NPN C-E Buffer Output in LT Spice

[eev_carl]

Hi,

(Big edit from the original post; had the wrong circuit breadboarded)

Is there a way in LT spice for me to AC couple the output as with my scope?  When I breadboard this, I get the same 3.4V offset but AC coupling brings this back to 0V.

Thanks,
Carl

[StillTrying]

The cap time constants might be much too long, change the 1000u to about 1n.

[Ian.M]

C2*R4 gives a time constant of one second.   Your sim runs for 0.1 seconds.   Don't use the .tran startup option unless your sim duration is greater than five times the coupling network time constant!

.tran startup prevents LTspice precalulating the steady state DC bias across the coupling capacitor, because it starts V1 at 0V.

Edit: struck out bad maths, see below. 

[StillTrying]

C2*R4 gives a time constant of one second.

I guesstimate it at 20 minutes! or 1ms with a 1n.

[Ian.M]

Why would you guestimate it?  The effective output impedance of the emitter follower is very small compared to R4, so its contribution can be ignored, and the component values of R4 and C2 couldn't be easier to multiply by mental arithmetic.

N.B. Five time constants is simply the time it takes for a RC exponential to settle to within 1% of its final value - good enough for practical work.   Add another five time constants settling time for every additional two digits of accuracy when taking a precision measurement.

[StillTrying]

"Why would you guestimate it?"

OP said:
"Is there a way in LT spice for me to AC couple the output as with my scope?  When I breadboard this, I get the same 3.4V offset but AC coupling brings this back to 0V."

I'm guesstimating that with 0.5M to 10M of scope impedance on point Vout in the 1st post, that the DC blocking of the 1000uF will appear to be not working, because it's taking so very long to charge the 1000uF.
Or something..

[Audioguru]

1000uF feeding 1Meg has the u and M cancel each other. Then the capacitor charges to 63% of a full charge in 1000 seconds which is 16.7 minutes. It takes 5 times as long (83.3 minutes or 1.39 hours to be almost fully charged but the capacitor leakage current will prevent that.

[Ian.M]

Audioguru,
Many thanks for spotting my three orders of magnitude brainfart.   Maybe I'm getting too old for math!   

[Zero999]

It's not what you asked, but the bias resistors (R1 and R2) are far too low. The input impedance will be equivalent to R1, R2  and the transistor's base impedance in parallel. The transistor's base impedance is equal to the Hfe * R3, which is a few Megs, so can be ignored, leaving R1 and R2 in parallel, which would give R = (1k2*1k)/(1k2+1k) = 1k2/2k2 = 0.545k = 545R

Assuming the Hfe is around 200, the base impedance will be 2M, so the impedance of the R1 and R2 in parallel can be ¹/₁₀ of that, which is around 200k, about 400 times the values of R1 and R2 in parallel, so they can be multiplied by 400, giving R1 = 470k and R2 = 390k as the nearest standard E24 values. The total input impedance will now be around 200k and the value of C1 can be reduced to reflect that, so use 100nF.