starts conducting. With Q1 fully conductive, the gate of Q2 is maintained at approx 15V through D1 (compensates leakage)c. the HILO switch - Q1 goes non-conductive over hundreds of nanosec. Once Q1's Vd(=Q2's Vs) goes over 15V, D2 becomes directly polarized and establishes a Vg=Vs, thus rendering Q2 non-conductive.The scheme looks so simple I'm afraid I'm missing something.Thanks in advance for comments.

author: bktemp
author_rank: Super Contributor
timestamp: April 02, 2017, 01:40:56 pm
content: Quote from: acolomitchi on April 02, 2017, 10:53:10 amThat does not work, because there is no path for discharging Q2's gatecharge.But there is a simpler solution: You don't need D1 and the 15V supply, because you already have a voltage divider providing an auxilary voltage. Just connect the voltage divider to the gate instead of the source of Q2.The capacitors need to be adjusted to provide just sufficient charge for charging/discharging the gate (probably a couple of 100pF, depending on the mosfet and operation voltage.)

author: acolomitchi
author_rank: Contributor
timestamp: April 02, 2017, 09:04:44 pm
content: Maybe if I'm making more evident the voltage values, there will be less potential of confusion. Say Vcc=330V (max rectified 230V mains), Vds(max)=300V for Q1/2.Quote from: bktemp on April 02, 2017, 01:40:56 pmBut... the Q2's gate doesn't need discharging! It has 15V potential and heading to a point where it should have Vcc/2 (same as the Q2's source).In initial H1 state, the Q2's Vs=0V (assuming Q1 ideal and strongly on) because Q1 shorts the lower resistor in the divider.As Q1 is switching off, the voltage on the centre point starts to raise towards Vcc/2- the lower resistor is less and less shorted.As it raises, is shouldchargethe Q2's gate, to keep it in sync with its source.What am I missing?Quote from: bktemp on April 02, 2017, 01:40:56 pmThe voltage divider is there to protect against unequal leak currents through the MOSFETs while they are switched off. It's exactly the same situation of using a series combination of diodes when their breakdown voltage is smaller that the voltage which needs to be blocked.If I'm attaching the divider to the Q2's gate, with a Vcc=330V, I'll finish by blowing the gate of Q2 - Vgs must be in the +/-20V range and switching Q1 will cause a variation of 165V.

author: acolomitchi
author_rank: Contributor
timestamp: April 02, 2017, 09:44:32 pm
content: Quote from: bktemp on April 02, 2017, 01:40:56 pmAh, I think I see. The Q2's gate forms a capacitor between Q2 gate andsource,notbetween Q2's gate andground.As such, any variation in the Q2's Vs will just translate as such in a corresponding variation of the Vg - sorta like a bootstrap/flying capacitor.Is it correct?Quote from: bktemp on April 02, 2017, 01:40:56 pmWhich means I'll need two dividers: one to keep the NMOS-off leakage current in check, the other to establish a "high" that is high enough for Q2's gate; D2 - the Zener - stays to take care of clamping, but D1's not needed.Is this what you suggest?Thanks.

author: Zero999
author_rank: Super Contributor
timestamp: April 03, 2017, 07:56:50 am
content: What you're trying to make is a circuit known as a cascode. A combination of a common source and common gate amplifier.How about replacing the upper transistor with a BJT? The lower transistor can then be a low voltage MOSFET, with a very low on resistance.http://electronicdesign.com/power/cascode-configured-gan-switch-enables-faster-switching-frequencies-and-lower-losses

author: bktemp
author_rank: Super Contributor
timestamp: April 03, 2017, 08:28:59 am
content: Quote from: acolomitchi on April 02, 2017, 09:44:32 pmYes, that's why your circuit won't work.QuoteYou don't need two dividers:If the gate-source voltage goes higher than ~3V the mosfet turns on, forcing the source to a voltage slightly below the gate voltage. If the voltage at the gate is lower than source, the external zener diode conducts. Therefore the voltage divider at the gate will also control how the drain-source voltage distributes between the two mosfets.Instead of using a voltage divider you could also use a zener or TVS diode + pullup resistor for setting the maximum voltage of the lower mosfet. Then instead of distributing evenly across both mosfets, the voltage at the lower mosfet will be fixed.Quote from: Hero999 on April 03, 2017, 07:56:50 amI wouldn't use a mosfet+BJT cascode. It works, but it can be quite difficult getting all the parameters right for a good switching behaviour. Today there are much better high voltage switches than BJTs.

author: David Hess
author_rank: Super Contributor
timestamp: April 03, 2017, 11:01:31 am
content: Quote from: bktemp on April 03, 2017, 08:28:59 amThe alternatives are also more expensive. MOSFET die size is roughly proportional to the square of the voltage rating so high voltage bipolar transistors are more economical assuming you can put up with their more complex drive requirements. IGBTs being minority carrier devices have the same advantage.SiC and GaN MOSFETs sure are neat though.

author: acolomitchi
author_rank: Contributor
timestamp: April 03, 2017, 11:04:06 am
content: Quote from: Hero999 on April 03, 2017, 07:56:50 amThank you for the references. much appreciated.For this case, I'd prefer to solve the problem inside "two identical mosfets in series", if only for the intellectual/academic challenge (practically, except for extreme cases, one can always find a beefier mosfet to simplify the design - e.g. for switching rectified mains in the 3-4kW range, a STP20NM60 seems cheap enough).

author: acolomitchi
author_rank: Contributor
timestamp: April 03, 2017, 11:27:31 am
content: Quote from: bktemp on April 03, 2017, 08:28:59 amI be damn'd! Of course you are right!!Now that you pointed it out, it seemssoevident.Thanks!

author: bktemp
author_rank: Super Contributor
timestamp: April 03, 2017, 12:09:56 pm
content: Here are two slightly differents variants I have seen.The IGBT one is being used in a 2 quadrant HV power supply for regulating and discharging the output voltage. It operates as a current sink, that's why there are resistors in the emitter/source lead. The capacitors have a rather high value, probably because the circuit operated in linear mode instead of fast switching.The other one uses a TVS diode to clamp the voltage of the lower mosfet to 400-450V for use with 600V mosfets.Series connection of mosfets/IGBTs/BJTs looks rather simple, but it needs carefully selected capacitors to make all mosfets switch at the same time: Without the capacitors only the lower mosfet switches instantly, because the large gate resistor for the upper mosfet (typically >100k) needs some time to charge/discharge the gate capacitor. The capacitors provide the current for charging/discharging the gates. Because of the high voltage swing, the capacitance can be quite small compared to the gate capacitance. If the capacitors are too large, all excessive charge has to be dissipated in the zener diodes protecting the gate.Before building the circuit it is probably a good idea simulating the circuit and adjusting the values for best performance.