Bipolar pulse generator - Page 1
Forum section: electronics / projects

author: ezalys
author_rank: Frequent Contributor
timestamp: February 07, 2022, 06:16:30 pm
content: I'm wondering if anyone knows a DC-coupled circuit for generating pulses from -10V to +10V that return to 0 volts. Generation from 0 to 10 volts is quite easy with a half bridge and some inverters. I just don't know how to design a bridge that will tolerate negative voltages.I understand that you could AC couple the output and just use a cap and invert the duty cycle. For reasons beyond the scope of the question, I can't do that. There's also the possibility to build an analog multiplier style circuit, but this would require a common to differential conversion in the end, and I'd like to avoid stuff with feedback, if only for the pedagogical challenge. I am, on the other hand, happy to AC couple or transformer couple the gate signal if that opens up some tricks.The next thought I have is basically a 2:1 analog mux, where the pulse switches from the 0V input to the pulse height voltage. How do such muxes work internally that can tolerate negative and positive input voltages?

author: TimFox
author_rank: Super Contributor
timestamp: February 07, 2022, 06:37:42 pm
content: I have seen circuits that use two matched current sources from a positive and a negative rail into a four-diode network that routes the current into a 50 ohm load resistor from one or the other based on the fourth terminal of the bridge. With low-leakage diodes, this can be very accurate.

author: ezalys
author_rank: Frequent Contributor
timestamp: February 07, 2022, 07:33:34 pm
content: Sounds really cool. I'm a bit confused how the voltage is set with the current sources, though.

author: ezalys
author_rank: Frequent Contributor
timestamp: February 07, 2022, 07:50:11 pm
content: Oh wait sorry -- it isn't... I guess this just ensures the current across the diode is constant.

author: TimFox
author_rank: Super Contributor
timestamp: February 07, 2022, 07:50:18 pm
content: Many pulse generators use an adjustable current source and a switch into the output jack. For 50 ohm systems, there is usually a switchable 50 ohm resistor inside the generator. My -hp- 8013B has separate negative and positive outputs, with a single front panel switch to switch both 50 ohm internal resistors. The output controls (separate for the two current sources) are calibrated for output voltage into either 50 or 25 ohms, where 50 can be either the internal or external load, and 25 is both loads together.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 07, 2022, 08:05:49 pm
content: So, half bridge on symmetrical supplies with tristate? Or three-level inverter if you need active driven zero?Here's a very crude waveform I made many years ago, just half-bridge into inductor with low duty. Is this the kind of symmetrical, RZ pulse you want..?Tim

author: ezalys
author_rank: Frequent Contributor
timestamp: February 07, 2022, 08:27:31 pm
content: Call it 3 level inverter with driven zero. I mean ideally it's two level -- one level is zero, and the other level is set by some supply that may be positive or negative. I was hoping there was some clever trick for doing that. The diode bridge sounds quite promising, though it seems like there may be some extra drops in there.

author: SiliconWizard
author_rank: Super Contributor
timestamp: February 07, 2022, 08:31:40 pm
content: You could of course use a single supply and a full bridge, if it's ok to have a differential output.

author: magic
author_rank: Super Contributor
timestamp: February 07, 2022, 08:43:13 pm
content: Okay, so you want a half-bridge which can either ground the load or connect it to a provided rail, which can be positive or negative, up to 10V.What's the control input signal?How fast it has to go? Propagation delay, transition time, permissible dead time?What's the load impedance?The diode bridge thing is basically like the transistor diamond buffer, but with diodes instead of transistors (think: transistors without current gain). Offset is close to zero if the diodes are matched.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 07, 2022, 08:45:07 pm
content: The solution for signal purposes, is the same as it is for power: pull it there with a bidirectional analog switch. (For power, we have to use back-to-back MOSFETs. For this, an analog switch ought to do.)Of course you ultimately need two input bits to drive the logic, which must count in sequence with little skew between the two bits, and you're only using 3/4 states so need to do something with the spare state (could be tristate, or degenerate with another intended state). Matching delays and edge rates between input and high/low side drivers/switches won't be the easiest thing, but such is the nature of signal creation.Put another way: it's a 2-bit DAC, which could be a direct (at scale) output, or created as itself an analog signal, then sent through traditional output buffer circuitry. This might be an easier way to control for edge rates, for example.It doesn't even need to be binary, it could be offset unary; in that case, rather than thinking in terms of pulling up and down, it's pulling up, or up more. (This is one resolution to the extra state: the 10 and 01 states give the same 1*Vref output.) Maybe the bits go to analog switches between 0/Vref (precision 1-bit DAC -- this cleans up the raw/dirty logic levels if needed) and then the analog signal path sums, offsets and buffers them.Whether it's current or voltage sources, I suppose doesn't really matter, if you're going to dump it into a termination resistor anyway. Norton or Thevenin equivalent, with the bandwidth limit given by shunt C (Ccb or whatever) or series L (the buffer will have a rising impedance at high frequencies, or stray inductance due to lead lengths) respectively.Note that you can be clever and do a hybrid between both, saving on supply voltage range -- set up the buffer as an opamp with joint voltage and current feedback, weighted for 50 ohms Zo -- which instead brings concerns of flatness (how much does it diverge from 50 ohms due to imperfect feedback near GBW?), and also, it's probably not the simplest thing to get current feedback in a non-isolated output stage. (So, would be interesting to do, but maybe not interesting enough to use.)Tim

author: TimFox
author_rank: Super Contributor
timestamp: February 07, 2022, 10:18:52 pm
content: If there are three selectable states (+, 0, and -) then two binary bits are required to select one of them.