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I cannot afford a function generator and oscope right now, so I'm trying to learn analog electronics with Spice (MacSpice, to be specific). Below is the Spice model for a simple emitter-follower circuit. Vcc is 15V, and the input at the base is a 5V 60Hz sine wave with a 0V DC offset. R1 sits between the transistor base and the sine input, and R2 sits between the transistor emitter and ground.

The output at the emitter has me a bit puzzled and I would appreciate it if anyone has an explanation for the behavior.

* Simple Emitter Follower
Vcc 3 0 DC 15
Vin 1 0 SIN(0 5 60) dc 0
R1  1 2 270
R2  4 0 3.3k
Q1  3 2 4 generic
.model generic npn
.control
tran 1ms 60ms
plot v(1) v(4)
.endc
.end

The output of the transient analysis plot command is below.

alt text

I was surprised by this output. As expected, there is a slight voltage drop in the amplitude due to the base-emitter junction of the NPN. However, the output is clipped and does not swing below zero. What would I need to do to catch the negative part of the sine wave?

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  • \$\begingroup\$ I though this was rather silly when I posted this: "Oops! Your question couldn't be submitted because: users with less than 150 reputation can't create new tags. The tag 'emitter-follower' is new. Try using an existing tag instead." So that is why it is posted under the ambiguous 'bjt' tag rather than something more helpful like 'emitter-follower'. \$\endgroup\$ – Dr. Watson Jan 6 '11 at 16:53
  • \$\begingroup\$ tag has been added! \$\endgroup\$ – stevenvh Jan 6 '11 at 17:14
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    \$\begingroup\$ If you're brave enough, you can use your computer as oscilloscope and function generator for low enough frequencies. I don't know if there's software for mac, but there is for windoze and GNU/Linux. \$\endgroup\$ – AndrejaKo Jan 6 '11 at 17:45
  • \$\begingroup\$ This online circuit simulator is easier to play with than Spice. It has emitter follower built in as an example under circuits -> transistors: falstad.com/circuit \$\endgroup\$ – endolith Jan 6 '11 at 20:31
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You need to connect R2 to a negative supply voltage with enough headroom for your signal, or bias the input waveform so that it doesn't clip.

The output voltage an emitter-follower is usually V_out = V_in - 0.7V, but there are bounds on where this function works. For the expected behavior, VCC + 0.7V > V_in > VEE + 0.7V must be maintained, where VEE is your negative supply. In your case, VEE is equal to 0V since you don't have a negative supply. When your input voltage swings below 0.7V, the transistor turns off, and your output stays at the negative voltage rail.

Modify these lines to add in a negative power supply.

R2  4 5 3.3k
Vee 0 5 DC 15

Modify this line to bias the sine wave for the existing circuit.

Vin 1 0 SIN(7.5 5 60) dc 0
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  • \$\begingroup\$ Thanks for the explanation, it all makes sense now. In a "real" circuit, would I have two separate DC supplies for Vcc and Vee, or would I just be hooking up the "red" end of a 15V supply to Vcc and the "black" end to Vee? I guess what I am asking is whether or not two supplies are required (like in the Spice model). \$\endgroup\$ – Dr. Watson Jan 6 '11 at 17:58
  • \$\begingroup\$ When you need to have positive and negative supply voltages, I've really only seen a two-supply setup. Single supply operation is also common, but usually you adjust the signals to fall between the power supply rails. \$\endgroup\$ – W5VO Jan 6 '11 at 18:10
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So if I am decoding your netlist correctly, you have the circuit on the upper left. The reason the output doesn't go below zero is that the thing pulling down on the output is connected to ground.

One solution is shown in "A", which is to have the pulldown pull from -15V instead of 0V.

Another solution is shown in "B", which is to replace the pulldown with a PNP transistor. This wastes less power, because there isn't the 3.3k resistor always dissipating. Class-B amplifiers form the basis of most audio amplifiers and op-amp output stages.

alt text

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  • \$\begingroup\$ Yes, the upper left is the circuit I was working with. Thanks for sharing the diagrams, especially the NPN PNP chain. When you say it wastes less power, where does the power go? The resistor likely dissipates it as heat, but what happens with the PNP? \$\endgroup\$ – Dr. Watson Jan 6 '11 at 18:01
  • \$\begingroup\$ @DrW: The pnp (while npn is conducting) or npn (while pnp is conducting) stay off and just don't dissipate any power. The whole circuit will take less power from the +/- 15 V supply rails. \$\endgroup\$ – zebonaut Jan 6 '11 at 18:17
  • \$\begingroup\$ Also, it's a bit odd that the base resistor is small and the emitter resistor is large. Usually, you use a transistor amplifier in this configuration to embiggen the input current, so you need a high input impedance (read: large base resistor) and a low output impedance (read: small emitter resistor). \$\endgroup\$ – zebonaut Jan 6 '11 at 18:21
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    \$\begingroup\$ You really don't need a base resistor at all in this circuit \$\endgroup\$ – markrages Jan 6 '11 at 18:25
  • \$\begingroup\$ Thanks for the info about the pnp and base resistor. I don't think this circuit is meant to be a practical circuit, it is from the Lab 2 section of "Student Manual for The Art of Electronics". \$\endgroup\$ – Dr. Watson Jan 6 '11 at 18:59
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You need to add a dc bias to vin to prevent the base-emitter junction from becoming reversed biased during the negative half cycle. When the base voltage is below the emitter voltage no current flows and the transistor is in cutoff.

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    \$\begingroup\$ Do you mean that Vin should be centered around a positive voltage larger than its amplitude, such as 7.5V (0.5 * Vcc)? \$\endgroup\$ – Dr. Watson Jan 6 '11 at 17:09
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    \$\begingroup\$ Yeah, that would be it. You could also use your original sine source (centered around GND), put its signal through a large enough capacitor and feed it tho the base from there, while setting the base to a positive voltage with a resistive divider. That would probably be the configuration that is most widely used because you can use just one supply voltage and you don't have to care about the DC part (offset) of the input voltage. \$\endgroup\$ – zebonaut Jan 6 '11 at 18:18

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