I know this question has been asked a million times, but I've read all the answers and still don't understand. I've been studying this off and on for years and can't seem to get it. I just give up for a few months and come back.
I keep reading that in order to properly bias a transistor, you need to first determine Ic then figure out Ib=Ic/Hfe. The problem is Hfe is all over the place and all the datasheets I look at give the transistor Hfe between 25 and 300. That doesn't help me. Maybe if it was between
50 and 60 I could average, but not 25 and 300. Plus, that data is for a 10V supply, which I might not want, and give no data for other supply voltages. I want Ic to be 100mA, but with no idea of what Hfe is, how do I go from there and be confident that my design will work and not explode in my face when I hook it up?
Ok, so Hfe varies with Ic. So the solution is to put a resistor before the emitter and have negative feedback stabilize the gain. But how do I determine the value of that resistor? And for that matter of the potential divider on the base or the collector resistor. I search and search and all I get are specific examples of biasing with no clue how the values were determined. Do I really have to trial and error this stuff? Is it even possible to plan a circuit on paper and not have it behave wildly different when it is built?
One more thing. The negative feedback is supposed to keep the gain of the amplifier constant regardless of the transistor inserted. But does that mean regardless of the differences between individual transistors of one type (like 2N222) or can I plug in ANY transistor of any type and the gain should stay about the same?
Any help is appreciated. Maybe it'll jar my brain into finally understanding...
Forget about Hfe of the transistor. It mostly maters only in that you have enough. The ratio of the collector and emitter resistors determin the DC gain of the circuit. The AC gain can be improved by a capacitor across the emiter resistor. Also do not try to calculate the gain to the smallest fraction or many decimal places as most formulars are only close aproximentations.
I think, we need "beta" as high as possible so we can use only resistors to determine the GAIN of the amplifier.
I'm confused with your explanation here, could you contrasted to using single base bias resitor only, without using Emitter resistor as negative feedback element.
------------------------------------------------------------------- Using potential divider bias, hence Thevenin source and Therenin resistor, I can't understand how that would stabilize Ic.
The general idea is to design for worst case conditions. If your minimum Hfe looks likely to be a ball park 25, then design to this value (or preferably less). In practice the Hfe value is only of interest when setting the base bias resistors and TBH it's usual to design most amplifiers without even looking at a data sheet; just assume all small transistors have a Hfe of 100 and all power transistors a Hfe of 10. Your Hfe estimates may be way off the mark but the circuit will still work, (that's why the biasing resistors are there). Only noticable change would be a small shift in the static collector voltage. The emitter resistor is simply set to a value that that will give as high a emitter voltage as you can get away with. The higher the voltage the more stable the biasing. A couple of volts or more is OK (say 22ohms at 100ma),
0.5V is pushing it a bit. The potential divider at the base is set to pass around 4x to 10x times the expected base current (your choice). The expected base current depends of course on that Hfe but is now of little consequency as the base current loading (ie due to Hfe differences) can vary over a massive range without upsetting the base voltage (hence emitter voltage, hence collector voltage) too much. And yes, plug in pretty much any transistor and it'll work the same. Which it must, seeing as you've just cornered the gain differences, so there's pretty much nothing left but basic 'transistor action'. Which is the same everywhere. Easy eh!. regards john
The point of view that makes base biasing sensible is to understand that junction transistors are turned on by voltage. That is, the base to emitter junction must be forward biased by some particular voltage to turn on some particular collector current. But while this bias voltage is doing its job, the base emitter junction is also leaking diode current. So there is an incidental ratio of collector current to base current (that varies with conditions and with different devices) and that ratio is called beta.
The resistor divider method is a way to produce the appropriate base voltage, while also making a voltage source that has a low enough effective (Thevenin) series resistance to provide the needed base current without much distorting that voltage. If the divider passes 4 to 10 times as much current as the worst case (lowest beta) base requires, then its voltage is sufficiently stiff to hold the bias voltage acceptably stable for devices of higher beta, as well.
The emitter resistor is a negative feedback mechanism that provides an other input voltage (remember that the input is the voltage between the base and emitter, so both those terminals are inputs) related to collector current. The difference between the voltage produced by the emitter resistor and the base divider voltage is the actual input bias voltage for the transistor.
Unless you want to measuer each transisitor you use, you can not use the single base bias resistor. Even then as the transisitor heats up the circuit will often bceome unstable (thermal runaway) due to the changing of the charistrics of the transistor. There are some circuits that try this with a diode mounted on the same heatsink as the transistor to compensate for the heating of the transistor.
The formular in the text book is for a "perfect" transisitor where the transistors parameters are known and do not change due to heat and other factors.
There are 3 common circuits that can be used to bias a NPN transistor. The single resistor from V+ to the base, The one that adds an emitter resistor, the one that adds a resistor from the base to the negative voltage. Each one is more stable than the previous. The single resistor to the base is way too unstable for mass production.
