# 10.4: JFET Biasing


There are several different ways of biasing a JFET. For many configurations, $$I_{DSS}$$ and $$V_{GS(off)}$$ will be needed. A simple way to measure these parameters in the lab is shown in Figure $$\PageIndex{1}$$. To measure $$I_{DSS}$$ we simply ground the gate and source terminals as this forces $$V_{GS}$$ to be 0 V. We insert an ammeter between $$V_{DD}$$ and the drain, and then set $$V_{DD}$$ to a value higher than $$V_P$$ (+15 VDC generally being sufficient). The resulting ammeter reading is $$I_{DSS}$$. Obtaining $$V_{GS(off)}$$ is only slightly more work. Leaving the ammeter in the drain, unhook the gate from ground and instead connect it to an adjustable negative power supply. Turn the supply more negative until the ammeter reads zero (practically speaking, < 1% of $$I_{DSS}$$). At that point the voltage source will be equal to $$V_{GS(off)}$$.

Figure $$\PageIndex{1}$$: Measuring $$I_{DSS}$$ and $$V_{GS(off)}$$.

## 10.4.1: DC Model

Before we begin examining the bias circuits themselves, we need a basic DC model of the JFET. A model sufficient for our analyses is shown in Figure $$\PageIndex{2}$$.

Figure $$\PageIndex{2}$$: DC model of JFET.

The model consists of a voltage-controlled current source, $$I_D$$, that is equal to the product of the gate-source voltage, $$V_{GS}$$, and the transconductance, $$g_m$$. The resistance between the gate and source, $$R_{GS}$$, is that of the reverse-biased PN junction, in other words, ideally infinity for DC. As a consequence, in most practical circuits we can assume that gate current, $$I_G$$, is zero. Therefore, $$I_D = I_S$$.

## 10.4.2: Constant Voltage Bias

The simplest form of bias is the constant voltage bias. The prototype is shown in Figure $$\PageIndex{3}$$ with current directions and voltage polarities shown.

Figure $$\PageIndex{3}$$: Constant voltage bias prototype.

This is a fairly straightforward design using only a couple of resistors and power sources. Figure $$\PageIndex{4}$$ shows the same circuit but with the JFET model inserted, ready for analysis.

Figure $$\PageIndex{4}$$: Constant voltage bias with model.

Ultimately, the goal here is to determine a means for finding the transistor's drain current and drain-source voltage, along with the potentials across any other components.

To begin, consider the gate-source loop. By KVL, the $$V_{GG}$$ source must drop across $$R_G$$ and the gate-source junction, $$V_{GS}$$.

$V_{G G} = V_{R_G} +V_{GS} \nonumber$

$V_{G G} = I_G R_G+V_{GS} \nonumber$

$$I_G$$ is approximately zero so this simplifies to

$V_{GS} = V_{G G} \nonumber$

Given the transconductance, $$g_m$$, we can find $$I_D$$. Alternately, $$I_D$$ may be found using Equation 10.2.1 along with the device parameters $$I_{DSS}$$ and $$V_{GS(off)}$$. For this circuit, the latter technique tends to be more practical. Once $$I_D$$ is found, the voltage drop across $$R_D$$ may be found, and then $$V_{DS}$$ is determined from KVL.

Example $$\PageIndex{1}$$

For the circuit of Figure $$\PageIndex{5}$$, determine $$I_D$$ and $$V_{DS}$$. Assume $$I_{DSS}$$ = 10 mA and $$V_{GS(off)}$$ = −5 V.

Figure $$\PageIndex{5}$$: Schematic for Example $$\PageIndex{1}$$.

