7805 with MOSFET as pass element? - Page 1
Forum section: electronics / beginners

author: fubar.gr
author_rank: Unknown
timestamp: August 28, 2014, 06:26:34 am
content: Is it possible to use a logic level mosfet as an external pass element with a 7805 regulator? All schematics I've found use a BJT instead.

author: Simon
author_rank: Unknown
timestamp: August 28, 2014, 09:09:23 am
content: Why not use a BJT ? the principle the BJT works on is that if the current through the resistor in series with the regulator causes a drop of more than 0.6/0.7V the BJT starts to conduct. The resistor is designed such that the BJT will start to operate once the maximum current you want drawn through the regulator flows. so if you want your regulator to put out 1A and then put the rest of the current through the BJT you choose 0.6-0.7R.How can you achieve that same effect with a mosfet ?

author: mikerj
author_rank: Super Contributor
timestamp: August 28, 2014, 09:24:39 am
content: Quote from: Simon on August 28, 2014, 09:09:23 amIt would certainly be possible to use a MOSFET as the pass element, but it would drive a lot more complexity in the design which ultimately makes it rather pointless.

author: Simon
author_rank: Unknown
timestamp: August 28, 2014, 09:32:13 am
content: Quote from: mikerj on August 28, 2014, 09:24:39 amThat was my point, the BJT method works using the inherent characteristics of the BJT, using a mosfet means dreaming up an entirely different and as you say more complex scheme. Using one versus the other will not affect cost or efficiency so best to use the simplest solution.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 28, 2014, 12:17:01 pm
content: Quote from: Simon on August 28, 2014, 09:32:13 amThere's nothing fundamentally different about a MOSFET in such an application. Just use Vgs(th) instead of Vbe.It will work just fine, but you burn another couple volts of dropout. Might as well use a switcher.Tim

author: DanielS
author_rank: Frequent Contributor
timestamp: August 28, 2014, 12:37:34 pm
content: Quote from: fubar.gr on August 28, 2014, 06:26:34 amTo use a MOSFET, you need enough headroom to accommodate the MOSFET's threshold voltage and even with logic-level MOSFETs, you are talking about a ~2V minimum loss there instead of ~0.7V with a BJT. If the reason you want to use a MOSFET is to avoid the base current "losses," keep in mind that the BJT's base current does not make much of a difference efficiency-wise for a 7805 boost design since it goes through the 7805 and then the load so the BJT's Ibe does not make linear regulation significantly more wasteful than it already is and certainly not as wasteful as the extra 1.5-2.5V Vds drop across the PMOS due to Vgs.With both the PNP and PMOS implementations all you need is a pull-up/bias resistor from supply to base/gate and 7805. If you want the 7805 to provide at most 500mA before the boost transistor kicks in, you will need ~1.5 ohms for PNP and ~3.3 ohms for the 2.5V PMOS assuming you can find a logic-level PMOS rated for the sort of continuous power you want to pass through it on its SOA graph. If your PMOS does not have a DC SOA curve, it cannot do linear regulation without considerable derating.PMOS as a pass element for the 7805 can be done but it is more wasteful and likely more expensive than a PNP.

author: David Hess
author_rank: Super Contributor
timestamp: August 28, 2014, 07:25:02 pm
content: The high MOSFET threshold voltage can be fixed with level shift but this represents extra complexity that the relatively low Vbe of a bipolar transistor neatly avoids. Further, the significant variation in Vgs(th) for different devices would require grading or adjustment.Offhand I do not know of any advantages which would make this worth doing. Efficiency will be the same and even without the extra complexity required for the MOSFET, bipolar transistors are less expensive.