3
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

I've been trying to study BJT Voltage Divider Bias Circuits. In particular, as they relate to their active region, amplification, and operation in getting a crystal to oscillate. I've followed a video and saw that the node labelled 2.452 below is really 2.6024 in calcs I match. LT spice says its 2.4922. Why all the variation, I am using a beta of 200. I followed equations in the following video...

https://www.youtube.com/watch?v=zTyuzHokWyA&index=4&list=PL4651816D92AB6B2B

I believe they are correct. I wish to get at exact currents and voltages but cannot match to ltspice or even the sim below. I wish to look at things theoritical and ignore any time or heat dependent ramp up time.

The video I watched (and hopefully learned correctly from) would break the circuit below to a thevenin voltage source and resistance on the circuit that biases the 3904 transistor.

It arrives at currents and voltages by the following breakdown...

Where Re is R4

Vth = R2*8/(R1+R2)

Rth = R1 parallel R2

Vth - IbRth -Vbe -IeRe = 0

Vth - IbRth -Vbe -(B+1)IbRC = 0

Ib = ( Vth - Vbe ) / ( Rth + (B+1)Re )

Vcc - IcRc - Vce -IeRe = 0

Ib = BIb

Ie = (B+1)Ib

He made the approx that Ie is equal to Ic in one of the later lines.

In any case, I've got very comfortable with his explanation, but not been able to duplicate the results in LTSpice or below.

Even something simple like the voltage created from the voltage divider. I wish to know what get at currents and voltages and why there is no match to LTSPice (or below)? Thanks,Jeff

\$\endgroup\$
  • \$\begingroup\$ The value 2.602 comes from assuming 0 base current in Q1; obviously that's not exactly correct. Now for the other two values, you're saying that CircuitLab and LTSpice give different values when using the same BJT model? \$\endgroup\$ – The Photon Aug 1 '17 at 16:56
  • \$\begingroup\$ Simply because LTspie is using more complicated BJT model then you in your hand calculations. Why are you worry about this? In real life, the result will also vary. \$\endgroup\$ – G36 Aug 1 '17 at 16:57
  • 1
    \$\begingroup\$ In LTSpice did you use the built in 2N3904 model, or did you adjust the model to have \$\beta=200\$. Because the built-in model has \$\beta=300\$ and of course that will slightly change the base current. \$\endgroup\$ – The Photon Aug 1 '17 at 16:58
  • \$\begingroup\$ I copied the one with 300 and then changed bf to 200 in the new model , all else equal \$\endgroup\$ – Jeffrey Edward Messikian Aug 1 '17 at 17:33
  • \$\begingroup\$ @ThePhoton I did some checking and found that my calcs would produce 2.494 and not 2.602. I get the 2.494 by taking the thevenin equivalent voltage and reducing it by the drop across the thevenin equivalent resistance. I had to solve for Ib. Ib is calculated as a function of the 1500ohm resistor. Is that video correct? Are the formulas I listed above the proper ones to get an exact answer? \$\endgroup\$ – Jeffrey Edward Messikian Aug 1 '17 at 18:13
1
\$\begingroup\$

I wish to know what get at currents and voltages and why there is no match to LTSPice

LTSpice will almost certainly give you the better picture but, the voltage on the base depends on the current gain for the transistor and the finite current gain means you get base current flowing that effectively loads the 27 kohm resistor.

So if you just took the simple scenario of no base current you would get a voltage of 2.60241 volts.

However, if you took a more sophisticated approach and projected the 1.5 kohm emitter resistor (times current gain) to be in parallel with the 27 kohm you would get 24.77064 kohm and that would make the voltage 2.453 volts i.e. pretty much what circuit lab produces.

However, if you assume that emitter resistor projected to the base also drew current through a lightly forward biased diode (maybe 0.5 volts) you would get a slightly more realistic answer as per LTSpice.

But, is this reality or is this still an approximation? Ask yourself how much does the current gain vary from device to device and how much error will this produce an any analysis?

\$\endgroup\$
  • \$\begingroup\$ Each transistor is different. Two, from the same batch, could have a significant difference in beta, or any other parameter. (This is where the term "matched pair" comes from - measuring transistors and picking two with similar characteristics.) While the knowledge to calculate all of these details is admirable, there comes a point where you should just test it (and get yet another result.) :) \$\endgroup\$ – rdtsc Aug 1 '17 at 23:42
2
\$\begingroup\$

The main reason why is because the spice model is much more complex than the simple large signal model you are using, they are different. The spice model includes temperature effects and other small details that the simple model you are describing above does not.

Here is a short description of what all goes into the spice model. In a spice model we want to model temperature effects, and other small effects. enter image description here Source: Spice NPN model ECEE Colorado

Another thing to realize is both of these values would differ from the values of a transistor in the real world, because the beta of a real 3904 transistor even at the same temperature can be 100 to 300.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.