First, because $$I_G \approx 0$$, the drop across $$R_G$$ is $$\approx 0$$ and $$V_{GS} = V_{GG}$$. Using Equation 10.2.1

$I_D = I_{DSS} \left( 1 − \frac{V_{GS}}{V_{GS (off )}} \right)^2 \nonumber$

$I_D = 10 mA \left( 1 − \frac{−2 V}{−5V} \right)^2 \nonumber$

$I_D = 3.6 mA \nonumber$

Looking at the drain-source loop, KVL shows

$V_{DD} = I_D R_D +V_{DS} \nonumber$

$V_{DS} = V_{DD} −I_D R_D \nonumber$

$V_{DS} = 25V −3.6mA \times 3.3k \Omega \nonumber$

$V_{DS} = 13.1 V \nonumber$

While the computation for the constant voltage bias is relatively simple, it does not exhibit a stable Q point. For example, if Example $$\PageIndex{1}$$ is repeated with another JFET, this one with $$I_{DSS}$$ = 12 mA and $$V_{GS(off)}$$ = −6 V, the results are starkly different: $$I_D$$ grows to 5.33 mA and $$V_{DS}$$ shrinks to 7.4 V. These are considerable changes given the relatively modest shifts in the device parameters. In this regard, the constant voltage bias is reminiscent of the simple base bias configuration used with BJTs.

To get a better understanding of the Q point stability issue, refer to Figure $$\PageIndex{6}$$.

Figure $$\PageIndex{6}$$: Variation for constant voltage bias.

Characteristic curves are plotted here for two different devices, one in green and one in blue. These represent the sort of device parameter variations we might expect to see across a product model. The fixed value of gate bias voltage is shown in red. From this graph it should be obvious that this form of bias will produce a wide variation in drain current, and thus, is not a good choice for applications that require a stable Q point. If the application does not have this requirement, constant voltage bias offers the advantage of requiring a minimum of components.

## 10.4.3: Self Bias

Self bias uses a small number of components and only a single power supply, yet it offers better stability than constant voltage bias. The name comes from the fact that the drain current will be used to create a voltage drop that sets up the gate-source, hence the circuit “biases itself”. It is also referred to as automatic bias. The self bias prototype is shown in Figure $$\PageIndex{7}$$.

Figure $$\PageIndex{7}$$: Self bias prototype.

Once again, we may assume that $$I_G$$ is 0. As $$R_G$$ is connected directly to ground, this means that $$V_G \approx 0$$ V. This being true, inspection of the schematic reveals that the magnitude of $$V_{GS}$$ must be the same as the voltage across $$R_S$$. Because $$I_D = I_S$$ then

$V_{GS} = −I_D R_S \label{10.6}$

This value of $$V_{GS}$$ is what generates the drain current. The definition is self-referential. This being the case, how do we analyze the circuit? A proper derivation of the equation for drain current is not trivial. We start with the characteristic equation (Equation 10.2.1) and expand it.

$I_D = I_{DSS}\left ( 1 − \frac{V_{GS}}{V_{GS (off )}} \right)^2 \nonumber$

$I_D = I_{DSS} \left( 1 − \frac{2V_{GS}}{V_{GS (off )}} + \frac{{V_{GS}}^2}{{V_{GS(off )}}^2} \right) \nonumber$

$I_D = I_{DSS} − \frac{2 I_{DSS} V_{GS}}{V_{GS (off )}} + \frac{I_{DSS} {V_{GS}}^2}{{V_{GS (off )}}^2} \nonumber$

Substitute using Equation 10.2.2

$g_{m0} =− \frac{2 I_{DSS}}{V_{GS(off )}} \nonumber$

$I_D = I_{DSS} +g_{m0} V_{GS} + \frac{I_{DSS} {V_{GS}}^2}{{V_{GS (off )}}^2} \nonumber$

Using Equation \ref{10.6} this can be expanded to

$I_D = I_{DSS} −g_{m0} I_D R_S + \frac{I_{DSS} {I_D}^2 {R_S}^2}{{V_{GS (off )}}^2} \nonumber$

Rearranging yields

$0 = \frac{I_{DSS} {R_S}^2}{{V_{GS (off )}}^2} {I_D}^2 − (1 +g_{m0} R_S )I_D +I_{DSS} \nonumber$

