Why mosfet is not used as current booster in Voltage regulators? - Page 1
Forum section: electronics / projects

author: Gurumurthy
author_rank: Regular Contributor
timestamp: May 16, 2017, 04:34:39 pm
content: Hi to all...Currently I am building a Lab power supply for my own purpose. All the series pass transistor fails at high temperature(2N3055/TIP142/MJ4502) with LM317 or LM723. But Increasing the temperature of an FET tends to decrease the mobility of the charge carriers in the channel, effectively reducing the current through the channel. Can i use mosfet as series pass Element(Souce Follower)? Please Suggest an Solution with a circuit.

author: Zero999
author_rank: Super Contributor
timestamp: May 16, 2017, 05:01:12 pm
content: You can but I don't know why you'd want to. The drop-out voltage will be much higher, than with a BJT because the threshold voltage for a MOSFET, is much higher than the base-emitter voltage required to turn on a BJT. And it isn't a source follower, but common drain amplifier. Another thing to bear in mind is that most power MOSFETs are designed to be used as switches, not analogue amplifiers.If your BJTs are burning out, it's because you're exceeding their maximum ratings and it's likely the same will happen with a MOSFET.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 16, 2017, 05:23:21 pm
content: MOSFETs also tend to be relatively nonlinear when used an an amplifier. So the circuit will behave different at low currents than at high currents. This is specially a problem with a regulator with a source follower configuration. This makes the current limit repose very critical, as the peak current on a short can rise very fast and high if you don't work hard against it. A BJT tends to have a lower current gain at high currents thus kind internally limit peak currents, while with a MOSFETs the trans-conductance goes up with current. They are also relatively slow at low current and fast at high currents which can be a problem. It might need a significant standing current to make them usable fast.

author: Gurumurthy
author_rank: Regular Contributor
timestamp: May 16, 2017, 05:24:30 pm
content: Dear Sir,As you said it is just acting as a switch and Supplies the extra current needed by the load. and Mosfet will not go to break down at high temperatures due to its properties. Then what Sir we can Go for MOSFET?If we want out to start from zero, Whether we can go for this circuit? If yes , what could become problem in circuit?

author: David Hess
author_rank: Super Contributor
timestamp: May 16, 2017, 06:02:13 pm
content: 1. Vgs of a MOSFET is higher than Vbe of a bipolar increasing dropout. Vgs is also not as tightly controlled.2. Most MOSFETs are designed for switching operation and have poor SOA curves at high voltages offering no advantage over a bipolar transistor. However this is not be a problem in a low dropout application.3. Power dissipation is limited by die size and MOSFETs are more expensive for the same area. Bipolar transistors are less expensive.

author: Audioguru
author_rank: Super Contributor
timestamp: May 16, 2017, 06:41:28 pm
content: Your new schematic has the Mosfet as a source-follower at the output of the voltage regulator. The gate-source voltage of the Mosfet might be 2V at low currents and be 10V at high currents then it has horrible voltage regulation.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 16, 2017, 07:15:55 pm
content: If you absolutely want a voltage regulator with a power MOSFET, use a LT1575 or similar chip. They are made to use an external MOSFET and at least works in some application. Still not sure about a difficult load and the required layout.For higher power dissipation at more than about 20 V, MOSFETs are usually not a good choice any more. It might have been an option some 25 years ago, but since then power-BJTs got faster and modern MOSFETs are generally poor SOA performance. So for linear operation one tends to look for kind of antique types.Already the combination of LM317 or similar and extra power device is a bad solution - It is one of those example circuits that might work under favorable conditions and with poor performance, but has never been meant to be actually used.

author: Zero999
author_rank: Super Contributor
timestamp: May 16, 2017, 10:33:41 pm
content: Quote from: Gurumurthy on May 16, 2017, 05:24:30 pmNo, that's even worse than the previous circuit. It is a source follower. The drop-out will be as high as the previous circuit with the added disadvantage of much poorer voltage regulation.

author: rstofer
author_rank: Super Contributor
timestamp: May 17, 2017, 02:52:23 pm
content: The 2N3055 has been used as a power device for at least 44 years that I know of. My first experience with them was in a seam welder for aluminum. There were a couple of dozen transistors on a massive heatsink.So, given that they have worked for decades, what is wrong with the circuit? Do you have enough transistors in parallel to keep the individual current down? Heat sinks? Too much Vce for the current? Current balancing resistors?The problem is in the circuit, not the device and changing to a MOSFET is unlikely to improve the situation.There are many example 2N3055 circuits on the web.The biggest problem with the 2N3055 is the limited gain. If you figure hfeat 20, you need to pump 1/2 amp into the base for 10A through the collector. At some point, you use another 2N3055 to provide the base current for the pass transistors.Under the best of conditions (ambient temperature and heatsink), you are limited to dissipating 115W. If you are pushing 10A through the device, the Vceneeds to be less than about 11V. And that is optimistic.