The Fairchild 2N3904 has a long data sheet so let's use that for this exercise. I'm interested in using an NPN in common emitter configuration --- it's for switching. It's my understanding that an NPN transistor is in saturation when collector-emitter voltage is equal to \$V_{CE(sat)}\$, which is about 0.25 V here. Say our load is simply a resistor, so I understand how to calculate \$I_{C(sat)}\$. My question is, how do I thence compute \$I_{B(sat)}\$? Do I just use the DC current gain (\$h_{FE}\$)?

Let's suppose for example that my \$I_{C(sat)}\$ = 90 mA. According to the data sheet would a conservative estimate for gain be \$h_{FE}\$ = 30? I say conservative because I want to choose a resistor \$R_B\$ such that my TTL signal is sure to drive the transistor into saturation. That is, my TTL signal will generate \$I_{B(sat)}\$.

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    \$\begingroup\$ You might want to read answers to this question and then let us know if you still have questions. \$\endgroup\$ – The Photon Jan 15 '15 at 21:45
  • \$\begingroup\$ Key point: Saturation happens when the external circuit can't supply enough current to maintain the linear forward-operating regime. \$\endgroup\$ – The Photon Jan 15 '15 at 21:47
  • \$\begingroup\$ If you look at Figure 3 of the datasheet, they are showing saturation as when \$\beta\$ = 10. So, \$I_B \cdot 10 = I_C\$. \$\endgroup\$ – caveman Jan 15 '15 at 21:52

Normally the saturation voltage is specified at Ic/Ib = 10, so that's what is guaranteed. Some transistors have it specified at different forced betas. Note that you're not computing it, you are specifying it as part of the design.

If you're not too concerned about how 'turned on' the transistor is, you can use \$h_{FE}\$, which is defined with \$V_{VE}\$ at 1.0Von this datasheet. It's also temperature sensitive so even if you use the minimum number, you should take temperature variations into account. That is not normally what you want to do when you are using the transistor as a switch. You want it turned on well and good, so it doesn't cause a malfunction or damage itself.

It's usually fairly safe to use Ic/Ib = 20 (on a 2N3904, at, say 20mA, or a 2N4401 at 100mA) if you don't mind a slightly higher \$V_{CE}\$ than with Ic/Ib = 10. If you're using a crummier transistor or if you're up near the high end of the current range, or it might get really cold etc. you may have to adjust that rule of thumb.

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    \$\begingroup\$ One point worth adding: That "slightly higher Vce" with Ic/Ib=20 is probably still < 0.2V. It's just not guaranteed to be < 0.25V across the temperature range for every batch of 2N3904s ever made. So... your choice. \$\endgroup\$ – Brian Drummond Jan 15 '15 at 23:06

If you go to your data sheet, and look at p. 3, you'll see a section called ON Characteristics, and there is a field for Vce(sat). It gives 2 examples, in both cases with a collector/base current ratio of 10. Likewise, figures 2 and 3 show saturation characteristics, and in both cases the upper left part of the figure specifies a beta of 10. Beta is also called hfe, and is (roughly) the ratio of collector to base current.

You can find a quick description of what transistor saturation means here http://en.wikipedia.org/wiki/Bipolar_junction_transistor. There is no hard-and-fast rule concerning the current levels which will mark the transition from active to saturation, (it varies with collector current and the transistor) but the rule of thumb is that a beta of 10 will always get you there. In fact, a beta of 10 will generally get you into hard saturation, where small changes in base current will produce small changes in Vce.

So the more detailed answer to your question, taking into account your supplementary information is this - First, calculate how much current you need to drive: for a TTL circuit, you need 1.6 mA per gate, or if you're looking at a resistor pulled up to 5 volts it's obviously about 5/R. You pretty much can't go wrong with a base drive 1/10 your specified load current. In your case, a 9 mA base drive will do the trick. You can do this in TTL by providing a pullup resistor of about 300 ohms and a base drive resistor of about 200 ohms, and this is (just) within the output capabilities of TTL. Note that, from figure 2 of the data sheet, this will typically produce a saturation voltage of ~0.15 volts. If you can live with a slightly higher output voltage you can provide slightly less base drive, and extrapolating from the data sheet (always dangerous) your worst-case output voltage with a 9 mA / 90 mA combination ought to be under 0.4 volts.


My question is, how do I thence compute \$I_{B}(\mathrm{sat})\$?

The same way you do for forward-active. Assume the b-e junction has 0.6 or 0.7 V, and let the bias circuit determine the current.

If there's no current-limiting element in the bias circuit,

$$I_b = I_s\exp\left(\frac{qV_{be}}{nRT}-1\right)$$

  • \$\begingroup\$ Thanks to the Internet's anonymity I feel comfortable stating that this answer means little to me (though I appreciate your time). What are the terms in this equation? What's a bias circuit? I was just thinking of a circuit like this: wps.prenhall.com/wps/media/objects/416/426356/19fig1.gif. \$\endgroup\$ – thehandoftheking Jan 19 '15 at 18:35
  • \$\begingroup\$ The bias circuit is the circuit that sets the bias point of the transistor. It's the circuit connected to the transistor, evaluated as a DC circuit. \$\endgroup\$ – The Photon Jan 19 '15 at 18:57
  • \$\begingroup\$ For the equation, see the Wikipedia article on diodes, section on the Shockley diode equation. (They use \$V_T\$ in place of \$RT/q\$ but it's the same thing) \$\endgroup\$ – The Photon Jan 19 '15 at 18:58

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