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Let say we have a very short pulse (100ns to 4000ns) and we want to make another wich last 1000 times longer so we can mesure it with a microcontroller.

We don't really care if the multiplication factor is accurate, neither if is really linear (vs logarithmic), but we want it to be coherent and repeatable : same input pulse should give same length output pulse.

I'm thinking of a capacitor A discharging in another capacitor B for mesuring the initial pulse, then discharging B trough a strong resistor so it discharge slowly.

Does this kind of timer works at ns resolution ? Is it reliable ? Is there a better way to do that ? (is it the worst ? :) )

---- Edit for good EM Fields questions

Accuracy of multiplier factor could vary between devices, something between x1000 and x5000 will be ok. But it must be stable for one given device within the same conditions (temperature, pressure...) over a 5s timespan. I want to be able to measure the initial impulse with an accuracy of 3ns, wich means 3 to 15µs jitter on the output pulse depending on multiplication ratio. It is more important to have a better accuracy on short pulses. I think and average of 1/500 of total duration is enough. Plus, maybe a kalman filter or something alike could smooth out this jitter.

We don't care about multiplication factor because the system will calibrate itself between every measures. There will be one known duration pulse, then one information pulse. We need to know the ratio duration_of_information_pulse/information_of_known_pulse.

The goal is to locate a device on a field using time of flight of a radio signal. It requires 2 measures. One position every 10s is enough. So we don't really care when the output pulse start.

------ Progress

I found this patent https://www.google.com/patents/US3712993. Looks like it uses Tcd and Tpd of cascading transistor. I don't think it is suitable for x1000 multiplication factor.

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  • \$\begingroup\$ One capacitor, different charge and discharge rates. Arrange two current sources, one with -1000* the current of the other, gated by the pulse. (Make that -1001 and you can start them both on the leading edge). The second turns off when V(capacitor) reaches 0. \$\endgroup\$ – Brian Drummond Oct 19 '15 at 15:17
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    \$\begingroup\$ You could charge a cap through series diode and resistor, then measure peak voltage on the cap. Then reset cap. Choose R such that longest pulse only charges cap to 1/2 the input voltage or so. (This assume all pulses are same height, voltage.) \$\endgroup\$ – George Herold Oct 19 '15 at 15:44
  • \$\begingroup\$ 1. I know you say it's unimportant, but there must be \$ \style{color:red; font-size:150%}{some}\$ limit on the inaccuracy of the multiplication factor, so how accurate do you need the multiplication factor to be? 2. For identical input pulses, how stable (in terms of jitter) would you like the output pulses to be? 3. With reference to the leading edges of the input pulses, when must the leading edge of the long pulses occur? \$\endgroup\$ – EM Fields Oct 19 '15 at 16:29
  • \$\begingroup\$ @EMFields thank for your good questions, I edited the question. \$\endgroup\$ – bokan Oct 19 '15 at 17:49
  • \$\begingroup\$ I think that it would be more straightforward to convert the original pulse to a voltage (for example by charging a capacitor) and then reading this analog voltage by the microcontroller. (The microcontroller would be triggered by an interrupt at the fall of the pulse). \$\endgroup\$ – Roger C. Oct 20 '15 at 16:31
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Could you do this all digitally, using ECL logic? The SY100E137 is a 8-bit counter that has a maximum clock rate of 1.8 GHz. The datasheet says " for counters in high-performance ATE time measurement boards. The counter costs $6.72 at Digi-Key. You would use two of them for a maximum count of 65535 ns.

Running at 1 GHz, you would start the counter on the rising edge of the pulse, and sample the incoming pulse train in 1 ns slices. When the trailing edge is detected, stop the counter, then keep reloading it into a count-down counter 999 times and delay the output until the latter counter goes to zero.

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  • \$\begingroup\$ Thank you verymuch, but the goal is precisely to avoid the cost of a counter and high frequency clock. It was not stated in the question, sorry. \$\endgroup\$ – bokan Oct 20 '15 at 17:12

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