CMOS vs NMOS inverter

Thread Starter

AlexX 52

Joined Mar 15, 2018
14
In my circuit I have a high-side PMOS to control a load. Its drain is connected to the main voltage rail (between 3.0V - 4.0V), its source to the load. Now I want to switch the PMOS with a uC pin (with HIGH a bit lower than the supply rail's voltage). Since the PMOS should be on when the pin is driven low, I need an inverter.

The first thing that came into my mind was a simple NMOS-inverter with a 10k pull-up. An alternative would be a CMOS inverter with an NMOS and a PMOS. The CMOS inverter has the advantage of only drawing current when changing states. So it seems like the CMOS option is better for low-power applications. Since I don't care about component count/size or switching speed, is a CMOS inverter better? I ask because lots of people use simple NMOS inverters or even N-BJT inverters, but I haven't seen a circuit using a CMOS inverter.
 

dl324

Joined Mar 30, 2015
16,839
As with most things, it depends.

If you have a spare inverter; use it. If you don't, you have to decide whether you want to add another IC to your bill of materials, use a discrete inverter that will dissipate more power when it's on, or change the polarity of the control signal from the uC.
 

crutschow

Joined Mar 14, 2008
34,280
You still have to be concerned about the CMOS inverter switching properly if its supply voltage is higher than the μC's output high (what is that difference)?

If the load is not being switched at a high frequency, you can save power with the single P-MOSFET by using a higher bias resistor value, such as 100kΩ or more.
 

Thread Starter

AlexX 52

Joined Mar 15, 2018
14
First of all thank you for your quick response.
You still have to be concerned about the CMOS inverter switching properly if its supply voltage is higher than the μC's output high (what is that difference)?

If the load is not being switched at a high frequency, you can save power with the single P-MOSFET by using a higher bias resistor value, such as 100kΩ or more.
The uC promises that a pin's HIGH voltage at VCC = 3V is greater than 2V with temperatures up to 105°C. Just to be safe, assume the pin's voltage is 1V lower than VCC.
Regarding your second sentence: Do you mean the resistor pulling the gate up? A switching delay of <100ms is definitely acceptable.
No. Swap the source and drain.

No, you don't.
Sorry, I realized I have some mistakes in my original question.
Of course, drain is connected to the load and source to VCC. I mixed NMOS and PMOS switching.
And when the uC's pin is low, the PMOS should be off. So it does require inverting.
 

Thread Starter

AlexX 52

Joined Mar 15, 2018
14
Then you should not use a CMOS inverter, but use a transistor driving a P-MOSFET.
A 100kΩ pullup should give you a switching time of <1ms with a P-MOSFET.
Do you mean an N-channel BJT or an N-channel MOSFET? I suppose BJT, but why not a MOSFET?
 
Last edited:

eetech00

Joined Jun 8, 2013
3,857
In my circuit I have a high-side PMOS to control a load. Its drain is connected to the main voltage rail (between 3.0V - 4.0V), its source to the load. Now I want to switch the PMOS with a uC pin (with HIGH a bit lower than the supply rail's voltage). Since the PMOS should be on when the pin is driven low, I need an inverter.

The first thing that came into my mind was a simple NMOS-inverter with a 10k pull-up. An alternative would be a CMOS inverter with an NMOS and a PMOS. The CMOS inverter has the advantage of only drawing current when changing states. So it seems like the CMOS option is better for low-power applications. Since I don't care about component count/size or switching speed, is a CMOS inverter better? I ask because lots of people use simple NMOS inverters or even N-BJT inverters, but I haven't seen a circuit using a CMOS inverter.
Hi
1. You can just use a single nmos connected with drain to load (other side of load to +supply), gate to uC, source to gnd.
I would also pull the gate down to ground with 10k resistor. When the uC output goes high, it turns on the nmos and energizes the load.

2. Or, you also could use a second PMOS to perform an inverting function before the PMOS driving the load.
The drain of the first pmos would connect to a 10k resistor to ground, gate to uC output, source to +supply. The gate of the second pmos would connect to the junction of the resistor and drain of the first pmos, drain to load (other side of load to gnd), source to +supply.
When the output of the uC goes high it turns off the first pmos, that then turns on the second pmos driving the load.
eT
 

Thread Starter

AlexX 52

Joined Mar 15, 2018
14
Hi
1. You can just use a single nmos connected with drain to load (other side of load to +supply), gate to uC, source to gnd.
I would also pull the gate down to ground with 10k resistor. When the uC output goes high, it turns on the nmos and energizes the load.

2. Or, you also could use a second PMOS to perform an inverting function before the PMOS driving the load.
The drain of the first pmos would connect to a 10k resistor to ground, gate to uC output, source to +supply. The gate of the second pmos would connect to the junction of the resistor and drain of the first pmos, drain to load (other side of load to gnd), source to +supply.
When the output of the uC goes high it turns off the first pmos, that then turns on the second pmos driving the load.
eT
Thanks, but I can't use an NMOS as this would invalidate any ground references of the load.
Concerning your second option: One PMOS is always turned on if I understand correctly. I can't do that since it continuously draws power.
 

eetech00

Joined Jun 8, 2013
3,857
Thanks, but I can't use an NMOS as this would invalidate any ground references of the load.
Concerning your second option: One PMOS is always turned on if I understand correctly. I can't do that since it continuously draws power.
Yes its on...but MOS devices are voltage driven, not current driven like BJT's. They don't use hardly any current. You could use a high value resistor for the first PMOS...may be 50k-100k.

What is the part number for the PMOS you are using?

eT
 

AnalogKid

Joined Aug 1, 2013
10,986
My second vote is for the n-channel small signal MOSFET as an inverter.

My first vote is for changing the logic polarity of the uC output pin.

ak
 
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