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To measure Beta (hFe) of the transistor you should zero u_ce (voltage across CE for the AC component in signal). To zero that component you put capacitor in parallel with transistor AND as shown on the picture "b" a source of amplified (varying) signal ("e_g") in series with U_CC.

enter image description here

If I use my DC power supply to set Q point of the transistor, I plug it to black and red terminals on my little test board, then I should have "e_g" plugged right before a switch, so that pressing the button would apply DC + AC signal.

enter image description here

I know you can't do just that, as the voltage sources will eat each other, so how do I achieve that? Is the best option to use an op-amp, are there any possibilities available which do not involve op-amps?

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  • \$\begingroup\$ WHich current gain test did you want? (1) hFE @ Vce(sat) , (2) hFE @ Vce=Vbat/2, (3) Hfe AC current gain vs Ic? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 24 at 22:09
  • \$\begingroup\$ (3) I believe. I measured current Ic and Ib for varied DC source and computed Beta from that, I got results like 1 to 400. Now I would like to check if I get same beta in test "b" shown in the picture, as this i show you should measure Beta. And what test says is to use UCC to set transistor's operating point and then apply "e_g" in series with base, this is what I want. \$\endgroup\$ – 4pie0 Feb 24 at 22:16
  • \$\begingroup\$ That would be apparent DC gain (2) with variable Ic unless you computed incremental changes \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 24 at 22:21
  • \$\begingroup\$ According to the book the circuit from picture "b" is how to measure h21e parameter from hybrid model of transistor in common emitter configuration == beta == hFe from datasheets, isn't it? \$\endgroup\$ – 4pie0 Feb 24 at 22:27
  • \$\begingroup\$ It seems some details are missing \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 24 at 22:42
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I realised this circuit and got superimposed signal S = DC + "e_g" = 1 + sin(2*PI*10^4*t) by connecting grounds of two voltage sources together and their positive terminals to ground and positive of the circuit.

I also put capacitor in parallel with transistor as shown, for varying capacitance: 1uF, 100uF, 1000uF. The observed result of that is decrease in collector current from 0.37mA to 0.322mA.

Capacitor value | Collector current | Base current | Beta = Ic/Ib

  • -,0.37mA, 1.2uA, 308
  • 1uF, 0.35mA, 1.1uA, 318
  • 100uF, 0.322mA, 1.1uA, 293
  • 1000uF, 0.322mA, 1.1uA, 293

All currents measured with DC current function of multimeter. Resistance is 430 kOhm (bbutton to base) and 100 Ohm (Vcc to collector), transistor is 2N3904 with hFe300 in datasheet.

enter image description here

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  • \$\begingroup\$ A po co w ogóle robiłeś takie pomiary ? I za "e_g" robiła microswitch? \$\endgroup\$ – G36 Feb 27 at 15:14
  • \$\begingroup\$ "e_g" = sin(2*PI*10^4*t) z generatora sygnałów, sinusoida 10kHz, 1V amplituda, 2V pp. Pomiary te wykonywałem w ramach ćwiczeń z tranzystorem \$\endgroup\$ – 4pie0 Mar 3 at 16:10
  • \$\begingroup\$ A to nie mogłaś zmierzyć bety (\$\beta = H_{FE}\$) z wykorzystaniem tylko prądu stałego? I tu sobie poczytaj forum.elportal.pl/viewtopic.php?p=87990#p87990 \$\endgroup\$ – G36 Mar 3 at 16:22
  • \$\begingroup\$ Chciałem ten dokładnie obwod zrealizować, również ze względu na problemy jakie napotkałem (źródła napięcia). Fajny, przydatny link - dziękuję bardzo. \$\endgroup\$ – 4pie0 Mar 4 at 1:16
  • \$\begingroup\$ A bardziej dokładnie z czym miałeś problem? \$\endgroup\$ – G36 Mar 4 at 16:09

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