Timeline for BJT switch vs amplifier mode
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 11, 2014 at 22:37 | vote | accept | joaocandre | ||
| Jun 11, 2014 at 21:13 | comment | added | Dave Tweed | The base current is set by the voltage on the Arduino pin divided by the combination of the two resistors, R2 plus the value of R1 multiplied by the current gain of the transistor. But whether or not Vce is less than Vbe is determined by the load (the LEDs in the backlight), about which we know very little. All the datasheet tells us is that the voltage drop will be about 4.2V at the maximum current of 130 mA. If the transistor is in saturation, then its current gain will be reduced, and the base current will be higher than the nominal value. | |
| Jun 11, 2014 at 18:57 | comment | added | joaocandre | Great answer, though I still have some doubts particularly the bit about saturation. From BJT 101 there is only saturation if Vbase>Vcollector. But in this case, with the current flowing from the PWM pin, what exactly sets/limits base current? | |
| Jun 11, 2014 at 17:43 | comment | added | Olin Lathrop | You're probably right. Never overestimate the intelligence of someone posting their home-brew circuit on the internet. | |
| Jun 11, 2014 at 17:39 | comment | added | Dave Tweed | @OlinLathrop: You're giving the author of the circuit too much credit; he just copied the calculation out of the LCD datasheet, and didn't know any better about where to place the resistor in the circuit. | |
| Jun 11, 2014 at 17:35 | comment | added | Olin Lathrop | 10 Ohms is a strange value on the emitter of Q1. I took it to mean that Q1 was intended to be a controlled current sink. R1 should be higher. With 10 Ohms for R1, I'm guessing R2 was found by trial and error to yield the desired maximum LED current. Like you said though, it's not a great design regardles of which way the LED driver was intended to operate. Given the digital output the same voltage as the LED anode, I'd lose the emitter resistor and put it on the collector instead to set the LED current more predictably. Current sink works well when ctrl V is less than LED cathode V. | |
| Jun 11, 2014 at 16:39 | comment | added | Spehro 'speff' Pefhany | +1 Perfect answer. The base resistor for Q1 should also be quite a bit lower, in order to guarantee saturation. | |
| Jun 11, 2014 at 16:20 | history | answered | Dave Tweed | CC BY-SA 3.0 |