This is a quadratic equation in the form $$ax^2+bx+c$$ and can be solved using the quadratic formula:

$y = \frac{−b\pm \sqrt{b^2−4ac}}{2a} \nonumber$

The positive option in the numerator may be ignored as this occurs for $$V_{GS}$$ beyond $$V_{GS(off)}$$. The result is

$I_D = 2 I_{DSS} \left( \frac{1+g_{m0} R_S −\sqrt{1+2 g_{m0} R_S}}{( g_{m0} R_S )^2} \right) \label{10.7}$

Although this is an accurate analytical solution, it's certainly not the sort of equation most people want to memorize or derive as needed. As the $$g_{m0} R_S$$ term is repeated in this equation multiple times, it is useful to plot this equation in terms of normalized $$I_D$$ versus $$g_{m0} R_S$$. This curve is plotted in Figure $$\PageIndex{8}$$.

To use this curve, the first step is to find $$g_{m0} R_S$$. The value of $$R_S$$ is determined by inspection and $$g_{m0}$$ may be determined by Equation 10.2.2, repeated below for convenience.

$g_{m0} =− \frac{2 I_{DSS}}{V_{GS(off )}} \nonumber$

The value of $$g_{m0} R_S$$ is found on the horizontal axis, traced up to the curve and then over to the normalized $$I_D$$ ratio. This number is multiplied by $$I_{DSS}$$ to determine the value of $$I_D$$.

Figure $$\PageIndex{8}$$: Self bias curve.

Example $$\PageIndex{2}$$

Determine $$I_D$$ and $$V_{DS}$$ for the circuit shown in Figure $$\PageIndex{9}$$. Assume $$I_{DSS}$$ = 10 mA and $$V_{GS(off)}$$ = −4 V.

Figure $$\PageIndex{9}$$: Schematic for Example $$\PageIndex{2}$$.

Using the graphical method, first determine $$g_{m0} R_S$$.

$g_{m0} =− \frac{2 I_{DSS}}{V_{GS (off )}} \nonumber$

$g_{m0} =− \frac{2 \times 10 mA}{−4 V} \nonumber$

$g_{m0} = 5mS \nonumber$

Therefore $$g_{m0} R_S$$ = 5 mS $$\cdot$$ 2.2 k $$\Omega = 11$$. The self bias graph yields approximately 0.12 for the normalized current ratio. Therefore

$I_D = 0.12 I_{DSS} \nonumber$

$I_D = 0.12 \times 10 mA \nonumber$

$I_D = 1.2mA \nonumber$

Using Ohm's law and KVL

$V_D = V_{DD} −I_D R_D \nonumber$

$V_D = 20 V−1.2mA \times 3.9 k\Omega \nonumber$

$V_D = 15.32V \nonumber$

$V_S = I_D R_S \nonumber$

$V_S = 1.2mA \times 2.2 k\Omega \nonumber$

$V_S = 2.64 V \nonumber$

$V_{DS} = V_D −V_S \nonumber$

$V_{DS} = 15.32 V −2.64 V \nonumber$

$V_{DS} = 12.68 V \nonumber$

An alternate technique is to make an initial guess for $$V_{GS}$$, typically one half of $$V_{GS(off)}$$. The value of $$I_D$$ is then computed from the characteristic equation (Equation 10.2.1) and compared with the Ohm's law relation, Equation \ref{10.6}, rewritten as $$I_D = −V_{GS}/R_S$$. Chances are, the two results will not agree so adjust the $$V_{GS}$$ estimate and repeat the process. If done properly, the currents should be closer. Iterate this process until you converge on the answer.

