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In cadence, that can become quite some task dependent on how well things are set up for you and your level of confidence with the tool. You would need to use, for example, the verilog-AMS simulator/netlister, which requires some extra licenses. My opinion might be a bit out-dated, but I think that Cadence has not really made life that simple for m
Multi-phase buck-boost is of course feasible and used in high power applications. It's however mostly using the topology boost-DCLink-buck. Paralleling multiple 4-switch/single-inductor buck-boost converter is nevertheless possible and e.g. supported by a sync feature of existing controller chips. I don't understand about "starts to get out of h
You need to perform the circuit analysis exactly, copying formulas from a similar looking structure (miller capacitor) doesn't help. The present circuit uses an OP, which has ideally zero output impedance and infinite voltage gain. Respectively you get different equations. The low pass filter circuit is nevertheless simple enough that any analog
That's an outstanding ignorant answer to your question, I think. May be from a sales guy... They actually don't deserve to earn money with technical products. You better email the UK manufacturer. Guess it's too late this week, but you can try. In fact, any definition how to talk to the device can be considered as protocol. Most likely it eit
If you have a ?C connected to the FPGA, you can simply pull down NCONFIG line to cause a reboot after updating the EPCS flash and don't need ALT_REMOTE IP. nevertheless it's possible to use the ALT_REMOTE reconfig command if you want to trigger it from the FPGA logic.
I conclude that the term "ethernet" in this thread actually means RJ45 connector, and that Omron sloppily designates a simple party line RS-485 topology "daisy chain". nevertheless the shown configuration should work.
The discussion on a 2 process and 1 process model has been discussed here many times, and will be done again in the future (because most people do not appropriately search the net for the answer the need). nevertheless, this article was published a couple of days and discusses bang-on, why a single process model. MicroZed Chronicles – Finite
Step response gives a mixture of linear low-pass with large signal behavior, e.g. due to slew rate limitations. nevertheless it can useful to test the circuit in time domain with signals similar to that expected in the final application. In any case it depends on the cut-off frequency to OP speed relation if you see an ideal low-pass response or
For my own information, what is the down side of using a gated clock like I'm doing? It's generally avoided when driving internal logic due to timing problems. I believe it's OK for external clock output with the suggested negedge triggered gating FF. nevertheless I prefer a straightforward single edge driven logic scheme.
No, it is not a typo, open loop TF= (130/s^2) *(s+1)/(s+10), closed loop TF = 130s+130/s^3+ 10s^2+130s+130 If you want someone else to understand what your are writing, use at least standard programming writing of mathematical expressions. Closed Loop TF (unity feedback) = (130s+130)/(s^3+ 10s^2+130s+130) Yo
There is a mistake in the book. Is it so? The 1st edition doesn't even mention the term transition frequency or fT respectively ωT in this context. Not sure about the 2nd edition. fT is the frequency at which the current gain of the MOS transistor becomes unity (at which Iout/Isig=1, see the figure). I see
As already clarified, the RS-232 standard doesn't provide a high-Z feature. As for standard RS-232 devices like PCs, the answer to your question in post #1 and #3 is clearly no. nevertheless some devices extend the RS-232 standard towards a multi-drop bus. There are RS-232 drivers with TX high-Z feature, and devices that use device addresses wit
As shown above, there is carrier leakage caused by a DC offset in I/Q signal. And, in terms of spectrum, the carrier leakage component is at the center of signal.This image shows a spectrum from a direct conversion modulator. nevertheless, 142208 As shown above, in terms of sp
There is a good story going back to Benjamin Franklin (famous for numerous accomplishments). He performed early experiments with electricity, and formulated theories. As he tried to figure out which direction electric current travels, he 'guessed' that it moves from a positive terminal to a negative terminal. His guess became the popular conventio
In the DC condition the feedback capacitor offers infinite resistance and so the integrator circuit will be like an inverting opamp amplifier with infinite feedback resistance (Rf = ∞). An amplifier with Rf=∞ would give a voltage gain of Av=∞. In practical terms, no real-world opamp has infinite gain, but nevertheless has a ver
That really sounds like a custom made transformer. I don't have even a remote idea of an application for of-the-shelf transformers with a similar specification. The specifications looks feasible nevertheless.
I agree that the datasheet lacks a clear specification of input common voltage range for G=0.1 configuration, similar to the figure for G=10. It's nevertheless possible to derive the information from the datasheet. You have the specification under maximum ratings Input Signal (G = 0.1), Voltage 11 ? Vs It can be read as a roug
A curious schematic nevertheless. It looks like the supply to the IR2153 is crowbar protected from the core voltage of the transformer rather than a winding on it. Brian.
3rd OT crystal operation depend on a circuit suppressing the fundamental wave. My first problem with the schematic is that L10/C51 doesn't seem to be tuned for 63 MHz which would be expected for an overtone crystal oscillator. It might be that the original circuit oscillates at 63 MHz though, possibly with only a small margin, but fails on a bre
Hello everyone, I realized a small printed circuit with the ARES software from PROTEUS. The worry is a mistake that persists despite that I routed via a wired bridging. On the attached image you can see the white link (error) and the wired bridge that is nevertheless well connected. Does anyone know how to solve this proble