# BJT switch vs amplifier mode

I'm getting really confused about a specific BJT circuit. I want to control LCD brightness and contrast as suggested here.

Apprently a capacitor is used to filter PWM output to the contrast pin, and an emitter-follower BJT topology is used to control brightness. However there are some aspects of this particular implementation that I've yet to understand:

1. Why isn't a simple capacitor is used in the brightness pin, similarly to the contrast pin?
2. In the emitter-follower, is the BJT operating in linear or saturation zones?
3. Why does the author discard the base emitter voltage (0.6V) in his calculations?
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Ah, the joys of finding a random circuit on the Internet that happens to have a slick presentation. This is actually a very poor design for several reasons.

First of all, the author should have put a resistor between D10 and the capacitor, which would have allowed him to use a much smaller capacitor to get the cutoff frequency he needed. The VEE pin of an LCD requires only a tiny amount of current. As it is, he's relying on the output impedance of the Arduino pin to limit the current into/out of the capacitor, which is very poor practice.

Secondly, the transistor is being used in a common-emitter mode, not emitter-follower. Using two resistors the way he does doesn't make much sense.

Why isn't a simple capacitor is used in the brightness pin, similarly to the contrast pin?

Two reasons:

1. The Arduino pin by itself can't handle the current required by the backlight. The transistor provides the necessary current gain.

2. In this case, the goal isn't to turn the PWM signal into a DC level, but instead, to use it to turn the backlight LED on and off rapidly to change the apparent brightness.

In the emitter-follower, is the BJT operating in linear or saturation zones?

Like I said, this isn't an emitter-follower. Because of the resistor in the emitter leg, however, it's operating on the cusp between linear and saturated zones.

Why does the author discard the base emitter voltage (0.6V) in his calculations?

I assume that when you say "discard", you mean "ignore". Good question, although it would really be the collector-emitter voltage in this situation. If he was using a circuit configuration in which the transistor would definitely be in saturation, this voltage would be relatively small (about 0.3V), but still significant.

The circuit would be better if the emitter of the transistor were connected directly to ground, and the resistor R1 were placed in the path between Q1's collector and the LED- pin of the display.

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+1 Perfect answer. The base resistor for Q1 should also be quite a bit lower, in order to guarantee saturation. – Spehro Pefhany Jun 11 '14 at 16:39
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. – Olin Lathrop Jun 11 '14 at 17:35
@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. – Dave Tweed Jun 11 '14 at 17:39
You're probably right. Never overestimate the intelligence of someone posting their home-brew circuit on the internet. – Olin Lathrop Jun 11 '14 at 17:43
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. – Dave Tweed Jun 11 '14 at 21:13

The contrast pin of LCD displays is usually a fairly high impedance node. It could be driven directly from a microcontroller PWM output, which is then low pass filtered with a resistor in series and a capacitor to ground. 100 µF as shown in your schematic seems extreme for this purpose, but we don't know what exactly is driving the D10 point.

A transistor is used to controll the "brightness" current, which is apparently really a LED backlight. That can take significant current, so just R-C low pass filtering a digital output isn't good enough. In this case Q1 and R1 appear to be used as a controlled current sink, although the addition of R2 is a dubious choice.

LEDs can be reasonably brightnes-modulated with PWM. If fast enough, like a few 100 Hz, your eyes will perceive the average brightness well enough. This simplifies the driving electronics which only needs to supply full current or no current. The duty cycle of the pulses then determines the perceived brightness. Low pass filtering these pulses, as a capacitor would do, would partially defeat the purpose by causes dissipation in the driving electronics. Pure switches don't dissipate power. Power is voltage times current. When the switch is open, the current thru it is zero. When the switch is closed, the voltage accross it is zero.

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Then, if I understood correctly, Q1 is switching between ON and OFF (and it does seems to me that the base-colector junction is forward biased). Then what is the purpose of Q1 vs the PWM input directly? – joaocandre Jun 11 '14 at 16:08
@joao: As I said, Q1 and R1 work together as a controllable current sink. The schematic is poorly drawn with Q1 upside down. It might help to draw it out more normally. The collector of Q1 goes to the cathode of the backlight LED, with the anode tied to 5 V. When Q1 is turned on, it lets a particular amount of current flow out of the LED cathode. – Olin Lathrop Jun 11 '14 at 17:27

I would suspect that the capacitor would not work as well due the the voltage drop on the LED for the back-light. Think of it like this: The dimming is achieved with the bjt by turning on and off the led at different rates while the caps method is to lower the voltage thus reducing the current. Now with the cap when you lower the pwm duty cycle the max and min voltage on the cap will lower and thus the brightness will drop, but I simply don't think you will achieve a large range of possible dim levels because the voltage drop of the diode will limit the capacitor method...The max voltage on the cap has to be enough to turn it on. Were as with the bjt it will have the full range of the pwm. As for the "discard the base emitter voltage (0.6V) in his calculations" I am not sure which calculation you are revering to on the site..Choosing R2?

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