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# in my engineering club i was asked to build a driver circuit push-pull emitter follower. Answered

I have no idea what this is. I have recently started an engineering club ( i am 14) and wanted some help with my homework. I am using my older brother account because he told me to found it out myself. i have until tomorrow to find the right circuit and then build it when i go to my club using a software called multisim.

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An emitter follower is simply a way to wire up a transistor so that the the output of it acts like a voltage regulator, and the voltage seen on the gate is what the output will be. w2aew on Youtube has a lot of great videos about the basics of transistors, and I have learned a LOT from him.

A will assume you know watts law and ohm's law, and the basics of electrical circuits. If not, check out my videos on the topic!

In order to understand how the emitter follower works, lets go over the other common configuration, the common emitter. The "common" just means ground, or what we measure all the voltages with respect to. So when we say "common emitter" we are describing the configuration where the emitter of the transistor is tied to ground (or the positive rail for the PNP types). As you probably know, BJT transistors are like switches, and will basically turn ON when there is more than 0.6V relative to the emitter at the gate**, causing several mA's of current to flow through it. So that means a voltage of just 0.6V relative to ground will turn it on, causing the collecter to be pulled down to ground for NPN devices on the low side, or in the case of a PNP on the high side w/ emitter connected to Vcc, the collector is pulled up to Vcc.

However, what happens when we try to use transistors in such a way, where the emitter goes to some load, and the collector is tied to the the power supply rails? A simple example would be a NPN device on the high side, with the collector connected to the Vcc. Lets say there is a lamp between the emitter and ground. Remember what I said about how to turn the transistor ON? The voltage on the base has to be 0.6V higher than that on the emitter. Lets apply about 5.6V to the base and see what happens. While the lamp is off, there is 0V at the emitter, so the transistor is turned on very hard and very fast. The voltage at the emitter will then shoot up and up, and as it does so, more and more current flows through the lamp. If we assume that the lamp is like a resistor, then when there is current going through it, there has to be a proportional voltage drop across it! That means the voltage seen at the emitter of the transistor is higher than it was before. In fact; it pretty much instantly went up to 5.0 volts, since that is when the transistor just starts to turn back off, because the difference between the base voltage of 5.6V and the emitter voltage of 5.0V is 0.6V, that is just what is required to keep the transistor ON~ish. If the voltage at the emitter goes higher for whatever reason, then the transistor is turns off, and no or less current means no or less voltage. Naturally, that voltage will just sit there, and not change much, even if the load resistance changes, or the power supply voltage changes. As long as the voltage seen at the gate of the transistor is 5.6V in this case, the output will follow that, minus 0.6V.

In reality, transistors are not that perfect, and they may require a lot more voltage on the gate to really turn on really hard, and the might just start trickling current through them at 0.3V or less! The exact numbers really depend on the transistor Hfe gain and stuff, all of which can be found in the datasheets!

Sorry if thats too long, hopefully I made sense of it. I may do a video of it, going over the ways to use transistors. Now I will hopefully explain what push-pull means.

There is one annoying disadvantage of using a single transistor to turn things ON, OFF, or anything in between: You need to know whether the load connected from the positive supply rail, Vcc, to your 'driver', or if the load goes from the common or negative rail, and the other side goes to your 'driver'. If you made a circuit with just a single transistor output, and put it into a box or housing, and come back to it, you may not remember what transistor you used for the output of the amplifier, and whether the output is relative to ground, or Vcc, or differential, etc.

When people talk about how some driver drives a load, we may mention that it either sources current, or sinks current, and you can imagine something that sources current is up high, at Vcc, and is like a faucet that can be turned on or off, and something that sinks current is like a plug for the drain. Either is lets the water flow, or it stops up and no water (current) can flow.

If you have dealt with an op amp, or a 555 timer, you have worked with a push-pull configuration for the output. You can easily add an between pin 3 of the 555 and the Vcc, or between pin 3 and ground. Either way, if you wire up the 555 correctly, you can get the LED to flash! No worrying about connecting it just the one way or the other!

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Now, what if we want a nice amplifier? We can just use the high-side NPN voltage follower, and have a smaller, less powerful, baby amplifier drive a that NPN. But then, electricity can only flow one way though the speaker, and that means that it can only push or pull the air to make sound. To avoid distortion and clipping, we bias the transistor so that with no signals, it is sort of halfway on, and then the signal can make it so that it goes all the way off or on, driving the speaker. However, this is really inefficient, since the transistor is always dissipating power, and there is excessive DC current going through the speaker, which can cause it to overheat. What if instead, we can make it so that no electricity flows through the speaker until a signal is applied??? That way all the power going into it is used to make just sound. Thats what a H bridge does, and a half-bridge driver circuit does.

Another complementary PNP emitter follower is added to the ground rail, and it's output is connected to the output of the NPN that is above it it terms of voltage. We can actually apply the same signals to both transistors, and both transistors will assure the output is always the same as the input. But this has the advantage that now it can either source current, or sink it. Make another "totem pole" circuit thats exactly the same, and drive it with the same signal, but flipped upside down (inverted) and connect the speaker between the two "outputs" of those totem poles, and you have an H bridge! Whats the benefit of that? Think about it! Now, power can flow back and forth through the speaker, and NO biasing is required to avoid clipping!!! That is the basic AB amplifier design. For complicated reasons, this will have crossover distortion (because all those NPN and PNP transistors dont always agree what the voltage should be) and the possibility of excessive current effectively shorting out the power supply, but built right, it can be considerably more efficient!

Hopefully I did not lose you in all of that BLAHH knowledge! :)

Than you once again. I will be using some of the information you gave me in my report that i will have to wright. ofcourse thats if you dont mind

Go ahead, I'd hate to have all that work fade into the 999th page of the 'ables question board, never to be seen again lol! Just make sure to cite your references, (typically with APA) for scholarly works and professional stuff! (not that I care to be cited or not, just for good measure and practice.)

Thank you very much -max- This was an in depth explanation. I feel that i understand much more about it now. Thank you very much once again.

A Google(r) Image search on the words "push-pull emitter follower" will give you a bunch of images, most of which are circuit diagrams.