Timeline for Resistance of mosfet's internal diode
Current License: CC BY-SA 4.0
27 events
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| Oct 4, 2019 at 11:34 | history | edited | Arundale Ramanathan | CC BY-SA 4.0 |
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| Oct 4, 2019 at 6:27 | history | edited | Arundale Ramanathan | CC BY-SA 4.0 |
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| Oct 4, 2019 at 5:16 | comment | added | Arundale Ramanathan | @BruceAbbott Battery is 9v. It is not rechargeable, so I am not supposed to be pumping current into it, which is why the mosfet is required. | |
| Oct 3, 2019 at 19:51 | comment | added | Bruce Abbott | What voltage is the battery, and exactly what voltage is the '9V' external supply? | |
| Oct 3, 2019 at 19:08 | comment | added | Russell McMahon♦ | @arun Dual MOSFETS add the Rdsons - so ABOUT 2 x as high - but still low for most purposes. As we do not know currents or battery voltage or ... we can't yet be sure whether this matters. || As shown so far (you and an answer) you will have problems with the FET turning on with no Vin but then being affected by the battery voltage that it supplies. This may not be an issue depending on Vbat and cct design but as we don't know Vbat or type of battery (affects range of voltage) or number of cells (affects ...) we can't tell. | |
| Oct 3, 2019 at 14:16 | vote | accept | Arundale Ramanathan | ||
| Oct 3, 2019 at 14:01 | comment | added | Hearth | Worth noting: If you plan to have significant current flow through the body diode, it's usually a good idea to add an external diode rather than relying on the parasitic body diode. | |
| Oct 3, 2019 at 13:54 | answer | added | Daniele Tampieri | timeline score: 3 | |
| Oct 3, 2019 at 13:18 | answer | added | WhatRoughBeast | timeline score: 3 | |
| Oct 3, 2019 at 12:04 | comment | added | Arundale Ramanathan |
@RussellMcMahon In your solution with two mosfets in series, will it have Rds(on) as low as around 10mOhms as that of one mosfet? I guess Rds(on) is what affects voltage drop.
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| Oct 3, 2019 at 12:00 | history | edited | Arundale Ramanathan | CC BY-SA 4.0 |
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| Oct 3, 2019 at 11:54 | comment | added | Arundale Ramanathan | @RussellMcMahon I am able to get it working if I reverse Drain and Source in the above circuit. But my worry is about the forward resistance of the internal diode, which will affect my backup time, and hence the original question. | |
| Oct 3, 2019 at 11:49 | history | edited | Arundale Ramanathan | CC BY-SA 4.0 |
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| Oct 3, 2019 at 11:39 | comment | added | pjc50 | @arun Quite often this problem is avoided by making the backup voltage lower than the battery voltage. E.g. if the battery is "12v" it may have a terminal voltage of 13V, and the input backup voltage would be 12V. So there would be no reverse voltage across the MOSFET. | |
| Oct 3, 2019 at 11:34 | comment | added | Arundale Ramanathan | @RussellMcMahon Thank you for the answer.. I will update my question.. | |
| Oct 3, 2019 at 11:33 | comment | added | Arundale Ramanathan | @PeterSmith Thank you for the answer.. The application is exactly what you have mentioned.. Battery backup system where I want to prevent the power feeding the battery through the mosfet. When the power is down, I want the mosfet to switch the battery to run the circuit.. | |
| Oct 3, 2019 at 11:25 | comment | added | Peter Smith | An ordinary high side switch can use a single P channel device without support circuitry other than a control signal and a resistor; other applications (such as battery back up systems) often use two back to back as described by Russell McMahon to prevent backfeeding power sources. There are other applications and every solution is application dependent so if you modify the question to show your application we will be able to assist. | |
| Oct 3, 2019 at 11:16 | comment | added | Russell McMahon♦ | ... and Vds polarity is not important as long as Vgs polarity is "correct". but THIS SOLUTION IS NOT USUALLY NEEDED. what IS YOUR ACTUAL APPLICATION? | |
| Oct 3, 2019 at 11:15 | comment | added | Russell McMahon♦ | For practical purposes ALL MOSFETs that you will encounter will have the reverse diode - it's caused by the structure of the device on a silicon substrate and is unavoidable without very special approaches. If you tell us what you are trying to actually achieve we can help provide a good solution. | A method to get low voltage drop and mono directional switched current flow is to use two identical FETs in series with one polarity reversed. Join s-s and g-g. One D is the new S and one D is the new D. Drive gg negative wrt to ss to turn on as usual. This works because MOSFETS are 2 quadrant ... | |
| Oct 3, 2019 at 10:24 | history | edited | Arundale Ramanathan | CC BY-SA 4.0 |
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| Oct 3, 2019 at 10:04 | comment | added | winny | @pjc50 You can model it as an ideal diode with series resistance if you want. | |
| Oct 3, 2019 at 10:02 | history | edited | Arundale Ramanathan | CC BY-SA 4.0 |
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| Oct 3, 2019 at 9:58 | comment | added | pjc50 | Diodes don't really have a resistance - the I/V graph looks completely different. | |
| Oct 3, 2019 at 9:58 | comment | added | winny | "•What would be the internal resistance of this diode?" It's rarely stated. You need to look at a I-V-curve for the diode, if supplied. | |
| Oct 3, 2019 at 9:57 | comment | added | winny | Vocab? You mean vocabulary? ;-) | |
| Oct 3, 2019 at 9:55 | review | First posts | |||
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| Oct 3, 2019 at 9:53 | history | asked | Arundale Ramanathan | CC BY-SA 4.0 |