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| Drive a 12V analog switch from logic level input |
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| hermitengineer:
I'm looking for a quick and dirty way to interface with a 4066 analog switch from a MCU. The 4066 is supplied with 12V, and if I read the datasheet right, that means it needs close to 9V to turn it on. So I need a logic to 12V level shifter (I picked 12V because there's a rail for that). One quick and dirty solution is to drive a PNP transistor with an NPN transistor. But I'd like to simplify that even more. Would an optocoupler be a viable way to supply the 12V? I'm thinking, MCU digital pin -> resistor -> LED side of optocoupler -> ground; 12V -> collector -> connector to switch. Possible IC would be VOM617A-3. |
| A Hellene:
How about just using a low power N-Channel MOSFET (e.g. the 2N7000 or its SMD version 2N7002) in common Gate topology, and a 1..22k pull-up resistor? Input will be the Source pin, the Gate pin will be connected to the low voltage Vdd supply rail, and output will be the Drain pin with a 4k7..12k pull-up resistor to +12V. -George |
| jhpadjustable:
Invert the control input and use just one transistor? :) Less dirty is to use complementary PNP/NPN transistors in 6-pin small packages. --- Quote from: A Hellene on November 22, 2019, 07:32:05 am ---How about just using a low power N-Channel MOSFET (e.g. the 2N7000 or its SMD version 2N7002) in common Gate topology, and a 1..22k pull-up resistor? --- End quote --- That level shifter design depends on the difference between the two rails being smaller than Vgs(th). 7V is way too much. |
| A Hellene:
Nah! The transistor doesn't care about Vdg being 7V, as long as its Source is tied to the Gate! It is a 60V device with a Vgs_th of 1.0 .. 2.5V, so a Vgs_th of 3.3V will open the channel all right in order to pass the 1..2mA pull-up Drain current. Also, being in common Gate topology, it does not need any inversion at its input: - When its input is logic-high Vgs == (Vdd-Vdd) = 0V, so the channel is closed and no Drain current flows; the output becomes +12V via the output pull-up resistor. - When its input is logic-low Vgs == (Vdd-Vss) = (Vdd)V, so the channel opens and Drain current flows; the output becomes 0V via the logic output transistor of the level-shifter driver and the 5Ω transistor channel. Actually, that circuit above, is a stripped-down version of the bi-direcrional level-shifter Philips proposed at their IIC protocol specification! -George EDIT: Clarifications and corrections. |
| magic:
You should probably use BSS138 instead of 2N7000, though. The latter is rather marginal at 3.3V. Not sure what to use in THT. |
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