9
\$\begingroup\$

I find in reference circuits that when BJT is used as switch when it will be used in saturation mode, the load is always at the collector. For NPN the emitter is connected to the ground, for PNP the emitter is connected to power supply like this:

enter image description here

  1. Why is the load always at the collector and not the other way around?
  2. Since the transistor is only acting as a switch can one also use an FET instead of BJT?
  3. if one is using BJT for multiplexing multiple 7-segment displays, the current of all the 7 segments shall pass through a transistor. So, when using discrete transistor per 7-segment unit in saturation mode, will the different current gain values of the different transistors lead to difference in brightness of the 7-segment displays?
\$\endgroup\$
  • 3
    \$\begingroup\$ Because that's the way you can switch the transistor on hardest, delivering virtually all Vcc to the load. And ... yes. And ... no, because you want to saturate the transistor. \$\endgroup\$ – Brian Drummond Nov 12 '16 at 19:26
  • 1
    \$\begingroup\$ So, it is possible to use it the othe around but there will be a huge voltage drop across the transistor collector-emitter junction which is what we are preventing? \$\endgroup\$ – quantum231 Nov 12 '16 at 19:29
  • 2
    \$\begingroup\$ I sense that there is a X-Y problem, how to drive a 7-segment LED under some voltage conditions. \$\endgroup\$ – Ale..chenski Nov 12 '16 at 20:40
5
\$\begingroup\$

It is not necessary to use a grounded emitter, but consider the alternative

schematic

simulate this circuit – Schematic created using CircuitLab

A transistor used as a switch (in saturation) will typically have a collector-emitter voltage of about 0.2 volts. Since the base-emitter voltage will be about 0.7 volts, Vs must be at least 0.5 volts above Vcc, plus whatever voltage is required across R2 to get the base current up to the level required. And that base current will be significant. Regardless of "ordinary" gain, an NPN transistor in saturation will display a much lower gain, with the typical rule of thumb being a gain of 10 to ensure low Vce. So the circuit as shown cannot be used without a second, higher power supply, which is not what you'd call convenient.

This, in turn, answers your third question. Since the transistor will be (by normal, linear standards) grossly overdriven, gain variations among transistors will typically have no obvious effect. In the circuit shown, a 50% voltage increase will cause the transistor voltage to increase from 0.2 volts to 0.3 volts, which will drop the load voltage from 4.8 to 4.7 volts, and for displays and LEDs and such this will be unnoticeable.

As to question 2, the answer is definitely yes. In many respects FETs and MOSFETs are easier to drive, since they require very little gate current (except during transitions). And, in fact, CMOS is the dominant technology for microprocessors and graphic chips, with potentially millions of transistors per chip. Well, actually, high-end CPUs and graphics ICs nowadays run between 1 and 2 billion transistors. Trying to do this with BJTs would simply be impossible due to the current requirements.

\$\endgroup\$
8
\$\begingroup\$

A simple reason to have the load on the collector is that it keeps the base current independent of the load. That makes it much easier to reliably keep the transistor saturated.

If the load is on the emitter, then the base current depends on the load. If the load is an LED, then the voltage you have to apply to the transistor base to reach the needed current goes up by the forward voltage of the LED.

If the load is a motor and it is connected to the emitter, then the base current depends on the motor, and will vary all over the place as the motor turns.

\$\endgroup\$
3
\$\begingroup\$
  1. Not always. There are circuits called "emitter follower". They don't amplify voltage, but they do amplify input current.

  2. Yes, for switching purposes FETs are used as well, n-channel for low-side switch, and p-channel for high-side switch.

  3. If you make a BJT into saturation mode, different current gains do not matter as long as you supply enough base current to keep the transistor in saturation for the lowest manufacturer's specified gain.

If you drive a 7-segment LED display, you don't control current by controlling the transistor. You control the current/brightness by using a calculated current-limiting resistor, and pulse-width modulation of saturated switches. This approach eliminates transistor's variability.

\$\endgroup\$
  • \$\begingroup\$ If I use the BJT to switch a 7 segment display, the brightness will be controlled by the collector current through the transistor. Are you trying to say that all transistors shall have same collector current in saturation.? \$\endgroup\$ – quantum231 Nov 12 '16 at 19:56
  • \$\begingroup\$ When transistor is used as switch in saturation mode, the load always seems to be at the collector. I am aware of the emitter follower configuaration. That is why I am confused and asked why not put load in emitter when using transistor as switch, \$\endgroup\$ – quantum231 Nov 12 '16 at 19:57
  • \$\begingroup\$ If your control signal has enough voltage swing, you can use the emitter follower to drive high side of 7-segment matrix with no issues. Or low side with pnp. But usually you have low-voltage control signal, and LEDs might need higher rail to operate efficently, so you have to use voltage-ampilfying circuits. \$\endgroup\$ – Ale..chenski Nov 12 '16 at 20:35
  • \$\begingroup\$ If you drive a 7-segment LED display, you don't control current by controlling the transistor. You control the current/brightness by using a calculated current-limiting resistor, and pulse-width modulation of saturated switches. \$\endgroup\$ – Ale..chenski Nov 12 '16 at 20:51
3
\$\begingroup\$

There are many cases where the load is better placed in the emitter. For example:

schematic

simulate this circuit – Schematic created using CircuitLab

Here a multiplexed set of LEDs is driven by emitter followers for the high side drivers. (with an 8-digit 7-segment + DP display you would have 8 high side, 8 low side and 8 resistors in series with the latter) There are no base resistors necessary, saving space and parts.

Or here:

schematic

simulate this circuit

Here a logic gate directly drives a 4.5VDC relay coil with no additional components required.

You don't get voltage gain with an emitter follower but you do get current gain, without inversion, and sometimes that's exactly what is required.

Emitter followers generally don't allow the transistor to saturate (it's possible by driving the base with a higher voltage than the collector, and adding a base resistor, but can't happen if the base is driven from the same voltage or less than the collector.

This means at least 0.6V drop across the transistor, which is not always all that bad, and because the transistor does not saturate it switches faster. Common emitter switch circuits can push the transistor deep into saturation, with maybe 1/10 the Vce, which minimizes heating.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.