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Design >> Mixed-Signal Design >> Basic Inverter Design
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Message started by hande.vinayak on Feb 14^th, 2007, 4:55am

Title: Basic Inverter Design
Post by hande.vinayak on Feb 14^th, 2007, 4:55am

Hello,
As I am beginner I have started with designing an Inverter.
I am having spec as IOH= 3mA.VOH =2.25 as Vdd is 2.5.
The VOHmin is the vtg where VILmax is applied.At that point I am getting
correct 3mA.
Is this correct way to calculate IOH?
Second thing,as I am not getting any load currect.
I am getting load current only when i/p's rise time and fall time.
Then how to decide above mentioned parameters?
As we have studied theoretically,nmos or pmas should ON / OFF periodically,
and current should get sink and source from load.
But in my case I could not find that load is sinking or sourcing any current except rise and fall time.
Plz,help me sort out the problem.
Thnx very much.

Title: Re: Basic Inverter Design
Post by Vabzter on Feb 16^th, 2007, 6:51am

Hi,
Are u not able to plot the current waveform? i am not sure if I got your question..
BR
Vabzter

Title: Re: Basic Inverter Design
Post by SRF Tech on Feb 16^th, 2007, 7:27am

Hande,
I am not quite sure I understand your question either, but I think what you are saying is that you have a IOH spec of 3mA, but you are not seeing any output current except during your transient. If this is the question, the answer is first that IOH and IOL specs are written for when the inverter is driving a resistive load, not capacitive. Basically, when your inverter attempts to drive a 2.5V high value across that resistive load, the actual output value is decided by the voltage division across the output impedance of your inverter and the load resistance. With your spec, it is your job to make sure that you can source atleast 3mA and still maintain a Voh=2.25V (essentially your ouput impedance must be less than 0.25Vdrop/3mA = 333 ohms).

You can test this in two ways: 1. Apply a fixed DC voltage of 2.25V at the output of your inverter and see how much current your inverter will source into the DC load when you try to drive it to 2.5V. 2. Or you can apply an ideal current sink of 3mA to the output of your inverter, thenm drive a high value and measure the voltage at the output of your inverter, it should be >2.25V.

I tend to use an ideal voltage source and measure the current (my preference)

Hope this helps,
-Stephen