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.
What causes the reflection? Why doesn't it occur with other
frequencies? (if it doesn't)
Reflections occur at all frequencies when there is a mismatch in impedances. At low frequencies, such as audio, these reflections are difficult to see but they are there all the same. Reflections are generally said to be significant when the frequency is high enough AND the interconnection between sender and receiver is long enough. Somewhere in the order of about a tenth of a wavelength or bigger is a general rule of thumb.
At 20 kHz, the wavelength (in 100% speed of light cable) is about 15 kilometres and if you had a cable of about 1.5 km length you might start to see the effect of reflections.
However, if you had a 100 MHz transmitter, you might see the effect of reflections at 300 mm.
Consider a battery and a lightbulb. The lightbulb is connected to the battery with a switch. The battery and switch are at one end of a lossless 10 km cable and the bulb is at the other end. When the switch closes, how much current is drawn from the battery? - how can the battery know how much current to supply in that instant? The answer is it can't - it supplies what the cable demands and, for a 50 ohm cable an appropriate current is supplied. If the voltage were 10V then the current would be 200mA.
This travels down the cable (at a power of 2W) until it hits the lightbulb. The bulb may have an impedance of (say) 100ohms - it only wants 100mA at 10V but it gets 200mA - there is a mismatch and the excess power gets reflected back up the cable to the battery and switch. This power can't be dissipated in the battery so it gets relfected back and forth. Of course, cable has real losses and these eat away at this reflection and the system stabilizes with 100mA flowing down the cable. This is a simplifed explanation.
Does this help you understand?
Best Answer
Another name for that config is emitter follower. It provides current drive ability with voltage gain of (almost) 1. It basically just buffers the signal.
Nope. The semiconductors would blow up. Assume that each semiconductor junction (diodes and transistor B-E) has a voltage drop of 0.7V when forward biased. Try and simulate it.
It may not be the best, but most simplest and practical two transistor push pull configuration that is good enough for whatever the puropose is - such as learning the basics from textbooks. If you need better performance for another purpose then you can easily find better but more complex configurations that will fullfill the requirements of those other purposes.