BJT switch vs amplifier mode

Question

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?

Answer 1

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 V_EE 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.

Answer 2

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.

Answer 3

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?