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I was reading that the MOSFET is more optimal for high power situations at low to medium supply voltages than the BJT. However, I'm questioning this, because MOSFETs and BJTs waste little to no power while off.Mosfets are more prone to damage due static electricity breaking down the gate source oxide layer. Is it true that BJTs are better at high junction temperatures because they dont have an oxide layer? .

Can anyone confirm whether or not this is true?

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  • \$\begingroup\$ MOSFETs are characterized by a low RDSon, which for low to moderate currents is less than the drop of a bipolar transistor. However, for very high power IGBTs are often preferred. \$\endgroup\$ – Chris Stratton Nov 30 '15 at 0:51
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    \$\begingroup\$ The answer may depend whether you are asking about switching or amplification. \$\endgroup\$ – Brian Drummond Nov 30 '15 at 0:52
  • \$\begingroup\$ There is a lot less consideration required for the current drawn by the Gate of a MOSFET than by the Base of a BJT. In general, the higher the Ice capabilities, the lower the hfe. Techniques like emitter-followers ('Darlington Drivers') help negate this, but a MOSFET doesn't have such requirements. \$\endgroup\$ – CharlieHanson Nov 30 '15 at 1:14
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    \$\begingroup\$ @MarkU -- this is generally not a concern for switching FETs. (They are driven hard enough that the FET's already all the way "on".) It is a concern for FETs operating in the linear region, however -- FET-based linear amps are the main things that run into this. \$\endgroup\$ – ThreePhaseEel Nov 30 '15 at 1:49
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    \$\begingroup\$ I'm voting to close this question as off-topic because the OP has made no attempt to research the "accusations" he/she raises. \$\endgroup\$ – Andy aka Nov 30 '15 at 8:52
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As usual, rules of thumb and knee-jerk answers can be misleading, especially when forgetting the constraints in the original guidelines that were then dumbed down to make "rules" for the non-thinking.

FETs are voltage-controlled, and BJTs are current-controlled. This alone leads to a whole set of tradeoffs between the two devices, having nothing to do with operating voltage, current, or power.

Both devices are capable of handling about the same power. Power dissipation is mostly a function of the package, and both devices are available in similar packages.

The advantages for voltage versus current control are not as simple and one-sided as others here would have you believe. The voltage control of FETs requires essentially no power to keep them in a particular state, but that ignores both the control circuitry and that changing the state often is necessary in many applications. A FET gate looks mostly like a capacitor to the driving circuit, so it takes current to change the voltage. That together with a typical 12 V gate swing over the full on/off range can lead to significant current and power. For example, let's say the total effective gate charge is 50 nC, and the FET is switched at 100 kHz (every 10 µs). That comes out to 5 mA at 12 V, or 60 mW. That's the same total control power into the device as a BJT with 80 mA drive at 750 mV. There are other concerns beyond these for driving FETs and BJTs, but I'm trying to point out that it's nowhere near as simple as "FETs take no power to drive".

In linear applications, the more predictable B-E voltage of a BJT can be advantageous over the D-S voltage of FETs. Constructs like emitter-follower generally have better characteristics than FET source-followers. Since BJT are both current in and out devices, they can be cascaded in ways that don't apply to FETs, like darlington pairs, or combined NPN-PNP devices. Another advantage of BJTs is the much lower voltage required to control them. You can control high current and high voltage BJTs with typical logic level voltages (3.3-5 V), which isn't possible with FETs.

Of course the voltage control and even the larger voltage control range of FETs can be advantages too. I'm not trying to make it sound like BJTs are better, just trying to point out some ways they can be more advantageous since the knee jerkers here seem to have decided FETs are "better" in broad classes of applications. FETs and BJTs are fundamentally different, so there are going to be various applications where one provides advantages over the other.

High current switching with low to medium voltage is one example where FETs are often used despite the generally more complex drive circuitry. This is because power FETs look like a low resistance when on, which can be 10s to single mΩ depending on how much money you are willing to spend. BJTs on the other hand look like a fixed voltage of maybe 200 mV to several times that, depending on how hard they are being pushed. At 10 A, for example, a 20 mΩ FET will have 200 mV drop, whereas a BJT will probably drop 2 to 3 times that.

