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.

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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.

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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?

  • \$\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 '19 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 '19 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 '19 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 '19 at 22:27
  • \$\begingroup\$ It seems some details are missing \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 24 '19 at 22:42

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.

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  • \$\begingroup\$ A po co w ogóle robiłeś takie pomiary ? I za "e_g" robiła microswitch? \$\endgroup\$ – G36 Feb 27 '19 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 '19 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 '19 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 '19 at 1:16
  • \$\begingroup\$ A bardziej dokładnie z czym miałeś problem? \$\endgroup\$ – G36 Mar 4 '19 at 16:09

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