Ron simulation problem of a simple switch

graph for ron resistance vs. voltage

How can i get the curve of Ron vs input voltage?
when i add a de sweep to input and get value directly on the output node,it seems not correct because there is no way to gnd.
is any risistor needed to connect out to gnd ? what's the Ron test circuit of switch like this one .

I appreciate of any response.

switch cmos simulation ron

There is no need to connect a resistor

there are different methods we can follow:

1) your method : applying dc sweep at "in" node and find out the current supplied by the source at "in" node. as u know V at node "in" V/I gives the on resistance , for this no need to connect resistor. this experiment does not give much information because ur on reistance will be fairly constant.

2) keep constant voltage(v1) at "in" and sweep voltage at n-gate terminal with vp(from 0 to VDD), at p-gate apply same vp or (VDD-vp) and find the current (I) supplied by source at "in" v1/I gives u the Ron. and this gives u lot of information about the characteristics of Tx gates

the v(in,out) is keep 0 when I add an dc sweep voltage form 0-vdd by method 1. yes,We can obtain much more information by method. But the curve which i need is like the followed one. I thing the mosfets work at either off or tran region, they won't be at saturation region. so how can i get the curve like this one?the vg on the graph must be Vin i think.

rock_zhu said:

How can i get the curve of Ron vs input voltage?
when i add a de sweep to input and get value directly on the output node,it seems not correct because there is no way to gnd.
is any risistor needed to connect out to gnd ? what's the Ron test circuit of switch like this one .

I appreciate of any response.

Click to expand...

You can fix your output voltage then sweep the input voltage.

Ron,nmos is equal to 1/(gdso,nmos), you can easily get the value in Hspice.

Ron,pmos can also does.

You can get Ronn by par('V(in,out)/i(Mn)') and Ronp by par('V(in,out)/i(Mp)').
The Ron=Ronn||Ronp.
You may add a load capacitor at the output of your switch.

Material said:

You can get Ronn by par('V(in,out)/i(Mn)') and Ronp by par('V(in,out)/i(Mp)').
The Ron=Ronn||Ronp.
You may add a load capacitor at the output of your switch.

Click to expand...

As i described in the method 2 short gate of p and n mos and dc sweep the voltage of this gates keeping "in" constant. than u can get Ronn, Ronp and Ron by above described way, but dont add cap at "out" it will give u zero current and u will not able to measure v/i. just short the out to ground. if u plot Ronn, Ronp and Ron u will get the required diagram which is shown

rock_zhu said:

But the curve which i need is like the followed one. I thing the mosfets work at either off or tran region, they won't be at saturation region. so how can i get the curve like this one?the vg on the graph must be Vin i think.

Click to expand...

I think, there is something wrong with the the graph. I guess it makes sense to check performance(Ron) of a switch as IN charges( or discharges) OUT. So 'vg' in the graph might be V(out) with V(in) fixed at 3.3V and the switching points be VTP and 3.3V-VTN.

Point me if i am wrong.

Regards

ipsc said:

rock_zhu said:

But the curve which i need is like the followed one. I thing the mosfets work at either off or tran region, they won't be at saturation region. so how can i get the curve like this one?the vg on the graph must be Vin i think.

Click to expand...

I think, there is something wrong with the the graph. I guess it makes sense to check performance(Ron) of a switch as IN charges( or discharges) OUT. So 'vg' in the graph might be V(out) with V(in) fixed at 3.3V and the switching points be VTP and 3.3V-VTN.

Point me if i am wrong.

Regards

Click to expand...

the graph shown is simulated

with v(in, out) =constant, v(out)=gnd, V(ngate, pgate)=0(shorted)and V(ngate(or pgate), out)=vp. value vp is DC swept from 0 to VDD

Hi,

as I mentioned in my earlier post, the graph is completely wrong. The graph is almost same as Fig. 12.16(b) in Razavi (pg:417). And the switching points are VTP and (VDD-VTN) unlike Vtn and (3.3-vtp) as mentioned in figure. 'Ron' is defined as the resistance from drain to source when source and drain voltages are ≈ equal, i.e. in this case "Ron is Rds at Vg=V(in)=V(out) for the gate voltages of PMOS and NMOS at 0 and VDD respectively."

Now to get the graph I guess these methods will work:

Method 1: Connect two different voltage sources of equal value to IN and OUT. Do ac analysis on one of sources and find ac resistance ro (=Ron). Repeat this for different voltages.
I think there is an option to cascade different analyses. With that you can be able to do this in one step by cascading ac and sweep analyses.

Method 2: Connect two different voltage sources to IN and OUT. Ramp these sources in such a way that one of them is always less than other by few or few 100's of uV. Do transient analysis. Ron=[V(in)-V(out)]/I.

Hope this helps.

the vg in the figure is not vin its the gate voltage of the nmos and pmos both shorted together

Hi,

you can check the attched file, which has some good circuits for testing CMOS switches.

Regards

u can add a voltage source at input and add a constant current source at output, then dc sweep the voltage source. use the voltage drop between input and output divide the preset current to get the switch resistance.