# Bypass capacitor in common emitter input impedance question

I can't seem to sort out how the bypass capacitor and the input impedance interact on this. I designed this simple CE amplifier stage (it doesn't do much of anything and it's mostly an example problem). I computed the input impedance using 10k||50k||hFE(re) and got 1408 ohms for Zin. [I used 150 for beta.]

I simply don't know how to find out if my calculations are correct. I imagine that I can't simply apply Ohm's Law to look to see if I'm getting the correct current through my source resistance (100 ohms), right? When I run the sim as pictured with a 20mV peak-to-peak signal input, that source resistor eats up 8 uA (peak-to-peak), but of course this changes quite a bit as I adjust the cap. Shouldn't my expression for Zin take that into account the value of the capacitor or its reactance at 1kHz somehow?

Does anyone have a clear source that will describe a good sizing method for the bypass capacitor if I'm building something to amplify 20Hz-20kHz? Also, can anyone tell me whether my 1408 ohms figure for Zin looks correct?

In case it's helpful, this is simply a learning circuit for me -- no real-world application and a lot of the component values made up to practice. My interest down the road will be in audio, so perhaps imagine this as a stage in an instrument amplifier.

• The bypass capacitor (bottom right) bypassess the resistor in parallel with it (1k, bottom right). It needs to be sized so that its impedance is much much lower than the resistor at the specified frequency range.
– AJN
Commented Jul 3, 2020 at 4:43
• It will help if you can name the components since some of them have the same value.
– AJN
Commented Jul 3, 2020 at 4:44
• The expression for input impedance doesn't contain 1k resistor or the bypass capacitor if they are designed to be much smaller than re at specified frequency range.
– AJN
Commented Jul 3, 2020 at 4:47
• To clarify, I have added in labels. I'm mostly concerned with sizing C3. Commented Jul 3, 2020 at 11:28

Here is how I compute Zin. [I discuss both 1,000 Hz Zin, and at high frequencies.]

Don't forget the transistor is in parallel with the 2 base biasing resistors

step A) Let us examine what BETA has to scale up.

At 1,000 Hz, that 10uF capacitor looks like

• 0.159 / (1,000 * 10 uf) = 0.159/0.01 = -j 15.9 ohms

What else? The Re of 1,000 ohm ---- we will ignore that because is so much greater than the capacitive impedance.

The small_signal emitter_base diode incremental resistance (computed as the transconductance, and inverting to get resistance) is

• 1 / (Iemitter / 0.026 ) = 0.026 / I_emitter

With I_emitter of 2 milliAmps == 2+volts / 1,000 ohms

the reac == 0.026 / 0.002 = 13 ohms.

Thus the Zin == beta * (13 - j15.9) == ~~ 100 * 20 == 2,000 ohms (at some phase shift) == [2,000/45 degrees]

step B) now compute the complete Zin

• [2,000/45_degrees] ohms in parallel with 10,000 and 50,000 ohms

=================================

For high frequencies, the C_Miller_Multiplication of base_collector capacitance becomes significant. (you can greatly minimize this, by adding a common_base transistor in series with your original transistor, with appropriate base biasing of the new transistor; this is called cascoding)

The (assume 10 pf base collector) is scaled up by the voltage_gain.

Tho beta will gradually roll off above Fbeta, which may be 3MHz for 2N3904; let us pick 1MHz as our "high frequency" and use BETA = 100.

At 1,000,000 Hz / 1,000 Hz, the value of Cemitter is << 1 ohm, so only the reac of 13 ohms (incremental diode resistance) gets scaled up by BETA.

Thus the previous transistor Zin is 13 * 100 = 1,300 ohms IN PARALLEL WITH C_Miller_Multiplication impedance.

We need to compute the voltage gain at 1MHz.

That is (Rcollector || Rload) / Remitter = (X || Y) / 13

Gain = 3K || 16K / 13 ohms ~~ 2,500 / 13 == 200X

The capacitance from collector side is JUST the 10 pF.

The capacitance from base side is (Av + 1) * Cob = (200 + 1) * 10pF = 2,010pf. Or 2 nanoFarads. This huge input capacitance must be charged from your signal source.

At 1MHz, using Zc = 1/ ( 2 * PI * F * C), we have Zc_miller == 75 ohms.

The requirement to CHARGE that 10pf cap, internal to the transistor, has caused the Zin to collapse to a mere 75 ohms.

Because of the capacitive nature, this Zin increases to 750 ohms at 100,000 Hz.

And becomes 7,500 ohms at 10,000 Hz. Thus audio design may be able to ignore C_Miller_Multiplication. Or not. High frequency distortion may result.

Notice the Zin now is very much dominated by the Input Capacitance, which is dominated by the large amount of CHARGE needed to charge and discharge that 10pF.

• Thank you for your answer! What I'm still wondering about is... is the word "lemitter" a typo, or some term I'm unfamiliar with? Commented Jul 3, 2020 at 11:48

Does anyone have a clear source that will describe a good sizing method for the bypass capacitor if I'm building something to amplify 20Hz-20kHz?

If you are building an amplifier with any half-hearted attempt at a decent level of quality, then you will never use an emitter capacitor like this; you will always put it in series with a resistor so that the gain is properly controlled and not maxed-out at medium to high audio frequencies. Without a resistor in series with the emitter capacitor you will get significant distortion and low input impedance.

So, it comes down to how much gain do you need from your circuit. At the moment, it is ill-defined because the capacitor acts like an AC short and $$\r_E\$$ is the only definer of gain. Unfortunately $$\r_E\$$ is highly affected by any collector/emitter current changes and so, it will distort the output signal because as the signal changes, so does $$\r_E\$$.

$$\r_E\$$ is also affected by temperature so, gain (albeit badly distorted) will vary as the circuit warms or cools. The low cut-off frequency is also poorly defined with adding a resistor.

Do your self a favour and put 100 ohms in series with the 10 uF capacitor and, to achieve decent low frequency gain control make the 10 uF into a 100 uF capacitor - this will produce a 3 dB point at around 16 Hz and much more compatible to a reasonable audio spectrum.

If you need more gain then use two stages of amplification.

• Thanks for the suggestion to use a resistor in series with C3. Can you look back at the original post and help me out with this? If we assume for the moment that the response of the posted circuit is exactly what I want using just re for gain (even though we know it's not and your suggestion is better), would I have computed the input impedance correctly at 1400 ohms? And, more importantly, how could I verify that I've gotten it correct on my own? MANY thanks! Commented Jul 3, 2020 at 11:52
• Yes, your calculation seems about right. 2.3 volts across Re means emitter current is 2.3 mA and $r_E$ will be about 10 ohms based on this. So, beta (150) x $r_E$ = 1500 ohms will be pure base input impedance and, that is in parallel with 8333 ohms = 1271 ohms. Then Rs makes that up to about 1400 ohms give or take. Commented Jul 3, 2020 at 12:11
• Thanks again. Is there a way for me to verify that I have the right input impedance as I work to design other circuits? I don't suppose I could do something like dividing the peak voltage across Rs by the peak current through it and hope to get the input impedance as a result, right? Commented Jul 3, 2020 at 13:13
• Sure, use a simulator is my strong advice - it's an independent means of verification. Commented Jul 3, 2020 at 13:19