:The resistor divider method is a way to produce the appropriate base :voltage, while also making a voltage source that has a low enough :effective (Thevenin) series resistance to provide the needed base :current without much distorting that voltage. If the divider passes 4
:to 10 times as much current as the worst case (lowest beta) base :requires, then its voltage is sufficiently stiff to hold the bias :voltage acceptably stable for devices of higher beta, as well.
for same type of transistor, would the one with higher beta draw more Ib compared to one with lower beta.
:The emitter resistor is a negative feedback mechanism that provides an
:other input voltage (remember that the input is the voltage between :the base and emitter, so both those terminals are inputs) related to :collector current. The difference between the voltage produced by the
:emitter resistor and the base divider voltage is the actual input bias
:voltage for the transistor.
wow....thats a clever idea to regards those terminals as inputs. I could never have thought that.
Based on your explanation I tried to simplify so my lame brain can digest it. I think, the variation on voltage across, BE junction, due to R-emitter, provide the negative feedback.
Yes. Last month I bought 200 off (cheap) BC557C transistors and in an idle moment measured the Hfe of one of them. It came out at a rather pleasing 450. Curious, I started to measure the others in the packet. Surprised to find all of them had values of 450 +/-5% !!. (still though design 'em in as if they have an Hfe of 100).
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The single resistor bias method is theoretical, only to be found in textbooks and is worthless in practice. (sorry :-)
John P' knows the 'current/voltage' dichotomy and his provided explanation would also be mine.
The gain of an amplifier using feedback is given by --- A' = A / 1+BA A' = Gain with feedback A = Gain without feedback (open loop gain) B = fraction of output fed back to the input (feedback factor)
Here are a few examples - If A = 50 and B=1% Then A' = 50/ 1+.05 = 47.6 If A = 100 and B =1% Then A' = 100 / 1+1 = 50 If A = 200 and B = 1% Then A' = 200/ 1+ 2 = 66.7
Note that with 1% of feedback, the gain with feedback varied by about 40% while the open loop gain varied by 400%. If this had been a transistor with beta varying from 50 to 200, by using feedback we could design a stage with a current gain that would fall somewhere between 47 and 66.
Now look what happens if we increase the feedback to 50% If A = 50 and B=50% Then A' = 50/ 1+ 25 = 1.92 If A = 100 and B =50% Then A' = 100 / 1+ 50 = 1.96 If A = 200 and B = 50% Then A' = 200/ 1+100 = 1.98
For all practicle purposes, with 50% feedback, the stage gain is 2 while the open loop gain varies by 400%. Again, if this was a transistor amplifier, any transistor in the lot would make an accurate current gain of
2
The downside to negative feedback is that we waste alot of gain but the upside is that we can make the gain very predictable. The reduction in gain is a direct indication of the effectiveness of the feedback. .
Current gain - It is typical to use enough feedback so that the stage current gain drops to
1/4 or less of the gain without feedback. All we need to know is the minimum beta to expect. For example, if you were using a transistor type, where beta varied from 100 to 200, obviously you would never design a stage having a gain of 150 unless you hand picked the transistors, but you could design a stage with a current gain of 20 and expect all of the transistors to work.
In a common emitter transistor amplifier - Rl is the collector load resistor Re is the emitter resistor Ra is the resistor from Vcc to the base Rb is the resistor from base to ground
The ratio Rb/Re is the stage current gain. This ratio is kind of equivalent to B. That is, as Re gets larger, or Rb gets smaller, we are using more negative feedback and the stage current gain is reduced. If we had a transistor with a minimum beta of 100 , it would be typical to set Rb/Re to about 25. A higher ratio is acceptable if you can tolerate the gain variations, or a lesser ratio, if really tight control of the gain is needed. It would also be possible to use a feedback resistor from the collector to the base or global feedback form another stage to stabilize this stage but that is a different story. For now we will look at biasing when the feedback is due to Re.
A typical transistor will have a maximum beta at a certain collector current, or a fairly constant beta over a range of collector current. This range of collector current is where the transistor was designed to operate. When Vbe is at .5 volts, the transistor collector current is at its lowest range. When Vbe is at .6 volts the collector current is just about right, and when Vbe is at .7 volts, the transistor is conducting heavily or near saturation. We don't know the exact value of IC for a given Vbe but we can get an idea of where IC is, over it's usefull range. I always shoot for a Vbe of .6 volts and then tweak as neccessary. If we have an emitter resistor of 1K and we want to operate at a collector current of 1ma, then the voltage across the emitter resistor is Ie x Re = 1 volt Note that Ie is Ib+Ic, but if Ib is a small part of Ie which usually is only about 1% of Ie, then for all practicle purposes Ie = Ic. The base voltage is .6 volts higher than the voltage across the emitter resistor or 1.6 volts. This base voltage appears across Rb. Ra then has to drop Vcc - Erb at a current of Irb. This gets you close to the value of Ra. Do not expect it to be exact. the reason being, is that Vbe increases 60mv for a tenfold increase in Ib. It is an exponential relationship and temperature dependent. Therefore the value of Vbe is somewhere between .5 to .7 volts over the usefull range of Ic. If you want an equation for Ra, try this approximation -
Ra = E/I = (Vcc - Erb) / Irb And Erb = Emitter voltage + .6 volts Emitter voltage = Ic * Er
Up to this point, you should have a grasp on the ratio Rb/Re to set current gain, and how to get a ballpark value for Ra which supplies the forward bias base current, but we have said nothing about the value of Vcc or the voltage gain. Providing we use enough feedback to reduce the stage current gain, the voltage gain will be Rl/Re. Our AC input voltage is divided up across the base emitter junction and Re, and the output voltage is across Rl. In terms of the AC signal voltage , voltage gain = E out/Ein = Collector Ac voltage / Base to Ground AC voltage. If the signal developed across the base emitter junction is small compared to the siganl developed across Re (which it will be with enough feedback from Re), then the Ac voltage gain is Rl/Re.
