# Determining gate voltage/drain current of voltage-divider JFET circuit using max/min q-point

#### bdn1790

Joined Dec 31, 2018
13
A basic common-source potential divider JFET circuit is designed with the following characteristics:

$$V_{DD} = 20V$$ (Power Supply)
$$I_{DSS} = 10mA$$
$$I_{DQ} = 3mA$$
$$V_{GS(OFF)} = -3V$$
$$V_{DS} = 7V$$
$$R_2 = 150k\Omega$$
$$R_L = 10k\Omega$$

The $$I_{DQ}$$ is not supposed to shift more than +-10% from the aforementioned original value for:

$$I_{DSS} = 10mA +- 2mA$$
$$V_{GS(OFF)} = -3V +- 0.5V$$

Assuming that I were to find the gate voltage (and the missing resistance/voltage values such as R1/RS/Vgs/etc) using the graphical approach only, I have figured that I need to first plot two transconductance curves in the same graph, one when:

[Minimum transconductance]

$$I_{DSS} = 8mA$$
$$V_{GS(OFF)} = -2.5V$$
$$I_{DQ} = 2.7mA$$

And the other when:

[Maximum transconductance]

$$I_{DSS} = 12mA$$
$$V_{GS(OFF)} = -3.5V$$
$$I_{DQ} = 3.3mA$$

With the given details, I can draw the transconductance graph like this as an example (where Q2 = max transconductance, Q1 = min transconductance):

From there, I learned that the slope of both the maximum and minimum transconductance curve will give me the value of the transconductance of the actual $$I_{DQ}$$ and $$V_{GSQ}$$ of the circuit ($$g_m$$ = $$\Delta I_D/\Delta V_{GS}$$ = $$(I_D_{MAX} - I_D_{MIN})/(V_{GS}_{MAX} - V_{GS}_{MIN})$$).

IDmax = maximum transconductance curve ID q-point
IDmin = minimum transconductance curve ID q-point
Vgsmax = maximum transconductance curve Vgs q-point
Vgsmin = minimum transconductance curve Vgs q-point

1. Is my assumption of the slope of both the maximum and minimum transconductance curve being the transconductance value of the actual drain current/gate-source voltage at q-point? If yes, then what are the following steps I should be looking at in order to find the gate voltage? If not, then where does my mistake lie at? Pointing me in the right direction would be very helpful.
2. My teacher advised me that the bias line does not necessarily have to intersect the Q-point of the maximum and minimum transconductance curve. Why is this the case?

Thank you for reading. This is an assignment, and I am not expecting any direct answer. Just a point in the right direction would suffice, thanks again!

Last edited by a moderator:

#### Jony130

Joined Feb 17, 2009
5,213
Hmm... I never heard "transconductance curve" before but I think I know what you mean by this.
What you should do now is to find (or read from the plot) the corresponding Vgs_min, Vgs_max values and solve for RS resistor value.
Because this red line on your plot represents RS resistor value. As for your Q-point, it should be equal to Id = 3mA (the middle transconductance curve not plot )

#### bdn1790

Joined Dec 31, 2018
13
Hmm... I never heard "transconductance curve" before but I think I know what you mean by this.
What you should do now is to find (or read from the plot) the corresponding Vgs_min, Vgs_max values and solve for RS resistor value.
Because this red line on your plot represents RS resistor value. As for your Q-point, it should be equal to Id = 3mA (the middle transconductance curve not plot )
Well, you can call it the Id vs Vgs curve, of which the slope of the curve would give you the transconductance value. I picked up that term from the online tutorials I've been referring to, do correct me if I'm using the wrong term.

Anyhow, I actually haven't figured out how the bias line will look like, or if it will intersect both the q-points. The bias line in the graph I've linked was from an example online. According to my lecturer, I should find the gate voltage first before plotting the final result.

EDIT: I can't edit my original post to include the notification regarding the curve plot (that the bias line shown there has nothing to do with my assignment).

#### Jony130

Joined Feb 17, 2009
5,213
The bias line in the graph I've linked was from an example online
Yes, I know that. But your case is exactly the same. And you got all the information needed to solve for RS. Because this bias line in the graph represent RS resistance (the slope of this line is RS). All you need to do is to solve for Vgs_min, Vgs_max and plug the numbers.

