Adolfo Garcia
Vice President - Applications & Marketing
After earning a BSEE from Santa Clara University, I spent the first 10 years of my career in high-speed bipolar IC design and, during that period, I earned an MSEE from UCLA. Since then, I’ve held various positions with increased responsibilities in applications, marketing, and business unit management at Analog Devices, Linear Technology, Micrel, and Advanced Analogic Technologies. My engineering background has primarily focused on data converter, amplifier, and power mgmt functions and that experience has lead to the definition and market launch of over 70 high-performance analog IC products.
AAGarcia
's contributions
- 04.15.2013
- A new 21st-century shootout at the O-K Corral?
- Thanks, Mr. McGilvra, for your comments. Yes, oftentimes to achieve high-performance in analog, the simplest approaches yield the best results. Also, thanks for your endorsement - please do keep posted as more Touchstone Semi analog products are on their way...
- 04.15.2013
- A new 21st-century shootout at the O-K Corral?
- Thanks for your comment, bikeron. You're right - white goods are just one example where SOCs with software dominate.
- 02.18.2013
- The engineering interview – interesting and challenging questions, Part 3
- Good Readers - Thanks to you who've "tweeted" and "liked" this week's blog post. Here're the answers to the 4 questions posed: 1) Below the LPF's cutoff frequency, the network analyzer would display a spectral density of 1mV/rt Hz. Why? A Gaussian noise source with a 1e-6 V/rt Hz spectral density and amplified by a noise-less x1000 amplifier would result in a 1e-3V/rt Hz or 1mV/rt Hz spectral density. 2) Since the LPF's cutoff frequency is 6.4kHz, you would have had to know that the spectral energy of a Gaussian noise source passed through a single-pole 6.4kHz LPF exhibits the same spectral energy as the same source passed through a brick-wall filter with a cutoff frequency of 1.57 x 6.4kHz. Therefore, the equivalent noise bandwidth of this example is ~10kHz. 3) The RMS value of the noise signal as displayed on the network analyzer would be 1e-3V/rt Hz x sqrt(10e3) or 100mVRMS - pretty basic stuff. 4) As displayed on an oscilloscope, the in-band peak-to-peak signature of the Gaussian noise, amplified by 1000, and passed through a 6.4kHz LPF would be ~660mVPP. Since Gaussian noise is statistical, a crest factor of 6.6 (ratio of peak-to-peak to RMS) indicates that the probability of peak-to-peak values exceeding this threshold is less than 0.1%. Ergo, the number of divisions needed to display 660mVPP signal would be 6.6 divisions since the vertical deflection was set at 0.1V/DIV. Thanks for playing...
- 12.10.2012
- Balanced (?) audio line receivers and the physics of fields
- Ok, Dear Readers - Now that a few days have elapsed since posting this blog, here are the answers to Qs 1 & 2 above: 1. For the circuit in Fig 1a, the 'differential' input currents are out of balance 3-to-1 (ib-/ib+). For the circuit in Fig 1b, the 'differential' input currents are out of balance 2-to-1. Reader Whitlock correctly pointed out that common-mode input currents are balanced; therefore, the common-mode input resistance is 25kohm in Fig 1a and 12kohm in Fig 1b. He also correctly points out that either level 'ain't' large enough. 2. Reader Whitlock was also correct in his suggestion to bring the input 'differential' currents back to symmetry by inverting the output signal and applying that result to the REFERENCE terminal in each circuit. No change to the input common-mode currents yet the circuit overall gain is reduced by one-half when implemented. Thanks for reading and new comments are always welcome!
- 12.10.2012
- Balanced (?) audio line receivers and the physics of fields
- Thanks, Videobub, for your proposed solution. Given that you're suggesting adding two additional components, what happens in production when from batch-to-batch the internal resistors (25kohm in Fig 1a and 12kohm/6kohm in Fig 1b) exhibit 20% tolerance? How does your external compensation network address that fact?
