Electronics > Projects, Designs, and Technical Stuff
Improve bandwith of peak detector
MasterT:
--- Quote from: radix on July 25, 2019, 12:36:26 am ---
--- Quote from: MasterT on July 25, 2019, 12:29:28 am ---100 nA is enormous current in such application.
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*proceeds to post a schematic with a schottky diode that has max. leakage specified at 250 nA at 3 V reverse voltage at room temperature*
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Good catch, means don't trust anyone. Though I tested circuits with DG419 instead of BAT17, it operates quite accurate - down to 18-bits of resolution. Offset is not an issue, as long as its constant over input common mode voltage.
Kleinstein:
The version I know of the simple 2 OP circuit uses an additional diode in feedback for the 1 st. OP, so that the OP only has to slew some 1.4 V when a new maximum is detected.
For long hold times, one can combine 2 such circuits in series.: a fast one with a rather small hold capacitor and a slower one to extend the hold time.
The fast part should than have a defined drift direction down.
The transistor circuits look good.
David Hess:
--- Quote from: MasterT on July 25, 2019, 12:29:28 am ---Probably, they don't test reverse current of the common BJT because they have no idea that someone is gonna to build a peak detector with B-E junctions. 100 nA is enormous current in such application.
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They do test it. They just only test it down to typically 25 or 50 nanoamps because the testing is fast at that level and that level is adequate for typical applications.
The least expensive commonly available low leakage diode is the gate junction of a 2N4117 which really is tested down to 10 picoamps. The 2N4117A is tested to 1 picoamp.
--- Quote from: radix on July 25, 2019, 12:36:26 am ---David Hess also stated that a BJT has lower leakage than the 1N4148 (specified at 25 nA at 20 V reverse voltage at room temperature).
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Sometimes designers get into trouble using 1N4148s as clamp diodes for differential stages and discover that even at zero volts, they have a low resistance associated with their leakage current.
I grade my own 2N3904s to use as low leakage diodes as needed but I have never found one which was much greater than 10 picoamps on either the emitter or collector junction. The emitter junction is very fast but has a low reverse breakdown voltage. The collector junction is slow and higher capacitance but has a high breakdown voltage.
MasterT:
--- Quote from: David Hess on July 25, 2019, 11:49:34 pm ---The least expensive commonly available low leakage diode is the gate junction of a 2N4117 which really is tested down to 10 picoamps. The 2N4117A is tested to 1 picoamp.
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I've checked on mouser, they have bav170 that has lower price and lower reverse current. Though data "typical" , but tested with high 70 Voltage.
Still if consider max value 5 nA and 2N3904 - 50 nA it's 10 times difference. Decent OPA has output current 20 mA, bringing into picture 200 mA transistor gets 10x faster charge time, AND 10x faster discharge. So, what is the point to bother with more complex circuits, except impress your professor.
marcoccio:
I don't think extreamly low leakage transistors are required since the input signal is permanent, meaning the hold time should be enough to mantain an stable output at low frequecy's and not much more since it will always be charging with the next peak of the input signal.
--- Quote from: imo on July 24, 2019, 03:55:35 pm ---I would use an AD8307. Works from DC to 500MHz. You get the LOG(input_peak_value) at the output (0-2.5V output).
https://www.analog.com/media/en/technical-documentation/data-sheets/AD8307.pdf
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I found abaout it in other post but it scapes my knowledge of how i should implement it, also it's expensive and where i live they don“t sell it. If i can't make this work i will look at it again and maybe i can get it shipped in a reasonable time, here customs take from two to five moths if you are a normal buyer like me.
--- Quote from: SiliconWizard on July 24, 2019, 05:32:34 pm ---
--- Quote from: MasterT on July 24, 2019, 04:52:50 pm ---To OP: What is the arm chip? If it has an ADC, may be you don't need peak detector after all, sampling input fast enough by 12-bits ADC and running peak detector in software.
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Well... you're right. The OP talks about 500kHz max. Many recent MCUs have several-MSPS 12-bit ADCs, so that should be doable in software without much problem.
Now if a 12-bit ADC would not have enough resolution for the OP's needs, then I think they would be dreaming if they think they can get better than the equivalent of 12-bit with a simple, fast and pure analog peak detector...
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That was my first thought, sadly, analog implementation is as must, teacher's requirement :palm: . We are using an F103C8T6 which has a 14Mhz ADC.. More than enouth to sample a 500Khz signal. If this gets too difficult maybe i can persuade him haha.
--- Quote from: Kleinstein on July 25, 2019, 09:39:33 am ---The version I know of the simple 2 OP circuit uses an additional diode in feedback for the 1 st. OP, so that the OP only has to slew some 1.4 V when a new maximum is detected.
For long hold times, one can combine 2 such circuits in series.: a fast one with a rather small hold capacitor and a slower one to extend the hold time.
The fast part should than have a defined drift direction down.
The transistor circuits look good.
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I tryed that technique, however, the problem here is the op not able to deliver the necesary current to charge the cap at high frequencies, so the primary diode gets replaced by the base-emiter junction of a bjt.
--- Quote from: David Hess on July 24, 2019, 03:19:48 pm ---
--- Quote from: David Hess on July 24, 2019, 03:09:41 am ---Replacing the diode with a transistor allows charging the hold capacitor with a much higher current without lowering the value of the pull-up resistance which applies to your existing circuit.
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In case I was not clear enough about this, replace your diode D2 with a transistor with its collector tied to the positive supply. Now the current available from R7 is multiplied by the transistor's current gain to charge the hold capacitor faster. The example I gave uses the same idea.
The reverse breakdown voltage of the base-emitter junction will limit the maximum range of operation to about 5 volts peak-to-peak however. If this is a problem, then a diode placed in series with the base can be added.
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Thank you for the clarification, i tryed this approach and found it works really well.... But (there is always a but :palm:) i have some doubts about the circuit i came up with. Replacing both diodes with bjt's works, however, now at low input voltages (1V) the diference in voltage between the inputs and the feedbak from the output needs to be almost 100mV higher for the LM393 to stop charging the cap and there is always a 50 to 80 mV diference in charged value at 5v, i can live with that even though it's higher than 2% error. Here is the circuit:
And here the simulated results, first at 1kHz and 1V peak, then 400kHz and 1V peak, finally 400kHz and 5V peak value.
Once again thank you all for the replys :-+
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