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V2 is turning the Q1 on very hard, such that it is in saturation, and Vc (product of Ic and R3) is quite large. If you don't want Vc to be so large, you have to reduce the base drive (V2/R1) of Q1 or divert some of that current away from the B-E junction.

What is this supposed to do?

Donald.

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In that case, you have to reduce the base drive. Separate bias and signals on the base would probably help.

I would change the series V2 and V3 into one current source - the DC offset is the bias and the Amplitude provides your signal. You can .STEP the DC offset to set different bias conditions.

Donald.

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A BJT is inherently a current-controlled and -controlling device. Ibe controls Ice, within the limits of the surrounding circuitry. I have no idea why you don't want to use current sources if you are after a device behaviour model. If there is another goal, it would be a good idea to tell us, otherwise we can lead you to a perfect but perfectly useless circuit.

To get the device in your earlier schematic into active mode, you'll have to drive the base with less current than I(e)/Hfe. This is known as "biasing a transistor" and is pretty standard electronic design. Most good biasing schemes that I know of use some form of current sources, since Vbe varies with temperature so much. Do a bit of net search for "transistor bias" and related terms; you should be able to come up with something useful.

In your original schematic (I'm not going to bother downloading another), note that V1 is 5V and V2 is -1V. The base current of Q1 will therefore be somewhere around ((5V) - (0.6V) - (-1V))/(100ohm + 1kohm) = 4.9mA. If Q1 has an Hfe of about 100 (just a WAG, understand), then Q1 will be capable of carrying? 490mA through R3 -- that WON'T happen given R3's value and the 5V from V1. To be in active region, Q1's base current has to be less than 0.01 * whatever current V1 can provide through R2 + R3.

If you _really_ want to use V2 and R1 to bias Q1, you'll have to increase V1 to a positive value near V1 - V(b,e): so somewhere around 4.0V to 4.5V. A base-emitter shunt resistor might also be helpful, but that will be left as an exercise for the student.

Donald.