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I have this datasheet of the BC546 BJT:

enter image description here

I need to find the exit-conductivity at the operating point of 60mA and 4V, which I marked as a blue point in the i-v-characteristic. But no curve crosses through there. Does somebody maybe know how to create a curve which crosses through the operating point, so that I can derive the slope? Thanks a lot for all answers and ideas.

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    \$\begingroup\$ Is your "exit-conductivity" dic/dVce at this point? \$\endgroup\$
    – Antonio51
    May 15 at 15:15
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    \$\begingroup\$ @n328 The word you're looking for is "output conductance", though it's more common to talk about it as output impedance. \$\endgroup\$
    – Hearth
    May 15 at 15:25
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    \$\begingroup\$ Your operating point is poor for gain. Why there? \$\endgroup\$ May 15 at 16:23
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    \$\begingroup\$ The datasheet graph is for a transistor that has "typical' specs. Your transistor might have minimum or maximum specs then it will have a completely different operating point. \$\endgroup\$
    – Audioguru
    May 15 at 16:51
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    \$\begingroup\$ The simulator link is very interactive. Feel free to ask more questions \$\endgroup\$ May 15 at 19:00

4 Answers 4

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You can numerically or graphically interpolate the curves.

Your Q point operates at 240 mW and has an Ro = 450 Ω from the plot shown.

from datasheet :

  • Thermal Resistance, Junction−to−Ambient R_JA = 200 °C/W
  • This means junction temp rise will be .24*200 = 48 'C above worst-case ambient.

If Vcc = 2Vce + 2V = 10V for Ve=2V then the load line indicates 10V/ 100 mA = 10 Ohms

Because it is on the knee of the slope THD will be high unless you use negative feedback on the base with an Rcb/Rb ratio some fraction of your open-loop gain. This lowers Zo, Zin and THD by the gain reduction.

To choose Av and THD and output R and output swing with Vcc , you need to define these specs.

[![enter image description here][1]][1]
What is the p

What is your purpose for this BJT?

For high current use the BC547B and high gain use the BC547C as the BC546 has a DC gain range from 110 to 450. with nominal 50% of max.

bonus

Alternative Q point with negative feedback if one wants to reduce THD nonlinearity in gain.

enter image description here

The above used hFE = 150 and changing it to 250 increase Acl= - 72 from -65 so that is much more dependent on R ratios and less dependent on hFE.

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You have several choices:

  • Estimate the performance by interpolating between values estimated from the curves provided at nearby bias points (250 and 300 uA).

  • Measure the response of your parts on a curve tracer

  • Find a SPICE model for this part and (after verifying that the model reproduces the response curves for known bias points) use it to determine the parameters at your bias point.

Given the limited accuracy with which you'd be able to estimate the slope of the curve from the printed chart, I'd think that even just using the curve for 250 uA without interpolation would provide comparable accuracy, assuming you can't find a SPICE model.

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If the spectrum of base current was continuous a curve would cross your operating point.From the chart the base current in your operating point will be between 250μΑ and 300μΑ.

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As @The Photon suggested, here an example of "spice model" simulated. microcap v12
NB: it can be a "little" different for some simulators.

enter image description here

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  • \$\begingroup\$ More than a "little different" AC \$h_{fe}=125~500\$. So this model uses 50% of Max AC small signal gain = 250. A much better choice is to use the binned versions with a suffix letter or the BC547C The tolerance error is that the hfe is rated at 2mA but here you are using 60 mA.. This looks like a modelling simplification that is not accurate \$\endgroup\$ May 15 at 16:58
  • \$\begingroup\$ In my case, hfe = ~260. But I was surprised that the curves are not really "similar" to the curves in the datasheet for low Vce :-( Should change some parameter ? \$\endgroup\$
    – Antonio51
    May 16 at 8:02
  • \$\begingroup\$ The above model does not show curvature due to saturation effects of hFE knee reduction and bulk resistance, Rce and is therefore inaccurate \$\endgroup\$ May 16 at 14:16
  • \$\begingroup\$ Rce is the far left slope is too steep, and hFE is too linear with even steps near saturation Vce<2V Vbe also has bulk resistance which affects why saturation in reality at high currents is Vce<2V and lowest current is Vce< 0.7V \$\endgroup\$ May 16 at 14:19
  • \$\begingroup\$ You can add estimate Rbe and Rce to the model in cct. All they are showing is a VAF Forward Early voltage slope, which is also dynamic from other effects including thermal which when hot increases hFE and shifts curve up and to the left. \$\endgroup\$ May 16 at 14:22

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