Differential Amplifier Common Mode Problem

[chaoseverlasting]

There are a couple of things I don't understand about the DA in general and when we use it in the common mode in particular (same input ac voltage to both input terminals). I am using Electronic Circuit and Theory by Louis Nashelsky and Robert Boylstad (p.591 for DA).

When we set the operating point, we do it using DC voltages which sets the values I_c, I_b and I_e.

In the ac analysis, however, we use the values of AC voltages to calculate the voltage across the output resistor. For the r_e model of the transistor, the value of r_e in the ac domain is calculated by r_i=\frac{26mv}{I_b} where I_b is the ac current.

However, r_e=\frac{26mv}{I_e}=\frac{26mv}{\beta I_b}=\frac{r_i}{\beta} is calculated using the DC value of I_e.

How is that possible? The DC and AC values of currents are bound to be different, but they are used interchangeably!

I thought that the net current would be the ac+dc values of the current (eg. I_{e net}=I_{e ac} + I_{e dc}).

Sorry for the long post, I couldn't help it.

 

[chaoseverlasting]

Anyone? Please?

 

[cabraham]

The Re value is dependent upon the dc value of current. The base-emitter junction, herein b-e jcn, has an logarithmic/exponential V vs. I (I vs. V) characteristic. Modeling the b-e jcn as a resistor is not valid under general large signal conditions. Re is a small signal value taken as a tangent to the I-V curve at a specific value of Ie. Ie, Vbe, & Ib denote dc values, & ie, vbe, & ib denote small signal ac values.

Since the curve is non-linear, the tangent slope changes depending on where you are on the I-V curve. Hence Re, which is the slope of V vs. I, or the inverse slope of I vs. V, varies with the dc value Ie. At larger values of Ie, the slope increases, and Re decreases.

Hence a larger Ie, emitter dc value or bias, results in a smaller Re which is an ac value. Have I explained it well? I used to be an adjunct professor of electronics, and I used Boylstad's texts.

Claude

 

[chaoseverlasting]

So we basically ignore the ac signal while setting the Re value, and this value depends on the dc bias.

Then the current we measure using the ac analysis is not Ib but ib (the ac current), and for this current Re is just a simple resistor. Also this current does not have any effect on Re whatsoever... is that right?

 

[cabraham]

Only a slight effect, too small to worry about as long as the ac deviation is small. When the b-e jcn deviates from its steady state operating point, the slope of the I-V curve changes, hence Re changes. If the ac signal is not small, but large, then Re changes with the signal, and this modulation of Re results in nonlinear gain and consequently distortion.

Claude