Timeline for How to drive power MOSFET by logic level signal without inverting phase?
Current License: CC BY-SA 3.0
14 events
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| Feb 12, 2021 at 19:51 | comment | added | pgibbons | @Lukasz why is R4 200 Ohms? It seems way too low, all you're doing is charging the gate of the MOSFET. The reason I ask is because I just built this circuit using 1/4 watt resistors and R4 got so hot that I burned my finger on it. I'm calculating 0.72 watts, I don't understand why it has to be that much, I'm basically going to have to use a 1W resistor. | |
| May 16, 2016 at 7:47 | comment | added | Lukasz | Great. I made this circuit in visual simulation tool: everycircuit.com/circuit/5233775657615360 I see the is a small current loss in the driver. Around < 20mA on the first transistor and < 60mA on the second (i changes value of the R5 resistor to 220Ohm as it is easier to buy). Just curious it there is an other way to do that without current loss (MOSFET driver)? | |
| May 16, 2016 at 7:31 | history | edited | Vladimir Cravero | CC BY-SA 3.0 |
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| May 16, 2016 at 7:30 | comment | added | Vladimir Cravero | Yeah I am sorry I did not really had time to think the values through. Fixing it now. | |
| May 15, 2016 at 23:29 | comment | added | WhatRoughBeast | @Lukasz - I also recommend reducing R5 to 1k. As shown, the gate voltage is not guaranteed to drop low enough to fully turn off the FET. R4 (200 ohms) will need about 60 mA, but the base current provided by 10k is only about 1 mA, and a gain of 60 is more than you should require under these circumstances. | |
| May 15, 2016 at 21:30 | comment | added | Lukasz | Let us continue this discussion in chat. | |
| May 15, 2016 at 21:23 | vote | accept | Lukasz | ||
| May 15, 2016 at 21:18 | comment | added | Vladimir Cravero | Your circuit was almost ok. It was not safe with the micro off. And please if you need infos on a resistor use its number, i.e. R1, R2 etc. | |
| May 15, 2016 at 21:12 | comment | added | Lukasz | So my circuit was actually OK? I see that comparing to mine you removed 1 or 2 resistors? Won't that cause transistor overheating (especially second one)? | |
| May 15, 2016 at 21:10 | comment | added | Vladimir Cravero | When the mos is turned on, i.e. its gate is high, the drain and the source get connected, since the drain is at 0V the source goes (almost) to 0V too. | |
| May 15, 2016 at 21:08 | comment | added | Lukasz | See the simulation plots (graph) in the questsion as simulation of your circuit shows the same results: When vin is off = 0V, Vdrain shows 12v, when vin is on = 5V, Vdraing = 0V. I just want to understand it - that's the most important reason I care ;-) | |
| May 15, 2016 at 21:04 | comment | added | Vladimir Cravero | What do you mean with inverted? And why do you care? | |
| May 15, 2016 at 20:59 | comment | added | Lukasz | Unfortunately V at drain is still inverted (even if Vgate and Vin are in phase). I have no idea what is going on | |
| May 15, 2016 at 20:52 | history | answered | Vladimir Cravero | CC BY-SA 3.0 |