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I would like to switch some N-channel MOSFETs (must have low RDS, <60mOhm) by the outputs of BCD decoder, like MC14028B for example. I was really surprised to find out that most BCD decoders have large supply voltage range (typically 2~20V) but very low output current (typically 0.5mA @ 5V). I am not sure I can directly drive FET gates with this, since FETs with low RDS(on) usually have high gate capacitance, >300pF is common.

Here is the desired set-up: MCU (3.3V) -> translating GPIO expander (3.3V -> 5V) -> BCD decoder (5V) -> MOSFETs (VGSth < 3V)

If I have to add some kind of a buffer I'd rather do it on 4 lines before the decoder than on 10 lines after. Maybe shifting gate voltage up where BCD decoder has better output drive (3.4mA @ 15V) can help:

MCU (3.3V) -> GPIO expander (3.3V) -> Level shifter (3.3V -> 15V) -> BCD decoder (15V) -> MOSFETs (VGSth < 10V)

There are many possible combinations and I am trying to find the simplest (less components, smaller footprint). Any suggestions?

Maple
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    What supply voltage will the FETs be switching and what are the loads? – Bruce Abbott Dec 31 '22 at 21:18
  • @BruceAbbott Still trying to get an answer from the manufacturer. Feels like pulling teeth. Per documentation 5.5mA max @ up to 34V, although actual measured voltage varies 7~10V between the devices. These FETs will substitute switches in the schematic shown [here](https://electronics.stackexchange.com/q/647226/187920) – Maple Dec 31 '22 at 21:31
  • @BruceAbbott My thinking is that since drain current is limited from the outside, it will unnecessary increase RDS(on). So choosing FETs with low VGSth and jacking up gate voltage is my way of making them act as close to the mechanical switches as possible. I can be way off base here, of course. – Maple Dec 31 '22 at 22:01
  • OK, so the device sources 10V through a 200 ohm resistor, and your FETs will simulate mechanical (?) switches connecting a chain of 34 ohm resistors to ground. This means the drive requirements are low because the Drain current is limited to a low value and fast switching isn't required. – Bruce Abbott Jan 01 '23 at 05:47

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Since Drain current is limited to 5 mA through a pullup to 10 V in the load, and assuming the FETs are replacing mechanical switches and so don't have to operate quickly, you should be able to drive the FET Gates directly from the 4028.

Each 4028 output has a buffer which isolates it from the internal logic, and at 5 V it can handle a continuous short to either supply rail. Therefore you won't even need series resistors to limit current. Short circuit output current on CD4000 series ICs is typically ~5 mA at 5 V.

Calculating the switching time is tricky because the buffer's output resistance varies with voltage. I simulated it in LTspice with an Si4920DY and got turn on and turn off times of 4.4 μs.

Bruce Abbott
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  • You are absolutely correct that fast switching is not required. In fact, we will be OK with total propagation delay as high as 50ms. The only critical requirement is getting each band resistance as precise as possible, since we could not get the input tolerances from the manufacturer. Hence low RDSon. I did not find that information about short current in the datasheet, so my main concern was current spike due to gate capacitance. I was afraid that it can damage the output of the BCD decoder. Do you know if the same is true for 4028 chips from other manufacturers, like CD4028BPW from TI? – Maple Jan 01 '23 at 09:04
  • Oh, and I don't think they have a simple pull-up there, more likely it is a current source. The current stays exactly at 5mA regardless of which band button we press. And it is the same on the devices that have only 7V on the inputs. – Maple Jan 01 '23 at 09:10
  • Any CD series chip with B after the number is buffered, and different brands should have similar characteristics. A few have higher output current (eg. CD4050B, CD4511B). I suspected it was a current source, which is why I didn't say pullup resistor. – Bruce Abbott Jan 01 '23 at 13:08