Timeline for Calculating collector current of NPN BJT with LED & resistor
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 20, 2018 at 8:55 | history | edited | Michel Keijzers | CC BY-SA 4.0 |
added 1 character in body; edited title
|
| Oct 20, 2018 at 8:53 | comment | added | D.A.S. | For 5mm, 10mA use 2V for UB Red, Yellow 5mm and 3V for W/ B LEDs which rises slightly with current due to 15 Ohm Rinternal which has a wide tolerance | |
| Oct 20, 2018 at 4:18 | comment | added | TimWescott | Lots of good comments. Something that I'm not seeing is that the transistor can't generate energy, it can only shove it around. So unless it's driven there by an outside source, the collector will never go below ground. 20mA through a 330-ohm resistor would cause a 6.6V drop -- that would put your collector at -1.6V, which is impossible. If you haven't figured it out yet, the key word for the newbie is saturation; it's a must to understand how it works in BJTs. | |
| Oct 20, 2018 at 3:56 | comment | added | jonk | But as others point out, probably better still to use a smaller saturation \$\beta\approx 10-15\$, I think. If you want some more explanation about putting a BJT into saturation, here is one link to read up on. This may help a lot in understanding what to do when trying to use a BJT as a switch, like this. | |
| Oct 20, 2018 at 3:54 | comment | added | jonk | \$\beta=\frac{6\:\text{mA}}{\frac{5\:\text{V}-700\:\text{mV}}{21.5\:\text{k}\Omega}}\approx 30\$ from that circuit, given your measurements. This means the BJT is likely in light saturation. This suggests that your LED has \$5\:\text{V}-2\:\text{V}-400\:\text{mV}\approx 2.6\:\text{V}\$ when running at \$6\:\text{mA}\$. I'd tend to go with \$3\:\text{V}\$ as an estimate for the desired drop across the LED. If you want to stick with the \$\beta=30\$, you should use about \$80\:\Omega\$ instead of \$330\:\Omega\$ and thereby also reduce \$21.5\: \text{k}\Omega\$ to \$5.6\:\text{k}\Omega\$. | |
| Oct 20, 2018 at 3:22 | comment | added | analogsystemsrf | Classic design method for good saturation is to assume Beta/10, or 10 for this circuit/transistor. Use a 2.2Kohm in the base. | |
| Oct 20, 2018 at 2:11 | comment | added | Andrew Debusiak | We weren't able to use the diode checker on our DMM to read the forward voltage of the LED so we were told to use 0.3V. I'm going to assume it's more around 3V, thanks for the help | |
| Oct 20, 2018 at 1:55 | comment | added | Transistor | Yes, the forward voltage should be more like 3 V for blue. Are you sure you didn't misread the saturation voltage (0.2 to 0.3 V) as the LED forward voltage? To turn the transistor hard into saturation you need to beef up the base current a lot more than the gain calculation suggests. Note also that the gain calculation breaks down once the transistor is saturated. The collector current is determined then only by the supply voltage and load. Stick at it! | |
| Oct 20, 2018 at 1:50 | review | First posts | |||
| Oct 20, 2018 at 5:04 | |||||
| Oct 20, 2018 at 1:48 | history | asked | Andrew Debusiak | CC BY-SA 4.0 |