To use this technique for the preceding problem we'd start by assuming $$V_{GS}$$ = −2 V (half of $$V_{GS(off)}$$). Using this in Equation 10.2.1 yields $$I_D$$ = 2.5 mA, while using Equation \ref{10.6} produces $$I_D$$ = 910 $$\mu$$A. Obviously the initial estimate was not correct. The second estimate for $$V_{GS}$$ needs to increase negatively as this will decrease the result from Equation 10.2.1 and increase the result from Equation \ref{10.6}, hopefully meeting in the middle. We might try −2.5 volts. This will yield 1.4 mA from Equation 10.2.1 and 1.14 mA from Equation \ref{10.6}. As the gap has narrowed, the adjustment for the third estimate will be smaller, so we could try −2.6 volts. This would be relatively close to the value as computed in Example $$\PageIndex{2}$$ ($$V_{GS} = −V_S$$).

This approximation technique also offers a clue as to how self bias gains stability over constant voltage bias. If for some reason $$I_D$$ was to increase, this would create a larger voltage drop across $$R_S$$. Because this voltage is the same magnitude as $$V_{GS}$$, this means that $$V_{GS}$$ grows negatively. A more negative $$V_{GS}$$ reduces $$I_D$$, thus opposing the initial change in drain current. This feedback mechanism is similar in function to the BJT collector feedback bias. The stability issue is visualized in Figure $$\PageIndex{10}$$.

Figure $$\PageIndex{10}$$: Variation for self bias.

Two device curves are plotted to represent parameter variation (green and blue). Equation \ref{10.6} shows the relationship between $$I_D$$ and $$V_{GS}$$. If we put this in the form $$y = mx + b$$, we find that the line goes through the origin and has a slope of $$1/R_S$$. This line is plotted in red. Where the line intersects the device curve yields the drain current and gate-source voltage for that particular device. Unlike constant voltage bias, self bias shifts some variation over to $$V_{GS}$$, making $$I_D$$ more stable. In fact, if there is a particular design target for $$I_D$$ or $$V_{GS}$$, a rearrangement of Equation \ref{10.6} can be used to find the needed value of $$R_S$$ along with the characteristic curve or equation.

$R_S =− \frac{V_{GS}}{I_D} \nonumber$

For example, if a certain $$I_D$$ is desired, this value could be used with Equation 10.2.1 to determine the corresponding $$V_{GS}$$. These values are then used to find the required $$R_S$$. Alternately, the normalized values could be obtained via Figure 10.2.4.

Example $$\PageIndex{3}$$

Determine a value for $$R_S$$ to set $$V_{GS}$$ = −2 V for the circuit shown in Figure $$\PageIndex{11}$$. Assume $$I_{DSS}$$ = 20 mA and $$V_{GS(off)}$$ = −4 V.

Figure $$\PageIndex{11}$$: Schematic for Example $$\PageIndex{3}$$.

We can determine the drain current using Equation 10.2.1.

$I_D = I_DSS \left( 1 − \frac{V_{GS}}{V_{GS (off )}} \right)^2 \nonumber$

$I_D = 20 mA \left( 1 − \frac{−2V}{−4V} \right)^2 \nonumber$

$I_D = 5 mA \nonumber$

$R_S =− \frac{V_{GS}}{I_D} \nonumber$

$R_S =− \frac{−2 V}{5mA} \nonumber$

$R_S = 400 \Omega \nonumber$

In sum, self bias is a minimal parts count circuit that offers modest stability. The stability can be improved with the addition of other components, as we shall see with the next bias configuration.

## 10.4.4: Combination Bias

The combination bias configuration (AKA source bias) is based on self bias but adds a negative power supply connected to $$R_S$$, hence its name. This will enhance the stability of $$I_D$$, $$V_{DS}$$ and $$g_m$$. The combination bias prototype is shown in Figure $$\PageIndex{12}$$.

Figure $$\PageIndex{12}$$: Combination bias prototype.

