What do we mean when we say a transistor amplifies current?

Question

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!

The current through a diode; it's very useful to understand the diode equation first, as it leads to an understanding of how the transistor is modeled. — Marcus Müller, Apr 27 '19 at 21:21
Can you post an example schematic that you're having trouble with, or a link. Five different ways to do analysis sounds about right -- different ways of doing the analysis have different advantages. Nothing simple will give you the full answer, nothing that gives you the full answer is easy enough for a human to reasonably do the computations (which is why we have simulation software). — TimWescott, Apr 27 '19 at 21:24
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. — Elliot Alderson, Apr 27 '19 at 21:25
@ElliotAlderson I'm a beginner so I was just assuming that the current should drop over a transistor because it is amplifying all this current — Resistor, Apr 27 '19 at 21:33
A linear mode transistor to control motor current will be very inefficient ( hot), Is it a small motor? WHat is it's DCR and V rating — Tony Stewart EE75, Apr 27 '19 at 21:39
@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. — jonk, Apr 27 '19 at 21:52
@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. — jonk, Apr 27 '19 at 22:42
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". — user57037, Apr 28 '19 at 20:26
In the circuit you have drawn, it is very easy to calculate the voltage drop from collector to emitter of Q1. It is just 5V because there is a 5V supply there. The voltage at the base will be unstable because it has a strong temperature dependence, and Q1 will be dissipating a lot of heat (Watts). Q1 will heat up and fail permanently probably within a few seconds. Certainly it will not survive for 1 minute. And so it is POINTLESS to do any calculation. It is unfortunate, but providing you with the information you seem to lack is beyond the scope of a single question. — user57037, Apr 28 '19 at 20:32

score 2 · Answer 1 · answered Apr 27 '19 at 21:18

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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.

answered Apr 27 '19 at 21:18

The Photon

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Thanks for your answer but how could I find out how much voltage drops from expected collector current? – Resistor Apr 27 '19 at 21:22
@Resistor, by analyzing the rest of the circuit. – The Photon Apr 27 '19 at 21:23
could you please give a basic example, like if I was amplifying current for a motor that needs 10A and 7.5V, where the battery only supplies 7.5V, the voltage would drop over the resistor, correct, not allowing the motor to turn on? But how much voltage would drop? – Resistor Apr 27 '19 at 21:35
@Resistor, if you're powering a motor, you want to use the transistor as a switch, not an amplifier. – The Photon Apr 27 '19 at 21:36
Sorry, if the battery only supplies 1A, then a transistor would be needed. – Resistor Apr 27 '19 at 21:37
In the example you added to your question post, the c-e voltage of the transistor is always 5 V exactly, because you connected an ideal voltage source from collector to emitter. – The Photon Apr 27 '19 at 21:37
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If the battery only supplies 1 A, then a transistor can't help you. – The Photon Apr 27 '19 at 21:38
How come? Doesn't a transitor amplify current? – Resistor Apr 27 '19 at 21:38
@Resistor, it allows a small current to control a large current. But it doesn't produce current itself. It needs to be used with a power source able to supply the desired current. – The Photon Apr 27 '19 at 21:40
So does a transistor only _**allow**_ more current to flow through, I thought it amplified it? – Resistor Apr 27 '19 at 21:40
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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. – The Photon Apr 27 '19 at 21:40
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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. – Resistor Apr 27 '19 at 21:43
Transistors need 5 to 10% base current to saturate. – Tony Stewart EE75 Apr 27 '19 at 23:04
@resistor- you are asking "What do we mean when we say a transistor amplifies current?". I know - many people will not be happy with my answer, because the have learned otherwise - but wrong): A BJT does not amplify current - and it DOES NOT ALLOW a small current to control a large current. Thecurrent (Ib) is just a small unwanted "byproduct" of the large current (Ic) with a fixed percentage beta=Ib/Ic. This ratio is called "current gain" - but it must not be interpeted as a cause-and-effect relation. Like the classical pn diode, it is just the VOLTAGE between B and E that controls Ic. – LvW May 01 '19 at 09:35
@LvW, LEDs are also voltage-controlled devices for the same reason. And yet practically we still prefer to drive them with fixed current sources rather than fixed voltage sources. Even if the physics says it's a voltage-controlled device, for circuit design purposes it's often easier to treat it (the LED or the BJT) as a current-controlled device. – The Photon May 01 '19 at 14:54
@The Photon, I know this argument (your last sentence). But surprisingly, all people claiming that the current-control approach would be easier during design, still apply the voltage-control method - because this is the only way to enable a desired Ic. Typical example: For finding the value for a "current- determining" resistor Rb between Vcc and base node they use - of course - the desired value VBE=(0.6..0.7) V, and are, thus, designing the voltage divider between Rb and the B-E path. Of course, with the right value for the current Ib through this "divider chain". Is this "current control"? – LvW May 01 '19 at 15:29
@LvW, When I bias a BJT, if I tried to bias it for exactly 0.65 or exactly 0.68 V, I might get vastly different behavior than I expected (for example, due to temperature changes). So I make a bias circuit that gives pretty much the base current I want,varying only a little bit depending on what the \$V_{be}\$ actually turns out to be. Since I'm setting the current (and I don't know what the voltage is going to be well enough to use my estimate of the voltage to predict the behavior), I'd call this current control, not voltage control. – The Photon May 01 '19 at 21:43
@The Photon, yes - of course, you are free to call this approach "current control". However, this does not touch the physical truth that the BJT internally is voltage-controlled (Ic is determined solely by Vbe). More than that, I am sure that no experienced designer will try to bias the BJT "exactly" for 0.65 or 0.68. This would be crazy. Of course, we all use negative feedback stabilization. (By the way, setting the base current you "want" (your term), did you ignore the huge beta variation ? ). – LvW May 02 '19 at 08:44

score 0 · Answer 2 · answered Apr 28 '19 at 01:47

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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}

answered Apr 28 '19 at 01:47

analogsystemsrf

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Tony Stewart EE75 · Answer 3 · 2019-04-28T19:22:33.303

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.

What do we mean when we say a transistor amplifies current?

3 Answers3