power consumption trade-off that you get to make as the engineer. That reduces your power consumption, but also reduces the logic speed as the smaller currents are able to charge the parasitic capacitances less rapidly. You can also scale the resistor values up, maintaining the ratio of base to collector current, but reducing the current overall. 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.Įvery transistor has a current gain, usually \$\beta\$ or \$h_ = 13 \mathrm k \Omega $$ I just want to make sure that I purchase the right resistors for the job. Like I said, it is working at the moment. 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. If you calculate the current for the base input with Ohm's Law, we get I = 5 / 10000 = 0.0005. I'm having a hard time understanding what that means exactly. It states that the Base-Emitter Saturation Voltage has the following specs: Simulate this circuit – Schematic created using CircuitLabĪlthough this works fine, I'm a bit concerned about what I've read on the data sheet for the 2N3904. 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: These transistors may come with straight or preformed leads.I have a ton of 2N3904 transistors and would like to use them for my RTL logic project.If you need a little beefier NPN transistor, look at our PN2222 which can support up to 600mA and if you need more than that, check out our TIP120 Darlington transistor which can handle up to 5A. Using two 100 ohm 1/4W resistors in parallel would be another approach. Power through the that resistor is P = 60mA * 3V = 180mW so a 1/4W resistor will work, though a 1/2W would be a better choice for long-term use. The example assumes an LED voltage drop of 2V, so R = (5V – Vled) / 60mA = 50 ohms. This value may vary a bit depending on what LEDs are being used. The 50 ohm resistor sets the overall current for the 3 LEDs. The 4.7K resistor on the base is there to limit the current out of the uC pin to safe levels and the value is not overly critical. This simple circuit allows you to drive all three LEDs off the same uC pin. ![]() You could use 3 pins on the uC to drive the 3 LEDs, but that is a waste of pins unless you have pins to spare. The indicator consists of 3 LEDs which each draw 20mA of current.Ī uC pin can easily drive an LED at 20mA, but this circuit will require 60mA which is outside the range of the typical uC drive capability. We have an LED indicator that we want to drive. NPN transistors are generally used in low-side switching applications where they are connected between the load and ground. These can be very handy when you need to boost the output of a uC pin to drive something that it can not drive directly. The 2N3904 are a good inexpensive general purpose NPN transistor for low power amplification and switching and should be part of every parts bin. Up to 40V collector-emitter and 60V collector-base.KEY FEATURES OF 2N3904 NPN GENERAL PURPOSE TRANSISTOR: Qty 10 – 2N3904 NPN General Purpose Transistors.The 2N3904 are a good general purpose NPN transistor for low power amplification and switching.
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