The analysis is similar to that of self bias but with one major twist: the source power supply increases the voltage drop across $$R_S$$. This stabilizes the voltage (and hence, the current) because it is no longer equal to $$−V_{GS}$$, but rather

$V_{R_S} = I_D R_S =∣V_{GS}∣+∣V_{SS}∣ \label{10.8}$

If $$V_{SS} \gg V_{GS}$$, then we can approximate $$I_D$$ as $$V_{SS}/R_S$$. As with self bias, an analytical solution for $$I_D$$ is possible. In order to do so, we would begin with the characteristic equation and Equation \ref{10.8}. The derivation is left as an exercise.

$I_D = 2 I_{DSS} \left( \frac{1+g_{m0} R_S (1+k )−\sqrt{1+2 g_{m0}R_S (1+k)}}{( g_{m0} R_S )^2} \right) \label{10.9}$

The formula is very similar to the self bias formula but with the addition of a factor, $$k$$. $$k$$ is a “swamping factor” and is defined as the ratio of $$V_{SS}$$ to $$V_{GS(off)}$$. If $$k = 0$$, there is no source power supply and the formula reverts back to the simpler self bias formula. On the other hand, if $$k$$ is very large, $$I_D \approx V_{SS}/R_S$$.

As was the case with self bias, we can plot Equation \ref{10.9} using the $$g_{m0}R_S$$ factor. A series of three plots for $$k$$ = 2, 3 and 4 are rendered in Figured $$\PageIndex{13}$$.1

Figure $$\PageIndex{13a}$$: Combination bias curve, $$k = 2$$.

Figure $$\PageIndex{13b}$$: Combination bias curve, $$k = 3$$.

Figure $$\PageIndex{13c}$$: Combination bias curve, $$k = 4$$.

Example $$\PageIndex{4}$$

Determine $$I_D$$ and $$V_{DS}$$ for the circuit shown in Figure $$\PageIndex{14}$$. Assume $$I_{DSS}$$ = 12 mA and $$V_{GS(off)}$$ = −4 V.

Figure $$\PageIndex{14}$$: Schematic for Example $$\PageIndex{4}$$.

Using the graphical method, first determine $$g_{m0} R_S$$.

$g_{m0} =− \frac{2 I_{DSS}}{V_{GS (off )}} \nonumber$

$g_{m0} = − \frac{2 \times 12 mA}{−4V} \nonumber$

$g_{m0} = 6mS \nonumber$

Therefore $$g_{m0} R_S$$ = 6 mS $$\cdot$$ 3.3 k $$\Omega$$ = 19.8. The swamping ratio, $$k$$, is $$V_{SS}/V_{GS(off)} = −8/−4 = 2$$. This requires the graph in Figure $$\PageIndex{13a}$$. This graph yields approximately 0.25 for the normalized current ratio. Therefore

$I_D = 0.25 I_{DSS} \nonumber$

$I_D = 0.25 \times 12 mA \nonumber$

$I_D = 3mA \nonumber$

Using Ohm's law and KVL

$V_D = V_{DD} − I_D R_D \nonumber$

$V_D = 24 V−3mA \times 4.7k\Omega \nonumber$

$V_D = 9.9 V \nonumber$

$V_S = V_{SS}+I_D R_S \nonumber$

$V_S =−8V+3mA \times 3.3 k\Omega \nonumber$

$V_S = 1.9V \nonumber$

$V_{DS} = V_D −V_S \nonumber$

$V_{DS} = 9.9 V −1.9V \nonumber$

$V_{DS} = 8 V \nonumber$

As a crosscheck, using Equation \ref{10.9} yields 3.028 mA for $$I_D$$. The deviation is no doubt due to inaccuracy in reading the graph. In any case, using this value of drain current we find $$V_S$$ to be 1.992 volts, a little higher than calculated above. This indicates that $$V_{GS}$$ is −1.992 volts (because $$V_G \approx 0$$ V). If we plug this value of $$V_{GS}$$ into Equation 10.2.1, $$I_D = 3.024$$ mA; an excellent match with the deviation being due to accumulated rounding errors.

In order to show the increased Q point stability of the combination bias, we'll repeat the preceding problem using a JFET with a significantly lower $$I_{DSS}$$.

