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I have a ton of 2N3904 transistors and would like to use them for my RTL logic project. Based on what I could figure out on the web, and the parts I had, I've gotten logic gates to work quite well with the following values:

schematic

simulate this circuit – Schematic created using CircuitLab

Although this works fine, I'm a bit concerned about what I've read on the data sheet for the 2N3904. It states that the Base-Emitter Saturation Voltage has the following specs:

Ic = 10mA Ib = 1.0mA Ic = 50mA Ib = 5.0mA

I'm having a hard time understanding what that means exactly. If you calculate the current for the base input with Ohm's Law, we get I = 5 / 10000 = 0.0005. Am I correct that this is 5mA? I replaced R2 with a 5K resistor and it switched the same, which would be 0.001 or 10mA.

Like I said, it is working at the moment. I just want to make sure that I purchase the right resistors for the job. I know that the goal is for the transistor to be fully saturated, however I don't know if this is how that is done or not.

Thanks,

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  • \$\begingroup\$ wouldn't one miliAmp be 1/1000A or 1e-3A? If so, then it is not 10mA as you have write, but 1mA. and also is not 5mA but 0,5mA or 500uA. agree? \$\endgroup\$ – Cristiano Dec 18 '16 at 16:51
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Every transistor has a current gain, usually \$\beta\$ or \$h_{fe}\$ in the datasheet. Typical values are on the order of 100. When the transistor is not saturated, then the base current and collector current are related by this factor:

$$ I_c = h_{fe} I_b $$

When the base current increases to the point where collector current can increase no more, the transistor is said to be saturated. The collector current can increase no more because it can't permit any more current -- the current is entirely limited by R1 in your diagram, and the voltage from emitter to collector is at a minimum.

When we design digital logic, we don't want to just barely saturate the transistors. We want to saturate them a lot. This provides some extra margin against variations in \$h_{fe}\$, and also takes into account that for higher frequencies (necessary for quick high/low transitions), \$h_{fe}\$ is effectively reduced.

Rule of thumb: in digital logic, design for a collector current 15 times greater than the base current.

So here, you've selected a collector resistor of 1kΩ. At saturation, the emitter-collector voltage is much less than the supply voltage, so we can estimate the collector current as:

$$ I_c = \frac{5\mathrm V}{1\mathrm k\Omega} = 5\mathrm{mA} $$

We want the base current to be 1/15th that (0.33mA), and the voltage across the base resistor will be the supply voltage, less about 0.65V from the base-emitter junction of Q1. So:

$$ R_2 = \frac{5\mathrm V - 0.65 \mathrm V}{0.33\mathrm{mA}} = 13 \mathrm k \Omega $$

Your selection of 10kΩ is close enough.

You can also scale the resistor values up, maintaining the ratio of base to collector current, but reducing the current overall. That reduces your power consumption, but also reduces the logic speed as the smaller currents are able to charge the parasitic capacitances less rapidly. This is a performance vs. power consumption trade-off that you get to make as the engineer.

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  • \$\begingroup\$ So, the ability to fully saturate the transistor is based more on the ratio of collector/base currents, than it would be the actual values of the transistor? In other words, I could probably use several kinds of transistors with a 1K (collector) 13K (base) ratio? \$\endgroup\$ – JohnnyStarr Jul 11 '14 at 21:04
  • \$\begingroup\$ How does this approach guarantee the transistor will be in saturation if you are designing only for a gain of 15? \$\endgroup\$ – sherrellbc Jul 12 '14 at 0:39
  • \$\begingroup\$ @sherrellbc because the gain of any transistor you are likely to use will have a gain of much more than 15. \$\endgroup\$ – Phil Frost Jul 12 '14 at 0:40
  • \$\begingroup\$ @JohnnyStarr it does depend on the gain of the transistor, but a typical, general-purpose, small-signal BJT (like 2N3904, BC547, 2N2222, etc.) will have a current gain on the order of 100. It actually varies by a wide margin even between specimens of the same part number, so the approach is to anticipate a gain that is surely less than the gain you will get so that you don't even run into a problem where your logic doesn't saturate the transistor. \$\endgroup\$ – Phil Frost Jul 12 '14 at 0:43
  • \$\begingroup\$ So then since the gain is likely much higher you would expect the transistor to saturate far earlier than with a base current of 330uA? The idea seems solid but that base current seems really small. If we are assuming the transistor can saturate at 1mA collector current then a rough estimate would but the associated base current at 10uA (~gain of 100)? And since we are designing for 330uA base current, saturation is guaranteed. \$\endgroup\$ – sherrellbc Jul 12 '14 at 0:44

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