**characteristics and configurations of JFET**in great depth. Now in this post, we are going to analyze the key differences between BJT and JFET.

But before that, I would explain to you some remaining concepts of JFET which we will be requiring for the comparison of

**BJT**and JFET.

While studying the characteristics of JFET we had seen that the output current Id was varying proportionally with both input(Vgs) and output(Vds) voltages i.e. **Id ****α Vgs, and Id ****α Vds****. **

**Id = Idss * [ 1 - (Vgs/Vp) ]^2**

where

**Id**:- Drain current which determines the Q-point

**Idss**:- Drain to source saturation current when Vgs=0v

**Vgs**:- Gate-Source voltage which determines the Q-point

**Vp**:- Pinch off voltage

**#** Now, On many websites or in some books you will see that** JFET is termed as a square-law device** but they don't provide the proper information that why it is termed so?

**@** Here is the reason for it, we studied the above equation and after analyzing it properly we saw that Id was proportional to Vgs square i.e. **Id **α** (Vgs)^2. **Now, **Id was on Y-axis** and **Vgs was on X-axis** so you can assume this equation as parabola or y=x^2 (Please don't get confused I had just given a random example of parabola so that you can understand the concept thoroughly). Since in the transfer characteristics, we got a curve, therefore, we can say that JFET is a non-linear square law device.

**# BJT vs JFET key differences with full analysis:-**

**1) Why Jfet is known as a unipolar device and BJT as a bipolar device?**

@ Before going into detail about JFET I would like to revise the basic term of the BJT, as we had studied in BJT that BJT is known as a bipolar device because there both holes and electrons were taking part in conduction. For example, consider an

**NPN transistor**, we are working with CE configuration and we know that

**emitter(output) current is the sum of base current and collector current**, and as I had mentioned earlier in my posts that

**base current is nothing but current due to minority carriers(here holes) while collector and emitter current is nothing but current due to majority carriers(here electrons).**Since the output current was the combination of 2 different carriers or charges

**that's why we termed BJT as bipolar transistors.**

**JFET**, the output current Id is only due to the flow of a single carrier no other charges or carriers take part in conduction, therefore, Jfet is known as a unipolar device. For example, in an

**N-channel JFET, the output current(Id) occurs due to the flow of electrons**from source to drain

**you won't see any holes taking part in conduction**. Similarly for P-channel JFET, the output current was occurring due to the flow of holes.

**Since in JFET, only a single carrier was taking part in the conduction therefore JFET is known as a unipolar device.**For your better understanding, I had inserted one video which will make your concept more strong.

**VIDEO:-**

**2) Why Jfet is known as a voltage-controlled current source and Bjt as a current controlled-current source?**

**@**To understand this concept you should know the input characteristics of BJT and transfer characteristics of JFET. In input characteristics of BJT, we saw that the

**output current(Ic)**was dependent on the

**input current(Ib)**. We also saw one equation

**(Ic =**

**β***

**Ib)**which gave us a clear idea that how they were related. Since for BJT, the input current was controlling the output current therefore BJT is known as a current-controlled current source.

**@** Similarly for JFET, as we saw above that the output current(Id) was proportional to Vgs square i.e. **Id **α** (Vgs)^2. **Since for JFET, the input voltage is controlling the output current, therefore, JFET is considered as a voltage-controlled current source. For your better understanding, I had inserted one video which will make your concept more strong.

**VIDEO:-**

**3) Why the input impedance of JFET is greater than BJT? **

**@** In BJT while studying CE configuration, I explained to you that the **base-emitter junction is in forward bias state** that means you can relate it with a p-n junction diode which is in a forward-biased condition and I hope that you know the working of the p-n junction diode. So **if the input voltage** of BJT is** greater than 0.7** then only it will** allow current to flow** through it and the common-emitter junction of BJT will **work as a short circuit**.

**@** Now, for JFET as I had explained in the earlier post that the g**ate-source junction is in the reverse-biased state** that means you can relate it with the p-n junction diode which is in reverse-biased condition. So you know very well that even if we increase the source voltage further the **current will not flow** **because the depletion region keeps on increasing** and the gate-source junction of JFET will work as an **open circuit**.

Now as you can see in the above diagram that the input current is zero. We know that by ohms law,

V = I * R;

since, I(input) = 0;

Therefore, **R** = ∞

Thus we can say that the input impedance of Jfet is infinite, but practically it is finite and maximum because the current is not exactly zero it is around in few nano amperes, and this is because of the minority carriers for your better understanding I had inserted the video below. This video shows how the reverse-biased p-n junction diode conducts current through it but it will also help to clear this concept as well.

**VIDEO:-**

**4)** **Why the power loss in BJT is high as compared to JFET?**

**@** As we saw above that the input impedance of BJT is less than that of JFET. So by ohms law, if resistance(R) is less than current(I) would be high and we know that power(P) is a product of voltage(V) and current, therefore power loss would be more.

Mathematically,

**Resistance = R(min)**

Current = V/R(min)

therefore, **I = I(max)**

**P = V * I(max) = P(max)**

therefore, **Power loss at the input = P(max)**

**@ **Similarly for JFET we saw that the input impedance was very high almost tending to infinity. So if resistance(R) is greater than current(I) would be less and we know that power(P) is a product of voltage(V) and current, therefore power loss would be less (ideally zero).

__NOTE__:- Resistance here is maximum not minimumMathematically,

**Resistance = R(max)**

Current = V/R(max)

therefore, **I = I(min)**

**P = V * I(min) = P(min)**

therefore, **Power loss at the input = P(min)**

**Ideally, R = **∞

Therefore, I = 0

Thus, **Power loss = 0**

**VIDEO:-**

**5)** **Why the switching speed of JFET is greater than that of BJT?**

**@ **Basically BJT is a current-controlled current device and Jfet is a voltage-controlled current device and it takes little time for current to pass from one point to another and voltage is like an influencer parameter that can control the output current quickly and easily. Since the power loss of BJT is greater than that of Jfet therefore BJT dissipates more heat than JFET. For your better understanding please do watch the video given below. For your better understanding, I had inserted one video which will make your concept more strong.

**VIDEO:-**

**Previous Post**

## Comments

## Post a Comment