Example $$\PageIndex{5}$$

Determine $$I_D$$ for the circuit shown in Figure $$\PageIndex{14}$$. Assume $$I_{DSS}$$ = 8 mA and $$V_{GS(off)}$$ = −4 V.

For this version we'll use Equation \ref{10.9}. First determine $$g_{m0} R_S$$.

$g_{m0} =− \frac{2 I_{DSS}}{V_{GS (off )}} \nonumber$

$g_{m0} =− \frac{2 \times 8mA}{−4 V} \nonumber$

$g_{m0} = 4 mS \nonumber$

Therefore $$g_{m0} R_S$$ = 4 mS $$\cdot$$ 3.3 k $$\Omega = 13.2$$. The swamping ratio, $$k$$, is $$V_{SS}/V_{GS(off)} = −8/−4 = 2$$.

$I_D =2 I_{DSS} \left( \frac{1+g_{m0} R_S (1+k) −\sqrt{1+2 g_{m0} R_S (1+k )}}{( g_{m0} R_S )^2} \right) \nonumber$

$I_D = 2 \times 8mA \left( \frac{1+13.2(1+2)−\sqrt{1+2 \times 13.2(1+2)}}{(13.2)^2} \right) \nonumber$

$I_D =2.906 mA \nonumber$

For the graphical method, a reasonable estimate for the normalized $$I_D$$ would be around 0.36, yielding a drain current of 2.88 mA. Stability is apparent because the drain current has dropped only a few percent in spite of the fact that $$I_{DSS}$$ decreased by 33%.

The graph of Figure $$\PageIndex{15}$$ illustrates nicely the increased stability of the Q point. Once again, we plot two representative device curves in green and blue. As was the case with self bias, a plot line can be drawn, the slope of which is equal to the reciprocal of $$R_S$$. This plot line does not go though the origin, though. Instead, the $$x$$ axis intercept is the voltage $$|V_{SS}|$$. Thus, the red plot line is shifted along the $$V_{GS}$$ axis.

As can be seen in the graph, the variation in $$I_D$$ is reduced (although at the expense of variation in $$V_{GS}$$). For large values of $$V_{SS}$$ with correspondingly large values of $$R_S$$, the bias plot line becomes nearly horizontal, indicating a very stable Q point. With two variables in play, this bias proves to be very flexible. It can also be realized by using a positive voltage divider at the gate and removing $$V_{SS}$$ (returning $$R_S$$ to ground).

Figure $$\PageIndex{15}$$: Variation for combination bias.

## 10.4.5: Constant Current Bias

The most stable bias for JFETs relies, oddly enough, on a current source made with a BJT. It is called constant current bias, yet another imaginative tag. Interestingly, although this will keep the Q point very stable, a fixed $$I_D$$ does not guarantee the most stable value of voltage gain. In fact, it might be easier to achieve that goal using combination bias. The prototype constant current bias circuit is shown in Figure $$\PageIndex{16}$$. An NPN BJT is used for an N-channel JFET and a PNP would be used with a P-channel JFET, typically driven from above (i.e., circuit flipped top to bottom).

Figure $$\PageIndex{16}$$: Constant current bias prototype.

Ignoring the JFET for a moment, the BJT is configured as in two-supply emitter bias. In this case the base is tied directly to ground, leaving the emitter at about −0.7 VDC. The remainder of the $$V_{EE}$$ supply drops across $$R_E$$, establishing the emitter current. As the collector is connected directly to the JFET's source terminal, this means that $$I_S \approx I_E$$. The source current winds up being just as stable as the emitter current, which we have already seen is very stable. The only requirement is that $$I_E$$ should not be programmed to be larger than $$I_{DSS}$$. This being true, $$I_D$$ will set up a corresponding $$V_{GS}$$. This also establishes $$V_S$$ because $$V_G \approx 0$$ V. Therefore, the source terminal will be a small positive voltage and this is precisely what the BJT needs in order to guarantee that its collector-base junction is reverse-biased.

