The offset voltage of a saturated transistor, makes bjt less attractive as a switch than mosfet. Why? Is there any other logics aside this?
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What you say is true only for low to moderate currents, up to, say, 100 A or so.
The VCE(SAT) of a BJT is more or less constant, regardless of current, assuming of course, that the transistor is robust enought to handle the current in the first place.
On the other hand, the channel of a full-on MOSFET functions as a constant resistance. This resistance can be very low (miiliohms), but at high currents, it can still result in a higher voltage drop — and higher power dissipation — than the equivalent BJT.
This is why the IGBT, which combines the best features of a MOSFET and a BJT, is so popular in the highest-power applications.
Dave Tweed
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There is also the fact that high-voltage BJTs are easier and cheaper to produce. – Connor Wolf Apr 15 '14 at 02:27
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4kV IGBT $100 rated for 30A 3.2V Vce(sat) at 30A. 4.5kV MOSFET $88 rated at 2A (20V drop at 1A). The IGBT is WAY better except speed is about half. – Spehro Pefhany Apr 15 '14 at 03:47
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Having a higher "offset voltage" (Vce(sat) for BJT and VDS(on) for MOSFETs) is undesirable for two main reasons:
- Power losses. The power lost at the transistor at its output stage is proportional to the current and the voltage drop. Higher losses means decreased efficiency and increased heat dissipation. This could meant that batteries don't last as long, and increased thermal management costs.
- Reduced dynamic voltage range. In certain applications you need to maximize your voltage range, and some of it will be gobbled up by the drop at the switch, and it is worse if you have both high and low side switches.
Like another answer already mentioned, the assumption of VDS(on) < Vce(sat) is contested for high currents.
apalopohapa
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what is the difference between static and dynamic power dissipation? – Quazi Mishkatul Alam Apr 15 '14 at 04:31