# Help me understand what happens with the circuit when an AC signal is applied

Consider this circuit. Transistor is 2N2222A. Ic is chosen to be 10 mA. Hfe is 225 for that current. Resistors are calculated to have Vbe = 0.7V and Vc = 4.5V.

I am trying to understand what happens when I inject an AC signal at the base.

Suppose an AC signal that is 1Hz, 0.02V of peak voltage (varies between +0.02V and -0.02V). The capacitor is huge (100F), just to be transparent to 1Hz.

What I think happens is this: as the input increases from 0 to 0.02V, base increases from 0.7 to 0.72V. So the base resistor is now connected between Vcc (9V) and 0.72V. Base current is now less than before, because the difference in potential across base resistor is less than before.

If base current is less, collector current is less, right? If collector current is less, the difference in potential across the collector resistor is less. If so, Vc increases, right?

My simulator says Vc (collector voltage) decreases...

What is wrong with my analysis?

YES, I know this transistor configuration for amplifiers is bad. I am just trying to see what happens to this circuit.

• Close, but try not to think in terms of base voltage but base current. – winny Sep 19 '18 at 19:24
• @winny - thanks. Can you expand your explanation or make it an answer? If I must consider the input current, how do I calculate it? – SpaceDog Sep 19 '18 at 19:33
• No, this seems like homework so I will only push you in the correct direction. You need to develop a method for solving this. :-) – winny Sep 19 '18 at 19:53
• Not a homework and I have no clue on how to solve that. Thanks anyway. – SpaceDog Sep 19 '18 at 20:05
• Keyword would be small signal analysis. – winny Sep 19 '18 at 20:37

## 2 Answers

The base-emitter junction looks rather diode-like. When forward biassed, it's a very low impedance. Increasing the voltage on the base by even a few mV increases the base current dramatically.

With a grounded emitter amplifier, driven by the voltage source you've shown, it's best not to think in terms of hfe, but of a linearised gm. That is a transconductance, a change of collector current with base voltage, centred around a mean bias point, for very small base voltage signals.

If this seems complicated, it's because you're approaching the single transistor amplifier using a difficult route.

• Thanks. I am sure your explanation is perfect but sorry, because I am still learning and 100 million light years behind you, your explanation sounds Klingon to me. :) ... – SpaceDog Sep 19 '18 at 19:45
• @SpaceDog Yes, I quite understand. Go back to your previous question on bias, and start from the circuit in Dan Mills' answer. That can be done entirely from hfe. You can use big voltage signals, and everything will behave. The problem with a grounded emitter amplifier is it's very non-linear, and strictly small signal, complications you don't need when you're starting out. Best think of the common emitter as just a saturating switch, not an amplifier (for the moment). Although Dan's circuit has 'more components', those components simplify the analysis. Trust me on that. – Neil_UK Sep 20 '18 at 8:02

Your capacitor is transparent at 1Hz, so current flows through the capacitor into the base. Yes, the current through R11 decreases a little, but the increased current through the capacitor is the larger effect. So, changes in the ac source voltage cause proportional changes in the base current.

• ah, do you mean that the current introduced by the signal will make the base current to increase, so the collector current? How can the effect be calculated? I men, how do I know how much current is coming from the input and the real base current? Simulator says the input current is 50uA but how do I calculate that manually? – SpaceDog Sep 19 '18 at 19:28
• Yes, exactly. To calculate the effect you really need an accurate equation for the base current as a function of base voltage. There is an exponential relationship between the two. – Elliot Alderson Sep 19 '18 at 19:37
• and how do I obtain that equation? – SpaceDog Sep 19 '18 at 19:42
• It's basically the same formula for the current through a PN junction (diode) as a function of the applied voltage. You can use your favorite search engine to find it. However, there is one parameter, the ideality factor, that varies roughly between 1 and 2. The manufacturer of your transistor measured some actual I/V data and fit a curve to the equation, adjusting the ideality factor for the best fit. You might be able to glean that information from the transistor's SPICE model. However, most of us would just run SPICE in the first place. – Elliot Alderson Sep 19 '18 at 20:04