Why is transconductance defined so?

Question

We know that transconductance is defined as:

$$g_m=\frac{\Delta i_{\text{output}}}{\Delta V_{\text{input}}}$$

My question is: why is such a quantity as transconductance defined in the way it is? How is it helpful to electrical engineers?

I ask this because I am having difficulty memorizing this formula. I always mess it up as \$g_m=\frac{\Delta i_{\text{input}}}{\Delta V_{\text{output}}}\$ instead of the correct one. I know conductance is inverse of resistance, so the current will always be in the numerator and voltage in denominator. But, I am not sure which one is output and which one is input. Perhaps, if I know the logic of this definition, I'll be able to recall better.

Sorry if this reason is apparent to you, but I am a beginner, and it is not at all apparent to me, so kindly explain in simple language. Thank you!

Note: Please restrict your answer to a Bipolar Junction Transistor only.

Answer 1

The ‘trans’ bit is short for ‘transfer’, and the transfer referred to is from input to output.

So think of transconductance as a transfer function, which is always in the form \$\frac{output}{input}\$, and ‘conductance’ is \$\frac{current}{voltage}\$, hence \$g_m=\frac{output\:current}{input\:voltage}\$

Answer 2

Transconductance by itself isn't related to inputs or outputs.

Transconductance is simply 1/resistance as you write.

In circuit design we can use transconductance or gm when there is a voltage to current relation present somewhere.

Only when that transconductance is present in a certain circuit (like the model of a BJT) can we talk about input and output.

Also input and output refers to the direction of a signal flow.

I can design a circuit where the signal goes in at the output of a transistor's gm and comes out at the input. Example: the input of an NPN current mirror.

So your "wrongly" remembered gm can be correct depending on the circuit!!!

Answer 3

It's the inverse of transresistance which is an incremental value, not just V/I

With an input voltage or current control and a gain of transresistance and a fixed drain or collector R you get gain.

For BJT's current gain depends on Iin which controls the gain in gm=.

$$g_m=\frac{\Delta i_{\text{input}}}{\Delta V_{\text{output}}}$$ But with no emitter resistor, the BJT becomes a VCCS.

For FET's voltage gain depends on Vin which controls the gain in gm.

$$g_m=\frac{\Delta i_{\text{output}}}{\Delta V_{\text{input}}}$$

It is the simplest way to define the gain or transfer function at a bias point.

How else would you define it?

Answer 4

I know you want to focus on the BJT.

For the BJT, the value of \$g_m\$ tells you immediately the highest possible voltage gain you can achieve by placing a resistor into the collector leg (assuming of course that the power supply rail voltages and other circuit details permit it.) And for the BJT, unlike a MOSFET, it's not based on construction and instead depends only on operating circumstances easily controlled by the designer using discrete parts.

When using the term \$g_m\$ for the BJT, it is a parameter that depends only on the dc collector current \$I_\text{C}\$: \$g_m=\frac{I_\text{C}}{V_T}\$. (This excludes the emission coefficient, \$n\$, which is almost always just 1 for BJTs.) \$V_T=\frac{k\: T}{q}\$ is a physical parameter based upon the equi-partition law of energy and the application of large population statistics. \$V_T\$ is not adjustable or controllable except by changing the temperature of the device.

(\$g_m\$ applies in active mode for the BJT, not saturated. This is because the collector operates like a current source (sink for NPN) in active mode, but more like a voltage source in saturation.)

For example, the highest voltage gain from a BJT operating in common emitter mode will be \$A_\text{V}=-g_m\cdot R_\text{C}\$. This can be degenerated by inserting an emitter resistor (and often is.) But that's the highest you can expect from a BJT.

The term is quite useful, though. For example, if you intend on operating the BJT at a quiescent collector current of about \$1\:\text{mA}\$, then you know that your maximum possible voltage gain using a common emitter configuration will be \$\approx 39\cdot R_\text{C}\$, if \$R_\text{C}\$ is expressed in thousands of Ohms (and the voltage supply can handle the drop across \$R_\text{C}\$.)

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For BJTs, \$g_m\$ only depends on the DC collector current, \$I_\text{C}\$. \$g_m\$ doesn't care about the construction geometry of the BJT (unlike the MOSFET.) (Except as the base-emitter junction area affects parameters that impact the total collector current -- but it is still about the collector current, even then.) In the BJT in active mode, excepting modifications like the Early Effect, \$I_\text{C}\$ is determined by \$V_\text{BE}\$. By comparison, the \$g_m\$ of a MOSFET depends on \$I_\text{D}\$, \$V_{OV}\$, and the ratio, \$\frac{W}{L}\$. So there are three different formulas used to express \$g_m\$ for MOSFETs. (One probably more comparable as apples to apples with the BJT, than the other two.) The \$g_m\$ of MOSFETs is generally considered to be smaller than for the BJT. (Because \$V_{OV}\$ is so much larger than \$V_T\$.)