- 12.10.2012
- Balanced (?) audio line receivers and the physics of fields
- Ahh - It's refreshing when fine details are slowly uncovered - thanks, Bill, for pointing out correctly that there are two mechanisms in play here - common-mode input current behavior and differential-mode input current behavior. When I posed the questions, I purposely left out that detail hoping that a reader would make the distinction (or find the discrepancy). You are absolutely right about the math with respect to common-mode input current behavior and what's really important in maintaining excellent common-mode rejection across the entire interface - when both inputs are driven at the same level, then the common-mode impedances (and therefore the input currents are matched to the accuracy of the input Rs (25kohms in Fig 1a and 12kohms in Fig 1b). However, differential-mode input currents remain unbalanced and, yet, there is still a fix out there that simultaneously satisfies both conditions...
- 10.29.2012
- Electronics engineering and the art of automatic transmission repair (or is it the other way around?)
- Greetings, Bill (aka Guru of Grounding) - From your picture, I knew that you looked familiar to me as I worked very closely with Walt Jung during my tenure at ADI and I knew that you and he knew of each other's work and may have communicated over the years. I distinctly remember attending an AES convention some time ago (I won't mention *how* long ago) where you presented (and I'm paraphrasing here) the "Pin 1 Problem." From that and work Walt and I had done on line-driver/receiver pairings using the SSM2142/SSM2143/SSM2141, I developed a real understanding of the underlying principles in improving CMR and a greater appreciation/understanding for the *right* way to execute signal grounding...I for one thank you for all your great work in the field. Adolfo
- 10.29.2012
- Electronics engineering and the art of automatic transmission repair (or is it the other way around?)
- Readers - From the comments below, some of us who have pursued a career in electronics may have done so from working on cars in their youth (and may still do). I wonder how many of us started working on cars as youngsters and then gravitated to electronics. I, for one, worked on cars in my youth (and still do), but I gravitated to electronics because of my interest in mid/high-end audio. Share with us your path to "enlightenment."
- 10.29.2012
- Electronics engineering and the art of automatic transmission repair (or is it the other way around?)
- Guru of Grounding - Thanks for your reply and how very cool to have worked side-by-side with your dad on "souping up" the family car. Pretty sharp guy - your dad - to have figured out the 3-dimensional operation of a hydraulic transmission and how to make it behave the way he thought. Back then, the only way a transmission shifted gears was based on manifold vacuum, thottle position, and gear selection. Add in yet another variable - modifying the valve body as your dad did - and the results can be completely different! In fact, it wasn't until your reply that I realize that an automatic transmission is really a mechanical analog computer. How very cool indeed!
- 10.08.2012
- The engineering interview – Interesting and challenging questions, Part 2
- First to All Readers: Thank you for your interest in this blog by the looks of the Facebook "like" counter. It's always reassuring that what we pen is timely, relevant, and interesting. To Readers alexsconway, chaka, and john.connell_#1: Nicely done as you all arrived at the correct answers. The circuit as posited contains 3 poles and two zeros - I used the term "primary poles and zeros" to make sure other readers - familiar with the design of discrete or IC op amps - avoid complicating their answers. The 1st pole is embedded in the op amp and the other two poles as well as the two zeros are generated by the feedback network. To make it easier, one could have stated that R1=R2=R and that C2 = C or C1=10C2=10C; thus, Reader alexsconway has the location of the feedback network's two poles and two zeros correctly located. The shape is indeed as he describes with the ckt's response initially starting at unity (all Cs are open circuits), breaking in an upward direction because of z1, leveling off at +2 because of p1, breaking in a downward direction because of p2, and finally leveling off at unity again because of z2 (above which all Cs are short circuits). This is all assuming that the break frequencies are completely enclosed by the op amp's loop gain response (for this academic exercise, a perfectly valid assumption). One would by inspection recognize this circuit as the combination of single-pole high-pass circuit and a single-pole low-pass circuit; thus, a bandpass response. A set of "well done" kudos to all!
