Wireline Transmitter

Posted on 2025-06-07 Edited on 2025-06-09 In analog

CML vs. SST based driver

Design Challenges Of High-Speed Wireline Transmitters [https://semiengineering.com/design-challenges-of-high-speed-wireline-transmitters/]

the resistance of MOS is not highly controlled -> \(R_T + Z_N\)

Peak power constraint of TX FIR

Due to circuit limitation, circuit cannot have arbitrarily large voltage on the output, i.e. a limited maximum swing. In order to create the high frequency shape, the best we can do is lower DC gain (low frequency gain < 1)

FIR is not increasing the amplitude on the edges
FIR is reducing the inner eye diagram

The maximum swing stays the same, \(\sum_i |c_i|=1\)

Circuit Insights @ ISSCC2025: Circuits for Wireline Communications - Kevin Zheng [https://youtu.be/8NZl81Dj45M?si=2a8FdfGNP6yBgIW8&t=829]

SST Driver

tx_leg.drawio

Sharing termination keep a constant current through leg, which improve TX speed in this way. On the other hand, the sharing termination facilitate drain/source sharing technique in layout.

pull-up and pull-down resistor

sst-evolution

Original stacked structure

Pro's:

smaller static current when both pull up and pull down path is on

Con's:

slowly switching due to parasitic capacitance behind pull-up and pull-down resistor

with single shared linearization resistor

Pro's:

The parasitic capacitance behind the resistor still exists but is now always driven high or low actively

Con's:

more static current

VM Driver Equalization - differential ended termination

\[ V_o = D_{n+1}C_{-1}+D_nC_0+D_{n-1}C_{+1} \]

where \(D_n \in \{-1, 1\}\)

vdrv.drawio \[ V_{\text{rx}} = V_{\text{dd}} \frac{(R_2-R_1)R_T}{R_1R_T+R_2R_T+R_1R_2} \] With \(R_u=(L+M+N)R_T\)

Normalize above equation, obtain \[ V_{\text{rx,norm}} = \frac{(R_2-R_1)R_T}{R_1R_T+R_2R_T+R_1R_2} \]

	\(D_{n-1}\)	\(D_{n}\)	\(D_{n+1}\)
\(C_{-1}\)	1	-1	-1
\(C_0\)	-1	1	-1
\(C_{+1}\)	-1	-1	1

Where precursor \(R_L = L\times R_T\), main cursor \(R_M = M\times R_T\) and post cursor \(R_N = N\times R_T\)

Equation-1

\(D_{n-1}D_nD_{n+1}=1,-1,-1\)

pre.drawio

\[\begin{align} R_1 &= R_N \\ &= \frac{R_u}{N} \\ R_2 &= R_L\parallel R_M \\ &= \frac{R_u}{L+M} \end{align}\]

We obtain \[ V_{L}= \frac{1}{2}\cdot\frac{N-(L+M)}{L+M+N} \]

Equation-2

\(D_{n-1}D_nD_{n+1}=-1,1,-1\)

main.drawio

with \(R_1=R_T\) and \(R_2=+\infty\), we obtain \[ V_M = \frac{1}{2} \]

Equation-3

\(D_{n-1}D_nD_{n+1}=-1,-1,1\)

\[\begin{align} R_1 &= R_L \\ &= \frac{R_u}{L} \\ R_2 &= R_N\parallel R_M \\ &= \frac{R_u}{N+M} \end{align}\]

We obtain \[ V_N = \frac{1}{2}\cdot\frac{L-(N+M)}{L+M+N} \]

Obtain FIR coefficients

We define \[\begin{align} l &= \frac{L}{L+M+N} \\ m &= \frac{M}{L+M+N} \\ n &= \frac{N}{L+M+N} \end{align}\]

where \(l+m+n=1\)

Due to Eq1 ~ Eq3 \[ \left\{ \begin{array}{cl} C_{-1}-C_0-C_1 & = \frac{1}{2}(n-l-m) \\ -C_{-1}+C_0-C_1 & = \frac{1}{2} \\ -C_{-1}-C_0+C_1 & = \frac{1}{2}(l-n-m) \end{array} \right. \] After scaling, we get \[ \left\{ \begin{array}{cl} C_{-1}-C_0-C_1 & = -l-m+n \\ -C_{-1}+C_0-C_1 & = l+m+n \\ -C_{-1}-C_0+C_1 & = l-m-n \end{array} \right. \] Then, the relationship between FIR coefficients and legs is clear, i.e. \[\begin{align} C_{-1} &= -\frac{L}{L+M+N} \\ C_{0} &= \frac{M}{L+M+N} \\ C_{1} &= -\frac{N}{L+M+N} \end{align}\]

For example, \(C_{-1}=-0.1\), \(C_0=0.7\) and \(C_1=-0.2\) \[ H(z) = -0.1+0.7z^{-1}-0.2z^{-2} \]

w = [-0.1, 0.7, -0.2];
Fs = 32e9;
[mag, w] = freqz(w, 1, [], Fs);
plot(w/1e9, abs(mag));
xlabel('Freq(GHz)');
ylabel('mag');
grid on;