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You can compute \$g_m\$ for the active mode BJT:

$$\begin{align*} I_C &= I_{SAT}\cdot\left(e^{V_\text{BE} \over V_T}-1\right)\\\\ \text{D}\left(I_C\right) &= \text{D}\left(I_{SAT}\cdot\left(e^{V_\text{BE} \over V_T}-1\right)\right)\\\\ \text{d}\:I_C &= I_{SAT}\cdot\text{D}\left(e^{V_\text{BE} \over V_T}-1\right)\\\\ \text{d}\:I_C &= I_{SAT}\cdot e^{V_\text{BE} \over V_T}\:\text{D}\left({V_\text{BE} \over V_T}\right)\\\\ \text{d}\:I_C &= I_{SAT}\cdot e^{V_\text{BE} \over V_T}\:{\text{d}\:V_\text{BE} \over V_T}\\\\ \text{d}\:I_C &= {I_{SAT} \over V_T}\cdot e^{V_\text{BE} \over V_T}\:{\text{d}\:V_\text{BE} }\\\\ {\text{d}\:I_C \over \text{d}\:V_\text{BE}} &= {I_{SAT} \over V_T}\cdot e^{V_\text{BE} \over V_T}={I_{SAT}\cdot\: e^{V_\text{BE} \over V_T} \over V_T} \end{align*}$$

But for almost all possible uses of the BJT:

$$I_C = I_{SAT}\cdot\left(e^{V_\text{BE} \over V_T}-1\right) \approx I_{SAT}\cdot\:e^{V_\text{BE} \over V_T}$$

Therefore:

$$g_m={\text{d}\:I_C \over \text{d}\:V_\text{BE}} \approx {I_\text{C} \over V_T}$$

Answer 5

Wikipedia:

Transconductance (for transfer conductance), also infrequently called mutual conductance, is the electrical characteristic relating the current through the output of a device to the voltage across the input of a device.

If \$I_{out}=f(V_{in})\$, then the transconductance becomes

$$g_m=\frac{\delta I_{out}}{\delta V_{in}}$$

It's simply the change in the output current due to the change in the input voltage. For a BJT working in the active region it becomes \$g_m=\frac{I_c}{V_T}\$.

In general, let's consider this parameter: \$Y_m=\frac{\delta Y_{out}}{\delta X_{in}}\$, where X and Y stand for the input current or voltage. Now we define the four possible parameters as follows:

• Transconductance: \$g_m=\frac{\delta Y_{out}}{\delta X_{in}}\$; Y=current, and X=voltage.

• Transimpedance: \$r_m=\frac{\delta Y_{out}}{\delta X_{in}}\$; Y=voltage, and X=current.

• Voltage gain: \$A_v=\frac{\delta Y_{out}}{\delta X_{in}}\$; Y and X=voltage

• Current gain: \$A_i=\frac{\delta Y_{out}}{\delta X_{in}}\$; Y and X=current

Once you determined the input and output terminals of the device you can find the above parameters. Each of these has meanings.

Usually the voltage gain most interests us. But notice this quantity can be written in terms of the transconductance: \$A_v=\frac{\delta V_{out}}{\delta V_{in}}=\frac{\delta V_{out}}{\delta I_{out}} \frac{\delta I_{out}}{\delta V_{in}}=g_m \frac{\delta V_{out}}{\delta I_{out}}\$. And similarly, for the current gain it can be written as \$A_i=\frac{\delta I_{out}}{\delta I_{in}}=\frac{\delta I_{out}}{\delta V_{out}} \frac{\delta V_{out}}{\delta I_{in}}=r_m \frac{\delta I_{out}}{\delta V_{out}}\$. So as you see \$g_m\$, which is the inverse of \$r_m\$, plays an important role in circuits. As an example, if you design an audio amplifier you are interested to see the amplification ratio, which is given by a multiple of \$g_m\$.

Answer 6

Conductance is a property of electricity while transconductance or "transfer conductance" is a circuit "device."

The property conductance, namely, how good the current is, is defined as the reciprocal of resistance, namely, how good the "obstacle to current" is and since it's a definition it means they are the same thing, i.e. conductance and the reciprocal of resistance are the same thing.

The device "transfer conductance", namely, to transfer or transform how good the current is, is the device on a circuit aimed to transfer a given voltage to a wanted current and since it's a device with the purpose of transfering a given voltage to a wanted current, it's actually the device with it's ability gm to transfer the given input voltage V_input by means of V_input times gm to get the wanted output current I_output.

Therefore, the output of the device must be current. So you must also be able to infer that its input must be voltage.