Ideally we can vary Vcc without significantly changing Ic provided that Ra is tied to a supply other than Vcc, because if we vary Vcc, then we vary Ib which will vary Ic. The value of Ra is probably best calculated after a value is chosen for Vcc. So far we can set current gain with the ratio Rb/Re and voltage gain with the ratio Rl/Re. To select Vcc we need to consider the Q point collector voltage and the transistor power dissapation. If we had Ic with no input signal = 1ma and Rl = 10K and Re = 1K then we drop 10 volts across Rl and 1 volt across Re. Lets assume Vcc = 24 volts. When the transistor is not conducting (open circuit), Vc = 24 volts. When the transistor is fully conducting (short circuit) then Vc = 2.2 volts. Visualize this as a simple voltage divider. When the transistor is an open switch, the collector voltage is Vcc and when the transistor is a closed switch we have a voltage divider of Rl in series with Re. The transistor will not be a perfect switch with 0 ohms but this is close enough for an estimate. Therefore our output signal can vary from 2.2 volts to 24 volts. We can have a peak to peak output signal of 24-2 or 22 volts. The largest voltage swing we can have at the output would occur when the collector Q point voltage sits in the middle of this swing or 2 +11 volts = 13 volts. This is class A operation. In a typical amplifier, we would not want to swing the out put signal this close to its limits because it would distort. If you needed a 22 volt peak to peak signal you would have to raise Vcc to provide headroom. But this example might be perfectly ok for say a 20 volt or less peak to peak swing. As the value of Vcc goes up, the transistor power dissapation also goes up which causes the transistor to heat up or smoke or both.
Hopefully this gives you a better understanding of the DC bias. Set the current gain Rb/Re, so it is small compared to beta. Set the voltage gain Rl/Re to any desired value, but remember that a high voltage gain means Rl has to be large compared to Re and this means that Vcc also has to be higher. The collector Q point voltage puts a limit on just how far the output signal can swing. The largest swing for class A operation is when the collector Q point voltage is midway between Vcc and Ve when the transistor is shorted. Typically, voltage gains max out at 20 to 30 in practicle amplifiers. Also when the voltage gain is high, the bandwidth is reduced and the output Z is raised, but these are compramises you don't need to deal with just yet.
For low frequency or DC operation, the input resistance is Ra and Rb in parallel and the output resistance is Rl.
Thank you to everyone that replied. I think my biggest problem was that I didn't realize that the transistor Hfe was not the same as the amplifier gain. That cleared a lot of confusion. Also, despite reading it over and over, it just never really sunk in that a diode drops about .7V regardless of the current through it. I was really mixed up with that when trying to set the base current. I've been able to design a few amplifiers on paper now and have them perform as I had calculated (or very nearly so). I feel much better now.
If there is no emitter resistor, both bases look like a diode to the emitter voltage, so, the base current would be independent of beta. What would change would be the collector current. If there is an emitter resistor, also, then as the collector current tends to rise, so does the emitter voltage, so the higher beta unit would draw less base current and the collector currents would be more similar (though the high beta unit would still have a somewhat higher collector current than the lower beta unit).
Right.
Except that there are two kinds of negative feedback possible. One is based on the collector current (That is mostly what the emitter resistor reacts to. Of course, the smaller base current also passes through it.) The other kind is based on the collector voltage. You can add the second kind by deriving the base voltage divider from the collector voltage, instead of from the supply voltage. That way, if the collector current increases, it lowers the voltage applied to the base divider and reduces the collector current increase.
You can use both types of negative feedback in the same circuit or one or the other. Current negative feedback raises both the input and output impedance, while voltage feedback lowers both. So you can combine them in various combinations to lower the gain (and stabilize it for beta variations) and also get the input or output impedance where you want it.
transistors from a single batch are likely to be well matched my (non authporitive) reference lists BC557 has Hfe 110-330, I think A,B,and C variants of a transistor model usually offer higher Hfe,
I use it here, well a single bias resistor from the collector to the base, it works great to amplify a surplus 600 Ohm dynamic microphone connected to a soundcard input that (I think) expects an electret microphone.
admittedly I did experiment with different base resistors until I found one that worked well.
+-[4M7]--- C -------------->
22uF | to soundcard || | Q1 --->>---||---+---- B BC547B +--------->
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