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OL is too high?

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alphadog

Banned
Hello,
I wanted to try this circuit as an inverter.
I was told that the VOL of this circuit is too high.
I have 3 question please:
1. des 0L region start when the VTC reaches a -1 slope?
2. Why is the VOL here considered high?
3. The VOH of this circuit is VDD - VT2. Is it considered a low VOH? (assuming VDD is 3.3V/5V).

I know I can use better inverters, but i want to understand whats wrong with this one.

Thank you.

 

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Sceadwian

Banned
Will those FET's even turn on at 3.3 volts?
You could replace Q2 with a 10k resistor and it will give you 0 volts when triggered from VIN but you'd have to keep the output very lightly loaded, you can reduce the value of the resistor if you don't mind the idle current. Two FET inverters are usually pmos on top nmos on bottom with both gates tied to vin.
 

alphadog

Banned
Thanks.
I know there are better invereters, i want to understand the issues i mentioned above.
 
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audioguru

Well-Known Member
Most Helpful Member
Both Mosfets will turn on at the same time which will short-circuit the supply.
 

Sceadwian

Banned
Yeah, I was wondering about that too, I thought it was just a simplified schematic.
 

Sceadwian

Banned
You're not supplying enough information alphadog. High compared to what? As it stands right now it's a circuit from a text book not something that works in the real world so I have no idea what you're asking. You have no specification for determining what the VOH or VOL limits might be.
 

Hero999

Banned
Hello,
I wanted to try this circuit as an inverter.
I was told that the VOL of this circuit is too high.
I have 3 question please:
1. des 0L region start when the VTC reaches a -1 slope?
2. Why is the VOL here considered high?
3. The VOH of this circuit is VDD - VT2. Is it considered a low VOH? (assuming VDD is 3.3V/5V).

I know I can use better inverters, but i want to understand whats wrong with this one.

Thank you.

It won't work, they're both N-channel, the high side should be p-channel.

Don't worry about it shorting the supply, if you're using low power MOSFETs and the gate voltage is just 3.3V the current will be very low.
 

alphadog

Banned
Why it wont work?
If VIN = 0L, then Q1 is in cutoff, and therefore also Q2 is in cutoff.
For Q2 to be in cutoff then VDS2 must be lower then VT2, otherwise it will conduct current, therefore it will be VDS2 = VT2.
Therefore VIN = 0V => VOUT = VDD - VT2.

If VIN = 1L, then Q1 is ON, lets assume its in linear region (because we want VOUT to be close to zero voltage, meaning VOUT = VDS1 < VGS - VT).
Q2 is always in saturation region (as long as VDS2 > VT2), therefore it conducts current.
Now we need to find the intersection between the curves of Q1 and Q2, and we shall find that they intersect in low value of Vo in the 0L region.
 
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Sceadwian

Banned
Not sure why, but it doesn't. I've simulated it on LTSpice with a few different mosfets and no matter what I pick or change the voltages to there's always a voltage on the Vout line. I'm not sure where it comes from. Someone else that knows more about mosfets might be able to explain it.
 

alphadog

Banned
Sceadwian
Thats exactly what i was asking about, that the VOL is too high, as you saw in your simulation.
1V is definetly considered high.
You say that it doesnt work just because of this VOL = 1V, and it doesnt mean that it doesnt work but it means that its not good enough.

kchriste
You are right.
Thats exactly what i've been told.
But they dont explain there why does it have high VOL.
Do you know why?

Thank you both pals.
 
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Sceadwian

Banned
Someone here has to know, it has something to do with the way mosfets work.
 

audioguru

Well-Known Member
Most Helpful Member
If the Mosfets are the same then turning on the bottom one will cause the top one also to turn on and while they are short-circuiting the supply, their output low voltage (VOL) will be near half the supply voltage.

If the bottom Mosfet is stronger or has more base voltage then it will conduct more which will cause the output low voltage to be a little less than half the supply voltage.

The output high voltage (VOH) will not be very high because when the bottom Mosfet is turned off then the top one turns on until its source voltage (the output) is a couple of volts less than the supply voltage when it begins to turn itself off.
 

Sceadwian

Banned
That would explain why in the simulation when I increased the VIN voltage the VOL dropped lower. If VOUT is loaded with a resistor the VOH should be pretty close to supply.
 
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audioguru

Well-Known Member
Most Helpful Member
That would explain why in the simulation when I increased the VIN voltage the VOL dropped lower. If VOUT is loaded with a resistor the VOH should be pretty close to supply.
Not if the load connects to ground.
If the Mosfet needs a gate-source voltage of 10V then as a source-follower its loaded output high voltage will be very low.

If the load connects to the positive supply then when the lower Mosfet is turned off the output voltage will be at the supply voltage.
 
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