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EE 105: Devices & Circuits

Wednesday, March 21, 2012

Midterm Friday; OBONNS.

Today: Graphing and calculating $G_m$, $R_o$, $A_v$, keep talking aboutfrequency response.

Graphings

So, the thing to do: you know you're going to have (if I've gotsome $A_0\sin(\omega t) \mapsto \abs{H(j\omega)}A_0\sin(\omega t + \angleH(\omega))$. If I give you a plot of angle or frequency, one way oranother, you've got to find the angle of $H(j\omega)$. Also, find max/minvoltages (determined by gain at various stages). Should make sure that thosepoints are marked (circled, even) so Pister can see that we know what's up.

If it's greater than 0, it's called "phase lead". You want to find andlabel and mark zero-crossings and min/max in time (which is really inphase). Points move left.

If it's less than 0, it's called "phase lag", and points move right.

Calculating $G_m$, $R_o$

There have been some questions about this on Piazza. There's the basics,like when you're looking into a collector or a drain, you're going to see$r_o$; when you're looking into an emitter or a source, and thecollector/drain are at some AC ground, you're going to see $\frac{1}{g_m}$,and then there's the blocks that you know and ought to recognize/use incircuits, which are: what happens if I have some emitter degeneration orsome source degeneration ($G_m \equiv \frac{g_m}{1 + g_m R_s}$ for MOS,$R_s \equiv R_E \parallel r_\pi$); if I'm looking into the source/emitter,then this looks like $\frac{1}{g_m}\parens{1 + \frac{R_D}{r_o}}$. Note thatfor a BJT, you have $r_\pi$ in parallel with this whole thing.

For example: steps for finding gain: $A_v = -G_m R_o$. In general, this ishow you'll find gain. $G_m = \frac{I_o}{V_i}\bigg|_{v_i=0}$. Straighttaylor approximation: this is why we set $V_o = 0$ when calculating $G_m$and $V_i = 0$ when calculating $R_o$.

When doing $G_m$ calculation, we just don't care about anything above thejunction.

Mentally, what people like Razavi do is say "aha! I've got a bias voltageon that gate; this thing now is one of my basic elements! Aha! I've got a$\frac{1}{g_m}$ to ground, and now I can draw my small signal model." Andso the current coming out is roughly $g_m v_i$. The only time you look upis if there's a current source up there that depends on one of the lowernode voltages.

The reason why we call it $g_m$ is because it's mutualtransconductance.

So now we're not so sure. What I see is $\frac{1}{g_m}\parens{1 + \frac{1 /g_m}{r_o}}$ when looking into source.

So now we've figured out what our $g_m v_i$ is. Current dividers.

So I wanted to do a couple more examples. For example, source follower(recall, measure output at source). We know $A_v = -G_m R_o$. So what is$R_o$? It's $\frac{1}{g_m} \parallel R_s$. $G_m$ is a little tricky. Drawthe small signal model, then do KCL at the output node. Note that sincewe're grounding $v_s$, $v_{gs} = v_i$, and no current is flowing throughthe source resistance (or the output resistance, even!). Thus $G_m \equiv-g_m$. What does this mean? We're shoving current out, instead of shovingcurrent in.

So finally $A_v = -G_m R_o = \frac{g_m R_s}{1 + g_m R_s}$

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