Electronic – Load line and voltage or current gain

bjtgain

Why does a steeper load line gives higher current gain, while a flatter line gives higher voltage gain? I read multiple tutorials where they only provide a graph without explanation and I can't visualize why it is.

For the upper load line, if a small or input signal fluctuates on a DC-biased base current (e.g. 40uA in the image) at a fixed operating voltage of \$V_{CE}\$, then the output amplified current, or \$I_C\$ fluctuates around 5mA. And for the lower load line, if small signal rides on the 20uA base current, output collector current fluctuates around 3mA with more of less the same amount. In either case, I don't see where \$ \partial I_c \over \partial I_b\$ differs for different load lines. Same question for the voltage gain.
http://hyperphysics.phy-astr.gsu.edu/hbase/Electronic/loadline.html

[EDIT]

Is that what it means by saying a steeper line gives higher current gain?

load line sample

Best Answer

This can be easily seen from graph below: enter image description here

There are two load lines: Green with smaller slope and Red with higher slope.
For a given change in \$V_{be}\$, lower slope (green line) will give a corresponding higher change in \$V_{ce}\$, simply because it has less slope. Hence voltage gain is higher, as it is just the ratio \$\frac{\Delta V_{ce}}{\Delta V_{be}}\$.
By the same reasoning, a line with higher slope (red line) will give more current gain simply because change in the output current \$I_c\$ will be more for higher slope.
This can also be seen analytically as follows: $$A_v = g_m(R_c||r_{out})$$ Here, \$r_{out}\$ is the output impedance of the transistor (along with its degeneration). Since, \$r_{out}\$ is very high, $$A_v \approx g_mR_c$$ Thus, higher \$R_c\$ implies higher voltage gain. $$A_i = \beta \frac{r_{out}}{r_{out}+R_c}$$ Clearly, if \$R_{c}\$ is very small, $$A_i \approx \beta,$$ which is the highest current gain.

In response to LvW's Comment
Assuming, \$R_E\$ is zero, the following is the relation between change in \$V_{be}\$ and \$I_b\$: $$v_b = \frac{(\beta+1)i_b}{g_m} \implies i_b = \frac{g_mv_b}{(\beta+1)}$$ Thus, change in base current is proportional to the change in base emitter voltage. Thus, the current gain can be compared just by the change in collector current since the change in base current is the same for the two lines.

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