Last edited:

#### bdn1790

Joined Dec 31, 2018
13
Yes, I know that. But your case is exactly the same. And you got all the information needed to solve for RS. Becausethis bias line in the graph represent RS resistance (the slope of this line is RS). All you need to do is to solve for Vgs_min, Vgs_max and plug the numbers.
Noted. After finding Vgs_min and Vgs_max, I also have to find the equivalent Id_min and I_d max for the slope, correct? Or as for what you are implying, I can find the slope using Vgs_min and Vgs_max directly?

#### Jony130

Joined Feb 17, 2009
5,213
Noted. After finding Vgs_min and Vgs_max, I also have to find the equivalent Id_min and I_d max for the slope
But you already knows the Id_min = 2.7mA and Id_max = 3.3mA , what is left to do is to find Vgs_min and Vgs_max.
Can you do just that?

#### bdn1790

Joined Dec 31, 2018
13
But you already knows the Id_min = 2.7mA and Id_max = 3.3mA , what is left to do is to find Vgs_min and Vgs_max.
Can you do just that?
The 2.7mA and 3.3mA is Idq, and not Id. Or do you mean the same regardless? Well, I am getting a negative value for RS. This is the procedure I followed:

1. I find the equivalent Vgsq_min and Vgsq_max for the given Idq_min and Idq_max (Vgsq_max = -1.66V, Vgsq_min = -1.05V)
2. I plot the maximum and minimum transconductance curve using the above values.
3. The slope of both the curves gives me the transconductance value ($$\frac{I_{DQ}_{MAX}/I_{DQ}_{MIN}}{V_{GSQ}_{MAX}/V_{GSQ}_{MIN}}$$). I take the absolute value, since I get a negative value from calculating this one.

Now, based on the guidance you have gave me, I did:

4. Using the transconductance value, I use the equation $$g_m = (\frac{-2I{DSS}}{V_{GS(OFF)}})*(1-\frac{V_{GS}}{V_{GS(OFF)}})$$ and rearrange it in terms of $$V_{GS}$$ So that I can solve for Vgs_max and Vgs_min. Assuming that I were to find the maximum value, I used the maximum Vgs_off and Idss. Same concept for finding the minimum value, which involved using the minimum Vgs_off and Idss. (Vgs_max = -3V, Vgs_min = -2.12V)
5. The equation of the slope is used in finding the gradient of the bias line $$\frac{I_{DQ}_{MAX}/I_{DQ}_{MIN}}{V_{GS}_{MAX}/V_{GS}_{MIN}}$$

Is step 4 proceeding in the right direction? Or have I veered off in any one of the above steps?

#### Jony130

Joined Feb 17, 2009
5,213

Slope = (1.66V - 1.05V))/(3.3mA - 2.7mA) = 1016.66667 Ω Hence, you should choose RS > 1kΩ

And now you can solve for Vgs at Idq = 3mA and solve for VG.

#### bdn1790

Joined Dec 31, 2018
13

Slope = (1.66V - 1.05V))/(3.3mA - 2.7mA) = 1016.66667 Ω Hence, you should choose RS > 1kΩ

And now you can solve for Vgs at Idq = 3mA and solve for VG.
Hi, I understand what you mean now. If I were to solve for Vgs at Idq = 3mA, I get Vg as 1.693V. I have a question though. Assuming that I were to use the slope equation again as you have described in finding Vg directly (At Id = 0A, Vgs = Vg), like:

Slope = 1016.6667 = (1.66V - Vg)/(3.3mA - 0)

If I were to use the above equation, then it gives me a negative value for Vg even though the magnitude is the same. Why?

#### Jony130

Joined Feb 17, 2009
5,213
Slope = 1016.6667 = (1.66V - Vg)/(3.3mA - 0)

If I were to use the above equation, then it gives me a negative value for Vg even though the magnitude is the same. Why?
Because as you can see on the graph the slope is indeed negative and Vgs is also negative.
Try this:
-1016.6667 = (-1.66V - Vg)/(3.3mA - 0)

#### bdn1790

Joined Dec 31, 2018
13
Because as you can see on the graph the slope is indeed negative and Vgs is also negative.
Try this:
-1016.6667 = (-1.66V - Vg)/(3.3mA - 0)
Okay, I get it. One final question I have is: how did you derive the equation for finding the slope Rs? Why is it not y2-y1/x2-x1? Even if I were to use the actual slope equation y2-y1/x2-x1, I am still able to derive the same Vg value. I have referred to a different book, and they have employed the same method you have used albeit without any given explanation (Integrated Electronics: Analog and Digital Circuits and Systems, Millman-Halkins 1971).

#### dl324

Joined Mar 30, 2015
12,418
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