FETs can also be more easily paralleled in high power applications because their on resistance goes up with temperature, unlike the BJT saturation voltage, which goes down with temperature.

For both BJTs and FETs, other characteristics become less desirable as the maximum voltage goes up. However, this happens more slowly with BJTs, so that above a few 100 volts, BJTs start looking like a good deal for power switching. In fact, this has given rise to the IGBT, which is FET and BJT working together. The FET is used to turn on the BJT, so doesn't need to handle as much current. The BJT then does the heavy lifting of switching the current and dissipating the power.

Again, different devices will have different tradeoffs, and devices as complex as transistors don't fall neatly into simple categories that lend themselves to rules of thumb. There is no substitute for actually understand what it going on, then weighing the tradeoffs for your particular application carefully to decide what parts to use.

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    \$\begingroup\$ +1 A very considered answer. I am in the middle of a high voltage (well, relatively - a few hundred volts) controller that uses FETs on one side of the control and bipolars on the other, due to the somewhat unusual requirements I have. The FETs are nice, but far more difficult to compensate in the control loop. \$\endgroup\$ – Peter Smith Nov 30 '15 at 13:26
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Generally MOSFETS are preferred for higher power applications (which are usually switching circuits).

This is because FETs are generally easier to drive (especially at high voltages) than BJTs -- because the BJT current frequently has to come from the same (high voltage) supply as the load; thus there can be a significant power loss in the base drive alone. At low - medium frequencies, the gate drive requirements of FETs are easier than BJTs.

At the highest power levels (say >> 10 kW), combinations (e.g. IGBT in electric vehicles and trains) of FET & BJT are used, and at extreme power levels (say >> 1 MW AC) (AC power transmission), the devices are bipolar (Triacs, SCRs).

IGBTs have some of the benefits of both types of devices -- relatively easy to drive, but high current capability (because of the bipolar device). The inherent VCE drop of ~ 1 V is not an issue in >> 100 V systems.

Triacs are often 100 mm in diameter and can carry 10 kA. They are completely bipolar devices.

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Consider a simple automotive DC application:

Your car battery supplies 12V DC. For reverse-battery protection (for when some genius decides to connect the battery terminals in reverse), you use a transistor in series with your input supply with its gate/base tied to ground through a resistor. This shuts off the circuit when the voltage on the negative (or ground) of your circuit rises above the positive voltage - the transistor (hopefully) opens up before the rest of your circuit is damaged.

Ok - now to your question about power.

Here are two cases (I will pick reasonable transistor values for each situation):

Case 1: Your load is a Bluetooth Low Energy device which draws a maximum of 10mA DC:

BJT Power: P = I * Vce = .01A * 0.2V = 0.002W = 2mW

FET Power: P = I^2*Rds-on = 0.01A * 0.01A * 10ohm = 0.001W = 1mW

Best Option: Who cares? Go for the cheapest.

Case 2: Your load is a fuel pump which draws a maximum of 20A DC:

BJT Power: P = I * Vce = 20A * 0.3V = 6W!!! Just for reverse battery protection!

FET Power: P = I^2 * Rds-on = 20A*20A*2mOhm = 0.8W

Best Option: FET. But consider the cost... a 2mOhm PFET won't be cheap.

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    \$\begingroup\$ Just to illustrate the previous answer -- the BJT in this case will also require base drive current. Say there is a forced beta of 20 -- this means 0.01/20 = 0.5 mA of base current is required. This consumes 0.5 * 12 = 6 mW from the battery also. It is easy to see that base current drive power can be a larger problem than BJT dissipation itself. Also note this current flows whether or not there is a load on the device itself. \$\endgroup\$ – jp314 Nov 30 '15 at 2:42
  • \$\begingroup\$ You got the basics down for a decent answer, but it could be improved a lot. This question was on the Hot Network Questions list, writing your answer more clearly would have it's benefits. \$\endgroup\$ – Mast Nov 30 '15 at 8:29

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