Hello,

I am working through the Art of Electronics using LTSpice to design and test circuits. I am trying to recreate the Emitter follower in figure 2.6. See the screen shot below for my attempt at this circuit. The pane on the right shows Vin at 10V and Vout .7V less than Vin with Vout at ground at 0V as expected. If I return the emitter resistor to a negative supply voltage, the output should be a replica of Vin less .7V as stated by the text. The result is in the pane to the left. Vout is not a replica less .7V, instead it is much higher than Vin. Any ideas what I am doing wrong with my circuit? I have tried every way I can think of to get this to work.

Thanks

 

Well, you have sine waves on both your input and for the supply voltage!

Try setting your supply voltage to being fixed, instead of a sine wave.

Have the emitter resistor grounded. Ground the - side of Vs, and set it to be 10v.
Set Vin to be a sine wave varying from 1v to 10v.

[eta] See the attached simulation.
In order to get a 9v p-p sinewave with the lowest point @ 1v, you need to specify it as having an offset of 5.5v with 4.5v amplitude

 

SgtWookie,

By fixing the supply voltage, do you mean change it to 10V Dc? When I change the supply voltage to 10V Dc, gound the - supply and emitter resistor, I get the same curves as in the pane on the right. The book states that the emitter resistor should have a negative 10V. I am not sure how to get a negative supply other than using Ac. I don't know how to vary Vin from 1 to 10V. I am new at LTSpice and have a lot to learn. I will try and figure out how to vary the supply voltage. Thank you for your suggestions

 

Look at your +10 and - 10 sources. Are you sure they are correct?

Here's what I got ...

 

OK, I've updated my simulation to show you how I get + and - voltages referenced to ground. It leaves little doubt as to what is where! Everything in SPICE needs to have a DC path to ground somewhere.

 

SgtWookie,

I saw your attachment and changed my circuit and it worked. Thank you! I am not quite sure I understand the dc offset of 5.5V and 4.5 amplitude. As far as the example in the book, it calls for a -10V supply to the emitter resistor which is what I am trying to re-create. I guess I really am trying to learn the basics of building a circuit in LTSpice.

JoeJester,

Thanks for the advice. I saw that my -10V source was set for output and I changed it to input. I ran the simulation with the original parameters reversing the -10V supply and still came up with something different from what you got. See the screen shot.

 

Two points are being made in the paragraphs concerning figure 2.6

One is the output will always be 0.6 (or so) less than the input, and the input must be above 0.6 volts for the transistor to conduct.

Look at Sgt's second post as it will help you with LTSpice's setup.

The emitter follower will work up to the source values ... minus 0.6 volts. (assuming Si transistors of course)

 

Ok, I got it now. SgtWookie, we must have been posting at the same time. I saw your simulation and think I now understand. See my screen shot below. I guess from now on when I see an illustration with a +10V or -10V it means it is coming from a Dc source. Now I understand how to run the negative Dc source. Thanks for the help.

 

Ok, I got it now. SgtWookie, we must have been posting at the same time.

That seems to happen frequently around here!

I saw your simulation and think I now understand. See my screen shot below.

By George, I think you've got it!

I guess from now on when I see an illustration with a +10V or -10V it means it is coming from a Dc source.

Absolutely! As you might've noticed, I like to keep my postive voltage supplies above the ground symbol, and negative supplies below the ground symbol. They take up less room on the simulation that way, and also take less time to find if there's a problem. Keeping schematics un-cluttered becomes very important when you start putting together larger projects, as it's easy to get "lost" in a cluttered schematic.

Now I understand how to run the negative Dc source. Thanks for the help.

That's what we're here for