How good a replacement is a rather complex question, mostly revolving around what you are looking to do with it. It looks from the datasheet that the 2n9013 has twice the Ic (continuous collector current) of 500mA, compared to a peak in the 3904 of 200mA. The 3904 is definitely intended to be a signal amplifier, and has a gain bandwidth product of 300MHz. The 9013 datasheet I found doesnt even have GBP, which to me tells me its much more intended as a current buffer, than as an amplifier. The fact that they tote the Hfe (essentially, but not the same as beta) as being linear suggests to me that it expects to have large signals fed into it, and not do much more than buffer an input to an output (hence the need for linear Hfe, but no mention of GBP). The other thing, that line 1W OUTPUT AMPLIFIER OF POTABLE RADIOS IN CLASS B PUSH-PULL OPERATION (I was as shocked to see the word potable is actually what they said on the datasheet. I'm like 90% sure they meant PORTABLE, not that its safe to DRINK the transistor). That line thats "worse than Chinese" means that its intended to be used with its PNP complimentary transistor the 2n9012 on the output stage (hence the class B push pull bit) of a power amplifier.
The long and short of it is this: its a good replacement as long as what you are replacing the 9013 with works the same for you as it did before. That is really all that matters. Generally tho, most designs are not THAT component parameter sensitive that they wont work with the wrong VAGUELY similar part.
Good luck!
You have a crude Class A amplifier there now.
Input to base.
Output from collector.
Gain is about Rc/Re = 10k/1k = 10.
Brief answer re base input current appears at "cut to the chase" below, but ...
Close enough,
- Ib = (Vdd x Rbu/(Rbu+Rbl) - Vbe) / Re / Beta
Don't even start to try and wonder about it or which resistor is which.
By the end it should make sense.
Calculate voltage at base point with transistor removed.
Call 110k = Rbu= R_base_upper.
Call 10k connected to base Rbl = R_base_lower.
Call Voltage where base connects Vb.
Call 20 V supply Vdd
Vb = 20v x Rbl/(Rbu+rbl) = 20 x 10/120 = 1.666V.
V base to emitter = Vbe
Vbe for an operating silicon transistor is about 0.6V
Can be somewhat different but use 0.6V for now.
As Vb = 1.666V then
Ve = Vb - Vbe = 1.666 - 0.6 = 1.066V.
Ve appears across Re (1K) so I_Re = 1.07/1000 = 1.07 mA.
We can call this 1 m or 1.1 mA close enough for this example. I'll use 1 mA for convenience.
Now "it happens" as a function of the formulae related to transistor action that the impedance of the emitter is 26/I for I in mA.
"Don't ask why for now" is good advice. The answer is - because as you will discover in due course, that's the way it is.
So at 1 mA Re =~~ 26 ohms. At 2 mA Re = ~= 13 ohms. At 0.5 mA Re ~= 52 ohms.
This is the effective resistance of the emitter junction to current flow. I'll call that Rqe rather than Re as I've already used Re as the external emitter resistor.
Call transistor current gain Beta, because that's what it is traditionally called for traditional reasons.
If you look into the base you effectively see Re multiplied by the current gain of the transistor. That's because for every mA that flows ij the emitter circuit you only need 1/Beta as much in the bases ciurcuit to control it so it APPEARS that the resistance is beta times as large.
Assume our example transistor has Beta = 100. This is well inside the range of normal for small signal transistors.
Looking into the base we see Beta x Resistance in base circuit =
- Rbase to signal = Beta x (Re + Rqb)
= here about 100 x (1000 + 26) = 102600 ohms or ~= 100 k ohms.
Note I said "to signal" as DC will or may have its own rules.
(All obey the same rules but other factors affect what is seen -
eg if we put a 10 uF capacitor across Re it is approximately 0 ohns to AC at audio signals so "vanishes". I said before that gain was ~= Rc/Re = 10
That was before we allowed for Rqe and before we bypassed Re to remove it for AC.
If we do the above gain becomes about Rc/Rqe = 10000 / 26.4 =~ 385
Cut to the chase:
Now, during the hand waving and mirrors we hid something. I said Vb worked iut at 1.66V. The current down the Rbu + Rbl string to ground will be i=V/r = 29/(110k+10k).
This current is just enough to set Vb = 1.666V as we calculated BUT with 1.666v on Vb the same current will flow via Rbl to ground. ie no base current will flow. Your original questiion was "how much base current" and that seems to say "none". However, with no base current the transistor will turn off, Ic will drop, Vre will drop and so Ve will drop causing more than 06V to appear on Vne so the ransistor will turn on and restore. Vb will fall just enough to draw the extra current needed fro Rbu and to reduce the current in Rbl. It will do this automatically and it will draw "just the right amount".
JTRZ (h=just theright amount is enough such that Ib = Ie/Beta.
So we see that is more and less to what happens than appared. The correct example is dynamic and needs load lines on a graph. But "bood enoug" result goes. Based on above.
Close enough,
- Ib = (V+ x Rbu/(Rbu+Rbl) - Vbe) / Re / Beta
After going through the above that should not be as scary as it would hev been previously.
E&OE - could easily have typo'd something there.
Please point out if errors seen.
Best Answer
If you look at the datasheets for the 2N3904 and BC107, the main specs to look for are:
So as you can see, most of the specifications are nearly the same. Some specs, such as V\$\small_{CE}\$(sat), f\$\small_{T}\$ and H\$\small_{FE}\$ are specified under certain test conditions, which are not exactly the same for both transistors. The minimum switching frequency supported by the 2N3904 is about double that of the BC107. Unless you are running at very high speeds, the two BJTs should be interchangeable.
Digi-Key uses I\$\small_{C}\$, V\$\small_{CEO}\$, V\$\small_{CE}\$(sat), I\$\small_{CBO}\$, h\$\small_{FE}\$, P\$\small_{D}\$, and f\$\small_{T}\$ as parameters in its selection tool.
As far as costs go, the 2N3904 is a very popular transistor, manufactured in many millions per year. The BC107 is much less common. Digi-Key currently has over 400,000 of the through-hole 2N3904 in stock and ready to ship at 18¢ (12 Rs.) apiece for singles or as low as 3¢ (2 Rs.) in quantity, but has only 830 of the BC107 at $1.91 (127 Rs.) apiece, or as low as 91¢ (60 Rs.) in quantity. So you got a really good deal on the BC107s at Rs. 15.
The prices of the surface-mount version of the 2N3904 at Digi-key are about the same as the through-hole, and they have 110,000 of those ready to ship. As far as I know, there isn't a surface mount version of the BC107, due to lack of demand.
In addition to the savings afforded by volume, the 2N3904 is manufactured in a plastic TO-92 case, whereas the ZBC107 is made in a metal TO-18 can, which would be much more expensive to make: