Can this circuit drive a logic level low here?

Thread Starter

matthej

Joined Oct 10, 2020
58
Hi,

I have this circuit and I wanted to see if this can drive a logic low here. The output is 2.5V LVCMOS and the input is 1.8V LVCMOS.

R1 is a strapping resistor used on power up to detemine the state of the interface (needed). R3 is an internal pulldown.

I added R2 and R4 to bring the level down from 2.5 to 1.8V.

Thanks!


Screenshot 2021-08-10 092459.png
 

AnalogKid

Joined Aug 1, 2013
9,494
Taking R1 out of the circuit, that leaves a 2.5 V LVCMOS output trying to drive a 95 ohm load, for an output current of over 26 mA. Without knowing more about the two devices, that seems unnecessarily high. Also, many LVCMOS devices have inputs that can tolerate higher voltages. Does yours?

ak
 

Thread Starter

matthej

Joined Oct 10, 2020
58
Taking R1 out of the circuit, that leaves a 2.5 V LVCMOS output trying to drive a 95 ohm load, for an output current of over 26 mA. Without knowing more about the two devices, that seems unnecessarily high. Also, many LVCMOS devices have inputs that can tolerate higher voltages. Does yours?

ak
The max input voltage is only +/- 125mV above the rail which is 1.8 so it looks like we cannot drive this input up to 2.5V
 

Thread Starter

matthej

Joined Oct 10, 2020
58
I agree that the extremely low values for R2 and R4 are a dead short to a low voltage Cmos output. Instead, use 27k and 68k or 270k and 680k.
So even with those values as suggest, wouldn't the voltage divider be such that it is 10K and 98K which would leave the input biased at 2.2V? Would the gate be able to drive it low? And the 2.2V would be too high for the input to handle with the 10K strapping resistor.
 

Audioguru again

Joined Oct 21, 2019
3,859
A Cmos input draws no current. Then the almost 2.5V high into the 95k voltage divider produces 1.78V at the input of the 1.8V powered device as a high and produces 0V as a low. The 10k pullup is not needed.
 

Thread Starter

matthej

Joined Oct 10, 2020
58
A Cmos input draws no current. Then the almost 2.5V high into the 95k voltage divider produces 1.78V at the input of the 1.8V powered device as a high and produces 0V as a low. The 10k pullup is not needed.
The 10K pullup is needed as a strapping resistor to tell the device which mode to operate in
 
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