transconductance frequency response problem

Dear robert21,

i have not changed the DC source ( it stayed 0.5 V DC) just to set the bias point for the output transistor.

After that i have defined an expression using calculator IF(In_node)/Vf(out_node) as shown bellow.

However i get a zero plot threw out frequency range.

Where did i got wrong defining the sources or expression in order to get the frequency response of Vout/Iin?

It is difficult to read mush detail from your posted schematic. However, if I understand from your comment "i have not changed the DC source ( it stayed 0.5 V DC) just to set the bias point for the output transistor." and zoom in on your schematic I can see voltage source V2 is connected to node Vtest and has a DC value of 0.50 V. You are also trying to measure the AC response at node Vtest in response to an 1 V AC applied to current source I5. However, your transfer curve shows 0. My question to you is, how do you expect to measure an AC response due to I5 at node Vtest when you have shorted node Vtest to ground with the ideal zero source resistance of voltage source V2. V2 forces the DC voltage of node Vtest to 0.50 V, but an ideal voltage source has a zero impedance for all frequencies and hence will set the AC output to 0 for all frequencies greater than 0 Hz - which is exactly what spectre is telling you.

If I recall, on a previous post of yours, you experienced the same problem and I think Andrew and I pointed out the issue to you. Does this make sense robert 21?

Shawn

Hello Shawn, Thank you very much for the help.

The main  conclution that it got is such,If we put on the output DC source just to maintain saturation on the transistor then it kills all other AC signal.(As you said by grounding Vtest)

So i put an idial  current DC source  on the output and connected a ressistor in series to create the BIAS point.

and i did measure an AC voltage response on the output node?

Is that a plausable solution with the issue,with respect of getting the correct signal?

Thanks

current source with a ressistor in series to make the same voltage on the drain of the transistor and it  worked.

Dear robert21,

robert 21 said:

So i put an idial  current DC source  on the output and connected a ressistor in series to create the BIAS point.

and i did measure an AC voltage response on the output node?

Is that a plausable solution with the issue,with respect of getting the correct sign

As long as the added DC current does not impact the device under test. An alternative to avoid that is to use an instance of a resistor from the analogLib and set its AC resistance to a very, very large value with a small DC resistance and connect its other terminal to an ideal voltage source. Please see the attached GUI of the component.

Shawn

Hello Shawn , I am trying to put a 0.7 Bias on DC and calculate impedance from AC

DC:

A simple calculation was made as shown bellow  and for some reason instead of getting 0.7 on the drain of the transistor i get 0.25V ,(as shown bellow)

I tried to reduce the voltage drop by reducing the resistance from 55K to 40K and its stuck on than 0.25V as if there is no kirchoffs law of voltage loop.

1.8-(20*10^-6)*R0=0.7

R0=55Kohm

AC:

I have defined my RES to be zero in AC resistance ,so we will see no extra ressistane on the frequency range and get the real impedance Z=V/I of the circuit.

I_AC=1 so Z=V_AC

Here  it gave me a very abnormal Voltage,as shown in the end.

Where did i go wrong?
Thanks


55KOhms:

40KOhm:

Dear robert21,

> AC

> :

> I have defined my RES to be zero in AC resistance ,so we will see no extra

> ressistane on the frequency range and get the real impedance Z=V/I of the

> circuit.

> I_AC=1 so Z=V_AC

> Here  it gave me a very abnormal Voltage,as shown in the

> end.

> Where did i go wrong?

Please read my prior response more carefully robert21. In the dialog box for the analogLib resistor, I noted that the DC resistance should be a small value (i.e., 1 milliohm as an example) and the AC resistance should be a very large value relative to your expected output impedance (i.e., set to >> 1e6 ohms for example). I also noted that the analogLib resistor should be connected to an ideal voltage source whose value is set to the DC bias you want to connect to your device (I believe you indicated 0.70 V). You do not need the resistor in series with the ideal current source. Please see the attached modified circuit diagram I created from your schematic.

Shawn

Hello Shawn , yes regarding biasing i put the resistor and voltage source as shown bellow and it worked fine.

But when i will need to find the output impedance (for bandwith manipulation) then AC_resistance =1M will cause a problem.

So for this purpose i have added a branch in parralel an I_AC=1A source as shown in the plot bellow  Z=V_ac/I_ac   I_ac=1  Z=V_ac  and calculated the imaginary part of the Voltage that it sees and extracted the capacitance from it as shown bellow.

1/(2pi*100*(10^6)*x)=49000

x≈3.24806006309990×10^-14

But when i tried to reduce the bandwidth by adding an 1n capacitor in series it hasn't changed at all.

Why my idial AC current source didnt see that the capacitance change. it gives me the same Imaginary part plot as before.

Where did i go wrong ?

AC Source:

Ressistor for Impedance calculation:

AADDING CAPACITOR TO MANIPULATE BANDWIDTH:

Dear robert21,

robert 21 said:

But when i will need to find the output impedance (for bandwith manipulation) then AC_resistance =1M will cause a problem.

I am not sure of your objective robert21. However, if you are just trying to use your ideal current source to measure the output impedance (amnd capacitance in particular) of your output node, the circuit I provided appears to be sufficient to me. In other words, you can simply change the AC resistance of the resistor in series with the 0.70 V voltage source to a value that is far greater than your expected output impedance. In your example, if I understand correctly, you expect to see a capacitance on the order of 32 fF at 10 MHz. This suggests an impedance at 10 MHz of 1/(2*pi*10e6*32e-15) = 497 Kohms. Hence, you can set the AC value of the resistance to, for example, 100 Mohms and it will not significantly impact your capacitance measurement.

robert 21 said:

But when i tried to reduce the bandwidth by adding an 1n capacitor in series it hasn't changed at all.

Why my idial AC current source didnt see that the capacitance change. it gives me the same Imaginary part plot as before.

Where did i go wrong

robert21, please think before you ask questions like this! Your 1nF capacitance is in series with the output capacitance of your circuit. Please take the time to compute the series capacitance of 1 nF in series with 32 fF. It is essentially 32 fF.

Shawn

Hello Shawn , Yes sorry about that 1nF in series with 32fF is 32fF at absolute proximity.

Unknown said:

the circuit I provided appears to be sufficient to me. In other words, you can simply change the AC resistance of the resistor in series with the 0.70 V voltage source to a value that is far greater than your expected output impedance

So my plot vary yours only by the resistor i put in series with the idial current source ,Which i need to remove :-)

Thank you very much.

Hello Shawn , Yes sorry about that 1nF in series with 32fF is 32fF at absolute proximity.

Unknown said:

the circuit I provided appears to be sufficient to me. In other words, you can simply change the AC resistance of the resistor in series with the 0.70 V voltage source to a value that is far greater than your expected output impedance

So my plot vary yours only by the resistor i put in series with the idial current source ,Which i need to remove :-)

Thank you very much.