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I'm learning about transistors so I build the circuit in the image below and plotted the voltage at Vc (collector) as a function of base voltage, marked VBE. The transistor was a 2N3904. I had two power supplies, one supplying the 5v rail and a second that I could adjust from 0 volts upwards (they had a common ground). I noticed that I couldn't get the base voltage to go much above 0.8 volts. If I pushed up the supply voltage to 5 volts I could get the base voltage to about 0.85 volts. I am sure there is a simple explanation but why does the voltage at the base max out at about 0.8v no matter what voltage I apply?

I know that the transistor is saturated by the time I get to 0.8 volts.

Update: I've also now plotted the base voltage as a function of the applied voltage via the 1K resistor.

enter image description here

enter image description here

Having now read the comments and thought more about this, the explanation seems straight forward. We see from the second graph that at a VBE around 0.7 v the curve starts to flatten out, this is of course when the transistor starts to open up and in doing so the BE resistance drops thus holding the VBE voltage constant after that. As others have pointed out this can't go on forever but will eventually start to rise again once the resistance stops dropping.

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  • \$\begingroup\$ What's the question? I love all the work you've put into this and the results look reasonable to me. But I'm just not sure what's bothering you about all this. Do you know what the Shockley diode equation looks like? Have you taken the derivative of \$V_\text{BE}\$ with respect to \$I_\text{C}\$? (That will be the dynamic resistance of the base-emitter junction.) There will also be Ohmic base resistance and Ohmic emitter resistance. If you play with the Shockley equation, all this will make sense. \$\endgroup\$ – jonk Nov 27 '19 at 7:30
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    \$\begingroup\$ FYI: Emitter resistance is 26/Ie_mA. eg 26 ohms at 1 mA, 2.6 ohms at 10 mA etc. Obvious once you know it :-) \$\endgroup\$ – Russell McMahon Nov 27 '19 at 7:34
  • \$\begingroup\$ $$\newcommand{\d}[0]{\mathrm{D}} \begin{align*}I_\text{C}&=I_\text{SAT}\left(e^{^\frac{V_\text{BE}}{V_T}}-1\right) & V_\text{BE}&= V_T\:\operatorname{ln}\left(\frac{I_\text{C}}{I_\text{SAT}}-1\right)\\\\&\text{where } V_T=\frac{k\,T}{q}&\text{or, } V_\text{BE}&\approx V_T\:\operatorname{ln}\left(\frac{I_\text{C}}{I_\text{SAT}}\right)\\\\&\text{Including Ohmic resistances }R^{'}_{_\text{B}}\text{ and }R^{'}_{_\text{E}}\text{,}\end{align*}$$ \$\endgroup\$ – jonk Nov 27 '19 at 8:06
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    \$\begingroup\$ $$\begin{align*}\therefore\\\\r_{_\text{BE}}=\frac{\d\: V_I}{\d\: I_\text{C}} &= \frac{V_T}{I_\text{C}}+R^{'}_{_\text{B}}+R^{'}_{_\text{E}}\end{align*}$$ \$\endgroup\$ – jonk Nov 27 '19 at 8:13
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    \$\begingroup\$ Have fun. tinyurl.com/sajn56r LEDs are like low voltage zeners. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 27 '19 at 9:54
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If you increase the voltage across the base-emitter junction, the base current increases exponentially, according to the Shockley diode equation

$$I_b = I_s\left(\exp(\frac{qV_{be}}{nRT})-1\right)$$

where \$I_s\$ and the other symbols you might not have seen before are parameters of the device.

But in your circuit, the more you increase the base current, the more voltage is dropped across the base resistor. Given a 5 V supply, even if \$V_{be}\$ stayed at 0 V, the most you could have delivered through that resistor is 5 mA. With 5 mA into the base, I'd expect more like 0.6 to 0.7 V \$V_{be}\$ (but I'm not specifically familiar with 2N3904). So it's likely you were actually using something more than 5 V to supply the base, and getting somewhere in the 20 - 100 mA range, where 0.7 or 0.8 V is a reasonable guess at the base-emitter voltage.

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  • \$\begingroup\$ I don't think I was supplying more that 5 volts to the base (via 1k resistor) and I've repeated the experiment many times. I understand your argument with a max of 5v I can only move a max of 5 mA through the resistor and hence into the base. In the new data I present, the base voltage increases linearly as I increase the applied voltage but then stops at about 1 volt. After than it's as if the base-emitter resistance is dropping to compensate as I increase the voltage further so that the voltage drop across the BE remains constant. \$\endgroup\$ – rhody Nov 27 '19 at 4:16
  • \$\begingroup\$ @rhody, that's reasonable. When \$V_{be}\$ is below threshold (about 0.6-0.8 V) the effective resistance of the junction is fairly high. You're seeing most of the voltage drop across the junction instead of the resistor under these conditions. Once the junction voltage gets to threshold, its differential resistance drops and any added source voltage mostly drops across the resistor, not the junction. \$\endgroup\$ – The Photon Nov 27 '19 at 4:21
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why does the voltage at the base max out at about 0.8v no matter what voltage I apply?

It doesn't. It only 'maxed out' at ~0.8V because you used a 1k resistor and stopped at 5V. However when voltage on the Base is increased the current increases exponentially, so if you applied 5V directly to the Base it would draw several amps and burn out the transistor.

This characteristic is shown in the Datasheet in the graph of Base-Emitter voltage vs Collector current (Base current = Collector current / 10). The curve looks almost linear, but that is only because the x axis is logarithmic.

enter image description here

This curve is described by the Shockley diode equation, but that doesn't explain why the transistor behaves this way. The real answer is buried in the physics of P-N semiconductor junctions. However unless you are designing the transistors themselves you don't really need to know what 'causes' it, just that the device does what its datasheet says.

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  • \$\begingroup\$ Well, a 2N3055 for example allows a base current of 7A at 7V Ube. \$\endgroup\$ – Janka Nov 27 '19 at 3:01
  • \$\begingroup\$ @Janka yep, if he had used a 2N3055 he might have been able to reach 5V - so long as the power supply was strong enough and he used a big enough heatsink. \$\endgroup\$ – Bruce Abbott Nov 27 '19 at 3:16
  • \$\begingroup\$ I increased the voltage to 15 volts (kept the main rail at 5v, I also increased it to 10 just to see, no difference) and the base voltage didn't go above 0.84 volts. I also doubled and halved the 1K resistor and got 0.8 and 0.7 volts max respectively. I can't seem to get the voltage at the base above 0.84 volts. Clearly I am missing something and will reread both answers to try to understand.I also updated the question with new data that plots the base voltage vs the applied voltage and one can see how the base voltage stalls at about 1 volt of applied voltage. \$\endgroup\$ – rhody Nov 27 '19 at 4:06
  • \$\begingroup\$ @rhody, reduce the base resistor to, say, 1 ohm, and you'll get more than 0.84 V across the b-e junction. You'll probably also blow up your transistor when you do it. \$\endgroup\$ – The Photon Nov 27 '19 at 4:23
  • \$\begingroup\$ @rhody: 10V/500Ω == 20mA. That's still nothing. You can see that from the graph. The exponential part of the curve begins at »100mA for that transistor, and it won't stand it. It's designed that way by purpose because non-linerarity means distortion. \$\endgroup\$ – Janka Nov 27 '19 at 11:47

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