I am design a transistor switch circuit.

Ic = 1mA-30mA Ib = 1mA Vce = 7V-18V

This is the transistor which I am going to use as an NPN Switch

Can someone tell me how to read the Fig 7 graph. In the graph, the gain value is fixed at 20. But according to my application, Max Ic/Ib = 30. How to take values from that graph. I just want to understand and ensure that my transistor will act only in the cut-off and saturation region.

Or in other words, In the below attached image, I would like to understand the gain values within the highlighted boxes. In other words, What would the gain range in the Cut-off region, Saturation region and active region. Can someone help me understand the gain with respect to the transistor regions.


Thank you.

  • 1
    \$\begingroup\$ Nothing inside of any of the boxes are "gain values". A gain is a ratio of two numbers. That graph is a family of curves that relate collector current to the collector-emitter voltage for a number of different base currents. What do you really want to know? \$\endgroup\$
    – TimWescott
    Aug 29, 2019 at 3:10
  • \$\begingroup\$ I want to know the gain value of the selected transistor with respect to my application mentioned at the start \$\endgroup\$
    – user220456
    Aug 29, 2019 at 3:35
  • \$\begingroup\$ which of the 3 regions do you wish to use, for your switch? have you thought about using a differential pair, using 2 transistors? \$\endgroup\$ Aug 29, 2019 at 3:55
  • \$\begingroup\$ Folks, I think the OP is a student and is supposed to work out \$\beta\$ (called 'gain' by the OP) from the curves as shown. For example, right at the marked Q-point on the graph (and with nearby varying \$V_\text{CE}\$) one might calculate \$\beta=\frac{40\:\text{mA}}{60\:\mu\text{A}}\approx 670\$. Other \$I_\text{B}\$ lines extrapolated back to the y-axis may (not sure, but may) provide differing values for \$\beta\$. Of course, that's only my guess. I've no real clues. \$\endgroup\$
    – jonk
    Aug 29, 2019 at 4:05
  • 1
    \$\begingroup\$ I suggest you edit your question to not use the word "gain", because you do not seem to understand it. Or, as part of your question, ask what values of gain may be extracted from that graph (like \$\beta\$, as @jonk pointed out). \$\endgroup\$
    – TimWescott
    Aug 29, 2019 at 5:01

1 Answer 1


I am design a transistor switch circuit.

OK, that then means that we are going to operate the NPN in saturation mode.

Why not in Active mode?

Because we want a switch, in active mode the transistor doesn't act as a switch it acts more like a variable current source.

In active mode we use the transistor's high current gain \$\beta = \frac {I_c}{I_b}\$. That then means that \$I_b\$ controls \$I_c\$. But we don't want "control", we want a switch so: on / off that's it.

That's where saturation mode comes in. In saturation mode we simply make \$I_b\$ so large that the \$I_c = \beta * I_b\$ becomes much larger than the actual \$I_c\$ that is flowing.

What then determines \$I_c\$? In saturation mode I mean.

The \$I_c\$ in saturation mode is then determined by the load. Your load wants 30 mA to flow so we need to make sure that we apply enough \$I_b\$ to the transistor so that the 30 mA can easily flow.

In active mode to make 30 mA flow when \$\beta\$ = 100 we'd need \$I_b\$ = 0.3 mA. But as I mention above, to make sure we're in saturation mode we need to apply a much higher \$I_b\$ than that!

How much more, well that is a choice, in the datasheet of the PDTC123J, figure 7 they chose \$\frac {I_c}{I_b}\$ = 20. I emphasize chose because it is a choice, they could also have chosen \$\frac {I_c}{I_b}\$ = 30. As long as the value is significantly smaller than \$\beta\$ (which is a factor 100 or more for this transistor) the transistor will be in saturation mode.

So for your load's 30 mA and chosing \$\frac {I_c}{I_b}\$ = 20 that would mean you need to make

\$I_b = \frac{I_c}{20}\$ = 30 mA / 20 = 1.5 mA

If you would chose \$\frac {I_c}{I_b}\$ = 30 then you would need \$I_b\$ = 1 mA and the transistor would be a little less deep into saturation. That would mean there would be slightly more voltage drop across the transistor when it is switched on.


You asked about Figure 6 which shows the DC current gain as a function of collector current

This plot is relevant as it shows what the transistor's \$\beta\$ is for a certain collector current. For your application we know that the collector current is 30 mA. From Figure 6 we can then see that \$\beta\$ (the plot shows \$h_{fe}\$ which is just a different name for \$\beta\$) is always more than a about 150.

As described above, for saturation mode we need \$\frac {I_c}{I_b} << \beta\$ and that condition is met in my examples above.

Note that you need a very small value of \$I_c\$ for \$\beta\$ to become much smaller. Experienced engineers know this so ignore the plot and just use the value of \$\beta\$ from the tables.

  • \$\begingroup\$ Thank you for the detailed answer. But all the calculations that you have made and arrived it, it was done without taking any values from that graph Fig.6, Gain vs Collector Current. What can I infer from that graph for my calculations? Basically, we have arrived everything without looking at the graph. So, does the graph serve no purpose? If so, how and when? \$\endgroup\$
    – user220456
    Aug 29, 2019 at 17:42
  • \$\begingroup\$ sweet answer, guy \$\endgroup\$ Aug 29, 2019 at 18:15
  • \$\begingroup\$ @Newbie I added a paragraph to my answer about your Figure 6 question. \$\endgroup\$ Aug 30, 2019 at 9:33
  • \$\begingroup\$ Thank you very much. Understood now \$\endgroup\$
    – user220456
    Aug 30, 2019 at 11:20

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