JFE150: Loop Analysis

Hi Yang, 

Thank you for reading my application note! To simulate gm you may use my simulation below. I have attached my simulations at the bottom of this post. 

The circuit for simulating gm is shown below. A constant current source is used to force a desired Ids current. A 1TF capacitor is connected to the source of the JFET such that at frequency the source is ground. This ensures modulation of Vgs. Given that the current source has an ideal infinite resistance we want to make sure we bypass that resistance at frequency in order to keep the source at AC ground. Otherwise the circuit acts as a source follower. A 1TF capacitor is added at the drain such that at frequency all the AC drain current flows through C3. A current meter is added in order to measure gm. The 2mA DC current flows through RD = 4k. RD ensures a desired VDS voltage is achieved. The DC parameters can adjusted as desired to determine different values of gm. 

gm is in decibels so I convert to mS. as shown below. 

yang fan said:

Why there is a 20dB/Dec from A to B? possibly because from A to B in frequency, RD and CD form a high pass filter. Capacitor CD can be viewed as a resistor whose value is larger than RD, ids flow into RD mostly to generate a voltage (of course it’s small signal), and impedance of CD decrease as frequency increase, which make OPA202 inverting amplifier configuration. At frequency above B, CD is shorted compared to RD, so RD is grounded and make no influence, ids flow through R1 to produce output of OPA202. Is it true?

Yes this is correct. 

yang fan said:

But why there is a -20dB/Dec from C? I think maybe R1 and C1 interact a pole, but 1/(2*pi*R1*C1) is about 16kHz, in fact point C frequency is 2.95kHz, so I’m confused. How to calculate point C frequency ?

This is a bit complicated. However I appreciate the question as it made me think a little deeper into the circuit. I was just looking at this and will try my best to explain. The output impedance of the JFET plays a role here. The ro of the JFET can be approximately calculated from the IDS vs VDS curve. The circuit is biased at 2mA of IDS. This means the Vgs is approximately -0.7V (see Figure 6-1 in the product datasheet). Looking at the curve below we can approximate the slope of the curve to determine 1/ro. 1/ro is the slope of the curve. So we can flip the slope calculation to determine ro.  

ro = 25/(1.6*10^-3) = 15.625k

I can now create a simplified model of the OPA202 transimpedance section of the circuit to analyze the feedforward gain of the composite circuit. I shorted capacitor Cd because it is 10uF and is a short at the frequency we are interested in looking at. It just forms the high pass filter section with RD as you mentioned. You can put it into the simulation to see. Below you can see the simplified model gives a corner frequency of 2.6kHz. This is fairly close to the corner frequency of 2.94kHz I get from the full composite circuit of the app note (see below). To be precise we would have to exactly know ro. The small discrepancy comes in because I am ballparking the number from the plot. I would call this close enough. 

Understand upper corner frequency in A.TSCJFE150 gm (1) (1).TSC

I hope you find this information helpful. 

Best Regards, 

Chris Featherstone

Hi Yang, 

To follow up on my last post. To be a little more precise I should have biased the RD resistor to the 12V supply and not 6. See the modification below. This gets us a little closer in the simplified transimpedance model. 

8623.Understand upper corner frequency in A.TSC

Best Regards, 

Chris Featherstone

Hi Yang, 

To be even more accurate albeit more complex you can use the two port network method from my hand written notes for the JFE150EVM. These notes were the very beginning of my analysis on this circuit a few years ago. Page 3 is the two port calculations. Basically I break the beta network into equivalent impedances hanging of the source node and output of the OPA202 in order to form the feedforward circuit. This still isn't perfect and there is probably some nuances I am neglecting in the sim. This does get us very close to the composite circuit frequency for the A curve in the app note. I now get 3kHz as opposed to the 2.94KhZ from the app note. The tina sim can be found at the bottom of this post. 

See page 3 from these notes

Two Port Network (2).pdf

Click the image below to see in better resolution

Understand upper corner frequency in A Two Port Model.TSC

If your are interested in my Matlab code it is here and can be adjusted to fit whatever circuit configuration you are using. 

Matlab Code:

s = tf('s');
Rs1 = 100;
Rs2 = 10;
Csource = 100*10^(-6);
Rf =1000;

% This section deals with the Beta newtork. If Aol is very large then the
% closed loop gain is approximately 1/beta
Zs = Rs2 + 1/(s*Csource); % This works
%bode(Zs)

Zsprime = Rs1*Zs /(Rs1 + Zs);

OneOverBeta = (1+Rf/Zsprime) %this is 1/beta fix this so it makes sense

Beta = Zsprime/(Zsprime+Rf)

%Acl = 1/Beta;

%bode(Acl)

%This is extra code
%B = (960*(s+100))/(106*(s+9.434))/((960*(s+100))/(106*(s+9.434))+10000);
%bode(1/B);
%This is the end of extra code

% This section deal with the gain circuit which we will call the Aol
% circuit of our system.

ZC10 = 1/(s*500E-12);
R10 = 1E6;
RD = 4E3;
ZCD = 1/(s*100E-6) ;
ZCG = 1/(s*100E-6) ;
gm = 10E-3;
R1 = Zsprime*Rf/(Zsprime+Rf);
R2 = Rf + Zsprime;
RG = 1E6;

% Feedback components around the Op Amp

Z10 = ZC10*R10/(ZC10+R10);

% Rg in parallel with R1 +1/gm
R1prime = R1 +1/gm;

Rin = R1prime * RG/(R1prime + RG);

% RD and CD in parallel

ZD = RD*ZCD/(RD+ZCD);

Aol = Rin*ZD/(R1prime * (Rin + ZCG)) * Z10/ZCD

opts = bodeoptions
opts.Title.FontSize = 14;
opts.FreqUnits = 'kHz'
opts.Title.String = 'JFE150EVM Bode'
opts.Xlim = ([0.001 10E6])
opts.Ylim = ([0 80])
bode(Aol, opts)
hold on
bode(OneOverBeta, opts)

hold on
ACL = Aol/(1+Aol*Beta)
bode(ACL, opts)

hold on

%h = gcr
%setoptions(h,'FreqUnits','Hz')
%h.AxesGrid.Grid = 'on'
%xlim([0 10E6])
%ylim([0 60])

Best Regards, 

Chris Featherstone

Hi Yang, 

This circuit is very complicated with two loops. You can decrease the frequency corner with passives however you wouldn't be able to extend the -3dB frequency in the feedforward gain curve with passive components alone. You have to choose a higher speed amplifier in order to extend the frequency. I will try to explain this reasoning below. 

In my application note I call the feedforward gain A. A is a combination of the JFET gain and the OPA202. There is another loop in the circuit that I did not break inside the circuit. The second loop is around the OPA202 alone. Below I have broken this loop to demonstrate what is causing this corner to occur around 3kHz. You can see that the OPA202 has it's own Aol that decreases over frequency. When Aol and 1/beta meet there is a -3dB point at roughly 3-4kHz. This is controlled by the OPA202 itself and the miller capacitance inside of the OPA202. In order to extend the frequency an op amp that is higher speed may be chosen.  The beta network of this loop is made up of the 1M resistor and the 10pF cap. These two components form a pole at 15.9kHz as seen below in the 1/beta curve. You may pull this corner in. There isn't a calculation to provide for the corner in the A curve. I hope this clarifies what exactly is creating this point however. 

Here is the tina simulation for this analysis. 

Second Loop Break Analysis.TSC

Best Regards, 

Chris Featherstone