Before we crack the lid on the op-amp a bit of transistor theory would be useful. In fact that’s pretty much all an op-amp is a bunch of transistors. By understanding the transistor you will get a deeper understanding of the op-amps “real world” limitations.
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The transistor comes in two flavours npn and pnp. The letters refer to the internal dropping of the semiconductor, but they can also be used to help remember the correct symbol:
NPN = Not Pointing iN
PNP = The other one… I know it’s rubbish but it helps, trust me.
PNP = The other one… I know it’s rubbish but it helps, trust me.
A PNP is basically the inverse of a NPN. NPNs are more commonly encountered so we will deal with this first.
The Ideal NPN
Like an op-amp, a transistor can be boiled down to a few basic rules. In fact the most basic two rules are the same as for op amps although they apply slightly differently.
“Keep your feet together and accept no currents“
But this time the “feet” are the Base and Emitter, and it will accept no currents only at its Base.
So let the base voltage be 7v, the Ve will also be 7v (feet together). 7v will therefore be dropped across the 1k resistor at the emitter. This causes a 7mA current to flow. As the current cannot be supplied by the base (accept no current) this must be supplied by the collector pin.
But the collector pin is tied to 10v which is too high, so it must drop 3v across itself. It does this by limiting the current, which can be thought of as varying its resistance. 7mA with 3v is equivalent to 428.6Ω. The transistor will have to dissipate 21mW (P = VxI = 3v x 7mA) of power to do this.
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The Less-Ideal NPN
In reality all transistors have a small volt drop between Vb and Ve called Vbe. For most bipolar transistors we can assume it’s around 700mV or 0.7V (we can deal with why its this another day).
Also in reality the base will accept some current. The Base current is related to the Collector current by the transistors current gain which is referred to commonly as hfe. Hfe for most transistors is around 100 but can vary from device to device between 50 and 300.
The Emitter current is the sum of the Base and Collector current but as Ic will be rough a hundred fold larger than Ib we can safely assume Ie = Ic.
Lets returning to the first circuit and assuming an hfe of 100.
We see now that if Vb is 7v then Ve will be 6.3v.
6.3v over the 1k resistor requires 6.3mA at Ie. Assuming Ie and Ic are equal then Ib = Ie/hfe = 6.3mA/100 = 63μA.
When you hear discussions about the transistor being used as a current amplifier this is what they are talking about. A small base current can drive a large collector current.
The Ideal PNP
The PNP is much like the NPN but in reverse.
It still aims to keep its feet together and accept no currents but the PNP is placed with its emitter (the leg with the arrow) facing the positive rail and so the currents flow towards the base and collector instead of away from them.
The Less-Ideal PNP
Because the current flow is reversed so too is the non-ideal volt drop Vbe. For a most bipolar transistors Ve will be 0.7V higher than Vb.
The current gain (hfe) of a PNP transistor is similar to that of a NPN transistor and will normally be stated between 50 and 300. As noted the currents of a PNP are reversed so a small current flowing out of the base will result in a large current flowing out of the collector.
Lets take an example of assuming an hfe of 100, Vb of 7v and an emitter resistor of 1k.
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We see now that if Vb is 7v then Ve will be 7.7v.
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This leaves 2.3v over the 1k resistor requires 2.3mA at Ie. Assuming Ie and Ic are equal then Ib = Ie/hfe = 2.3mA/100 = 23μA.
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We an see here that if Vb were to decrease, Ve would decrease, and Ie would increase. This is the opposite to what wwould happen with a NPN. Taken to an extreme position a Vb of 0V would result in a PNP pulling maximum current (Fully ON) and a NPN drawing no current (Fully OFF).