Computation of circuit currents and voltages is straightforward and does not involve the use of graphical aides. The first step is to examine the BJT's emitter loop and determine $$I_E$$. Once this is found, $$I_S$$ and $$I_D$$ are known, and all remaining component potentials may be found using Ohm's law and KVL.

This technique does not involve the calculation of $$V_{GS}$$. In fact, because $$I_D$$ is very stable, $$V_{GS}$$ will show the widest variation of all biasing circuits when the JFET is changed. If $$V_{GS}$$ is needed, it can be determined via a little algebraic manipulation on Equation 10.2.1.

Example $$\PageIndex{6}$$

Detemine $$I_D$$, $$V_{DS}$$ and $$V_{GS}$$ in the circuit of Figure $$\PageIndex{17}$$. $$I_{DSS}$$ = 15 mA and $$V_{GS(off)}$$ = −3 V.

Figure $$\PageIndex{17}$$: Schematic for Example $$\PageIndex{6}$$.

We begin by finding $$I_E$$.

$I_E = \frac{∣V_{EE}∣−0.7V}{R_E} \nonumber$

$I_E = \frac{10V −0.7V}{3.6k \Omega} \nonumber$

$I_E = 2.58mA \nonumber$

$$I_E$$ is the same as $$I_S$$ and $$I_D$$, therefore

$V_D = V_{DD} −I_D R_D \nonumber$

$V_D = 20 V −2.58 mA \times 4.7 k\Omega \nonumber$

$V_D = 7.87V \nonumber$

To find $$V_S$$ we note that $$V_S = −V_{GS}$$ and rearrange Equation 10.2.1.

$I_D = I_{DSS} \left( 1 − \frac{V_{GS}}{V_{GS (off )}} \right)^2 \nonumber$

$V_{GS} = V_{GS (off )} \left( 1 − \sqrt{\frac{I_D}{I_{DSS}}} \right) \nonumber$

$V_{GS} = −3 V \left(1 − \sqrt{\frac{2.58 mA}{15 mA}} \right) \nonumber$

$V_{GS} =−1.24 V \nonumber$

Therefore $$V_S = 1.24$$ V and

$V_{DS} = V_D −V_S \nonumber$

$V_{DS} = 7.87 V −1.24 V \nonumber$

$V_{DS} = 6.63 V \nonumber$

We turn next to a computer simulation of a similar circuit to validate our methodology.

## Computer Simulation

A constant current bias circuit is entered into a simulator as shown in Figure $$\PageIndex{18}$$.

Figure $$\PageIndex{18}$$: Constant current bias circuit in simulator.

A cursory estimate shows that $$I_E$$ and $$I_D$$ should be around 4.3 mA. Also, $$V_D$$ should be approximately 20 V − 4.3 mA $$\cdot$$ 2.2 k$$\Omega$$, or about 10.54 volts. The results of a DC operating point analysis are shown in Figure $$\PageIndex{19}$$.

Figure $$\PageIndex{19}$$: Constant current bias DC operating point simulation results.

The drain voltage (node 3) is just over 10.6 volts, agreeing with our estimate. Also, note the minuscule gate voltage (node 1) of 12 $$\mu$$V which verifies our continuing assumption in these circuits that $$V_G \approx 0$$ VDC. Finally, we see a modest potential of about 1.5 volts at the source terminal (node 12). This shows the proper reverse-biasing of both the gate-source and collector-base junctions.

Finally, we can examine the Q point variation using Figure $$\PageIndex{20}$$. Here, the plot line is perfectly horizontal and all device variation is manifest in $$V_{GS}$$.

Figure $$\PageIndex{20}$$: Variation for constant current bias.

## References

1We could add a third axis for $$k$$ and plot a surface, and while it might be pretty, a 3D plot like this rendered onto a 2D surface, such as a page in a textbook, is of marginal utility.

This page titled 10.4: JFET Biasing is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by James M. Fiore via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.