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No site, seems to explain this super well. For example, if current amplifies current then shouldn’t there be a larger voltage drop after, the current, per se, is amplified by 100? But the voltage drop seems to only be very small

The question I have arises from this, what do we really mean when we say a transistor amplifies current?

So, I am looking for a clear explanation, thanks so much!


When we say that a transistor amplifies current, we mean that it allows much larger current to flow from a much smaller current, it doesn't amplify the current in a circuit. The circuit doesn't make sense and should have another power source that the transistor is acting upon, a transistor is a switch after all.

There are equations that calculate the voltage drop but it is a tiny amount, and almost any circuit will still work with the voltage drop across a transistor, because there it is so tiny, that is.

This is all thanks to The Photon clearing this up, so thank you The Photon!

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  • \$\begingroup\$ are you familiar with the current through a diode? \$\endgroup\$ Apr 27, 2019 at 21:19
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    \$\begingroup\$ Your question doesn't make much sense to me. Transistors are not used by themselves, they are used in circuits with other components, and there are many common transistor circuits. So when you talk about "voltage drop" it is unclear where this voltage drop occurs. \$\endgroup\$ Apr 27, 2019 at 21:25
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    \$\begingroup\$ @Resistor Think of the transistor as more like a valve attached to a lever. Consider a garden hose (water source) hooked to the collector and a tiny aquarium hose (also a water source, but much less water per minute) hooked to the base. The emitter is the outlet. So long as you don't squirt water through the aquarium hose, the lever is held back by the force of a spring so the garden hose water doesn't flow. But if you start water through the aquarium hose, this pushes upon the lever and allows lots of garden hose water to flow. The valve isn't the source of current. That comes from elsewhere. \$\endgroup\$
    – jonk
    Apr 27, 2019 at 21:52
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    \$\begingroup\$ @Resistor Good. Hopefully, you can now see that the source of collector current is the voltage source you have hooked up to the collector. The transistor is just a valve. When the base has no current (isn't hooked up, for example, or is otherwise prevented), then the collector just remains closed and blocks the voltage source from providing current towards the emitter "outlet." But if you supply a little base current, then this fact opens up the collector "valve" somewhat. The lever advantage ratio is called "beta" and may be around 100 to 300. So a little bit of base current goes a long way. \$\endgroup\$
    – jonk
    Apr 27, 2019 at 22:42
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    \$\begingroup\$ The bottom line is this: you want a simple answer. But the answer is not simple. So your question, in effect, contains a fallacy built-in to it. It is like saying "Can't someone just give me a simple equation that tells me how far a car will go on a tank of gas.... I don't want a lecture, I just want one simple equation". \$\endgroup\$
    – mkeith
    Apr 28, 2019 at 20:26

3 Answers 3

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For an NPN BJT in forward active mode, the voltage drop across the base-emitter junction is typically 0.6 - 0.7 V.

The voltage drop from collector to emitter is whatever it takes to pull the expected collector current from the rest of the circuit. (That means, to calculate the drop you treat the collector-emitter branch as a current source and calculate what voltage that produces at the conductor due to the rest of the circuit; then check to be sure all voltages and currents are consistent with your assumption of forward active operation)

If the c-e voltage gets too low, then the gain (whether you want to quantify it as a current gain \$\beta\$ or a transconductance \$g_m\$) falls until at about 0.2 V from collector to emitter you end up in saturation operating mode rather than forward active.

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  • \$\begingroup\$ Thanks for your answer but how could I find out how much voltage drops from expected collector current? \$\endgroup\$
    – Resistor
    Apr 27, 2019 at 21:22
  • \$\begingroup\$ @Resistor, by analyzing the rest of the circuit. \$\endgroup\$
    – The Photon
    Apr 27, 2019 at 21:23
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    \$\begingroup\$ If the battery only supplies 1 A, then a transistor can't help you. \$\endgroup\$
    – The Photon
    Apr 27, 2019 at 21:38
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    \$\begingroup\$ If you had a battery capable of supplying 10 A, you could control the motor from (for example) a microcontroller output only capable of providing 10 mA (probably requiring either a MOSFET instead of a BJT, or a two-transistor BJT circuit). This is what we mean when we say the transistor amplifies current. \$\endgroup\$
    – The Photon
    Apr 27, 2019 at 21:40
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    \$\begingroup\$ Oh, that clears up a lot, I had thought the whole time that the transistor amplifies current in a circuit, and that another use was to be a switch, but now I understand that it is used to allow small current to allow much larger current to flow, how much longer is relative to the current from Base to Emitter multiplied times gain, and then there is actually only a very little voltage drop, thank you! You have cleared it up. \$\endgroup\$
    – Resistor
    Apr 27, 2019 at 21:43
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here is a one-transistor amplifier; voltage gain is about 10 (R4/R3)

schematic

simulate this circuit – Schematic created using CircuitLab

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A transistor is just a regulator of current between supply and load with rules on the base Voltage and/or current depending on how you want to control the load.

  • In linear mode it can amplify base current
    • with the load on the emitter , it's voltage follows the base voltage (Emitter Follower)
      • as long as supply is adequate and base current is adequate Ib=Ic/hFE) The AC emitter voltage out is non-inverting.
    • with load between collector to supply, the Ic is modulated by Ib*hFE again except now the collector voltage drops with rising base current so the AC voltage is said to be inverting.

\$ΔVce = - Rc*(ΔIb*hFE) \$

When Vce drops well below 2V towards Vce=Vce(sat), the current gain, hFE drops towards 10% hFE_max or actually hFE=10 by definition in the datasheet, if given as Vce=Vce(sat) @ Ic , where you are responsible for choosing Rb to create Ic/Ib=10 to 20 if you want it saturate at the rated Vce(sat).

In Saturated Mode, the Vce(sat) and Ic/Ib=10 to 20 the Vce will rise with load current on the collector and the C-E junction then has an equivalent resistance Rce such that;

\$ΔVce/ΔIc = + Rce \$

e.g. ZTX1049A TO-92 hFE= 300 to 1200 @ Ic= 1 A ( very special) $0.58 (500 pc)
Vce(sat) = 75 mV @ Ic = 1 A , Ib= 10 mA , Rce= 75 mΩ

e.g. PN2222A TO-92 hFE= 100 to 300 @ Ic = 150 mA $0.10 (500pc)
Vce(sat) = 1 V @ Ic = 500 mA, Ib = 50 mA , Rce= 2 Ω

So you may conclude that devices with very high hFE may have Rce(sat) at high Ic/Ib ratios, but also cost more. However, Rce generally drops with bigger junctions rated for higher power.

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