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With a 2N3905 PNP Transistor, with base (input) and collector (output) resistances of \$R_b=1k\Omega\$ and \$R_c=200\Omega\$ (Load to Ground), connected to \$VCC=5V\$ and \$GND\$ respectively, I got \$I_b=7.5mA\$, \$I_c=46.6mA\$, \$V_{eb}=0.91V\$ and \$V_{bc}=0.75V\$ (edit. Rechecking the circuit I have \$I_b=4.2mA\$, \$I_c=24.5mA\$, \$V_{eb}=0.85V\$ and \$V_{bc}=-0.76V\$)

Hence, by switching base from \$0\$ to \$5V\$ I obtain \$0\$ to \$4.85V\$ (edit. \$0\$ to \$4.91V\$).

Is there a BJT or other transistor circuit for a better switching, closer to \$0\$ to \$5V\$, keeping the \$200\Omega\$ as load?

enter image description here

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    \$\begingroup\$ 1) include a schematic (even though it is common and obvious what you're doing). 2) You're using an NPN for switching and all BJTs will have some voltage drop. If you want lower, use an N-channel MOSFET like AO3400: aosmd.com/pdfs/datasheet/ao3400.pdf at less than 33 mohm at Vgs = 4.5 V you should get less than 1 mV instead of 150 mV. \$\endgroup\$ – Bimpelrekkie Jul 1 '19 at 21:04
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    \$\begingroup\$ I guess a mechanical relay would have smaller voltage drop than a transistor. 150mV loss is quite small. If you really need more accurately 5.0V to the load, consider a regulating circuit which is based on feedback control. If you want a transistor switch, you can find some higher price power fets with substantially lower voltage drop. \$\endgroup\$ – user287001 Jul 1 '19 at 21:07
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    \$\begingroup\$ Note that if you follow Bimpelrekkie's advice and use an N-MOS you will want to put the transistor on the other side of the load. You could also use a P-MOS. \$\endgroup\$ – evildemonic Jul 1 '19 at 22:16
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    \$\begingroup\$ 150mV or 3% loss is small compared to possible 5% resistor tolerance. Losses are a part of any circuit. \$\endgroup\$ – StainlessSteelRat Jul 2 '19 at 2:47
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    \$\begingroup\$ I wired up your circuit on a breadboard using a 2N3905. The transistor dropped 0.027V between Emitter and Collector, and the 200 Ohm resistor got 4.973V. \$\endgroup\$ – Bruce Abbott Jul 2 '19 at 6:54
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YES Diodes Inc. Has superBeta transistors rated by Rce.

Meaning if you wanted 25mA across 200 Ohms and chose one ($$) with 50 mOhm Rce or even 20 mOhm.

I say expensive for high volume as these are $0.65(1) instead of $0.10 but for you, cheap!

  • you can expect ( in theory) Vce(sat)= 25m*50m= 1250 uV +/-(EST.)50% using the correct Ic/Ib ratio suggested by datasheet in this Ic range.
  • Thus 0 to 4.998 But I would debate this hFE has a wide tolerance over the range of Ic and they only rate Vce(sat) at a few higher currents where it ought to be better at your lower current, but only in theory and no guarantee which is always defined by the tables ( worst case), not the curves (typ).

E.g. https://www.diodes.com/assets/Datasheets/ZXTP23015CFH.pdf

Also for both NPN and PNP we use the convention Vbe and Vce(sat) @ Ic, where PNP have -ve values

  • and not Veb, Vcb = ....
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    \$\begingroup\$ Thanks for the information and the conventions... \$\endgroup\$ – Brethlosze Jul 3 '19 at 19:21
  • \$\begingroup\$ @Brethlosze These are also called SuperBeta transistors because I know but few realize that saturated is often defined with Ic/Ib ratios of 20:1 or 50:1 which are close to 10% of average hFE in linear range. this is true for almost every transistor when “saturated” and begins <=1 V for low currents and <2V=Vce for max currents in a linear amp with no negative feedback where gain is controlled by hFE Rc/Re \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 3 '19 at 19:29
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Your measurements are not correct. 1) 4.3V/1k= 4.3mA but you wrongly measured 7.5mA. 2) 5V/200 ohms= 25mA but you wrongly measured 46.6mA. The transistor is switching with a normal loss of 150mV but it could be 400mV.

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  • \$\begingroup\$ Values rechecked... \$\endgroup\$ – Brethlosze Jul 3 '19 at 19:17

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