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Designing differential Signal modulator.

Alex_Bam

Sep 28, 2020
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I would like your suggestion on my first design Differential signal modulator [simulation attached].

Before jumping into design let me highlight my objective. I am differentiating a control signal of 1MHz and then modulating it with carrier frequency 15MHz(not limited to this specific value but for testing). The whole modulator needs to be integrated on a silicon substrate, in other words, the design should comply with IC design regulations i.e. occupy less space, power-efficient, and simple. The current design consumes 5.6mW power.

So I would request suggestions who had experience in a related field, that

1-How can I further improve this design?

2-Is there other alternatives design that would be much better than this one?

Your suggestions and comments will be highly appreciated. Thanks
 

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Harald Kapp

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I am differentiating a control signal of 1MHz
I wouldn' call this differentiating. You're using an inverter to create a signal 180 ° out of phase (inverted) to the input signal.

As far as I remember having resistors in an integrated circuit is usually not optimal. resistors can be created on-silicon only with comparatively low accuracy.
Capacitors can be controlled better. Even better is to use capacitor ratios as the determining elements. Capacitor values may vary quite a bit. Capacitor ratios can be reproduced with high fidelity.

Power dissipation may be reduced by redcuing the currents through the various stages. E.g. by reducing the w/l ratio of the transistors.
 

ratstar

Aug 20, 2018
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Harold...
So how bad is this resistor error on ICs? 1 ohm or 10 ohms? if its 10 ohms id be a little bit worried, but 1 ohm wouldn't worry the circuits I make. But I guess maybe keeping that in mind is important, always pays to be cautious in the unknown, but that would be in design time, and giving yourself an ohm error your allowancing for.

It could be a form of boolean differentiating, things boil down to something simple. multiply is and, and or/xor is like add a bit, but it is a truncation of what they actually do for real.

Alex_Bam. so your going to print these switches yourself? Ive got no idea how to make them, I just stick to caps, resistors & spark-gaps cause the concepts are simpler, and is alot less toxic and dangerous? But im only guessing, maybe the stuff on the net, alot of it is lies to scare ppl out of doing it.

I did make one semiconductor type thing, its called a wet-diode, I found it on the internet, and its when you stuff up trying to make some kind of water super capacitor you end up with a mulfunctioning half diode half capacitor. :) Its actually pretty cool apart from the horrible response, if only when it dried out it didnt stop functioning. (they only last till the water dries out.)

Having a diode would be really useful, alot of my stuff works better if only I had homebaked diodes, avoiding using diodes can be a poor practice, it makes the circuits bigger, things arent possible, more energy wasted, all sorts, I wonder if I ever learn how to make them (Diodes.), its one step closer to the transistor itself.

But my IC im putting together is a Marx Generator, and its only res&cap&gap's, so I continue forth with it, with my anycubic photon by my side to get me some nice exact printouts, helps a great deal.

Doing it with carbon ink is only 6,400 times worse conductivity than copper! =)
I think it compares a cm wire with 100 micrometre wire of copper, then its equal, but it doesnt go around bends as sweet.
 
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AnalogKid

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I wouldn' call this differentiating. You're using an inverter to create a signal 180 ° out of phase (inverted) to the input signal.
Maybe. My read of the schematic is two separate oscillators that are gated on and off with the two phases of an input square wave. To me that doesn't match the written description, but that might be a translation problem.

However ...

Note that because you are gating the oscillators on and off, the first half-cycle of each one will be longer than the following half-cycles (and cycles). This is because the charge on the timing capacitor (C1 or C8? fuzzy image) is greater after the oscillator has been inhibited for a while than it is at the beginning or end of any cycle when the oscillator is running. This error term grows the longer the oscillator is off, reaching a maximum after the time equivalent of 5 cycles. The amount of error is dependent on the difference between the high-to-low and low-to-high transition voltages of the NAND gate input.

ak
 

Harald Kapp

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So how bad is this resistor error on ICs? 1 ohm or 10 ohms? if its 10 ohms id be a little bit worried, but 1 ohm wouldn't worry the circuits
That is not the way to look at it:
  • If you need a 10 Ω resistor and the error is 1 Ω, that's 10 %.
  • if you need a 100 Ω resistor and the error is 10 Ω, that's 10 %, too
always pays to be cautious in the unknown
Heed that quote!
 

ratstar

Aug 20, 2018
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That is not the way to look at it:
  • If you need a 10 Ω resistor and the error is 1 Ω, that's 10 %.
  • if you need a 100 Ω resistor and the error is 10 Ω, that's 10 %, too

Heed that quote!

If it scales up with the resistance thats probably ok.
 

Harald Kapp

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control signal of 1MHz and then modulating it with carrier frequency 15MHz
Usually the control signal modulates the carrier, not vice versa.
wo separate oscillators that are gated on and off with the two phases of an input square wave
@Alex_Bam : a bit more of explanation will certainly not hurt. Help us understand in more detail what you're up to.
 

davenn

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I am differentiating a control signal of 1MHz and then modulating it with carrier frequency 15MHz


That doesn't even begin to make sense
your use of terms is incorrect a carrier signal IS NOT a modulating signal

The definition of a carrier signal is that it carries the modulation provided by another signal
 

Alex_Bam

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@AnalogKid @davenn @bertus
@Alex_Bam : a bit more of explanation will certainly not hurt. Help us understand in more detail what you're up to.[/QUOTE]

Thanks for your suggestion. Maybe I did not translate my issue properly. I am trying to implement the idea mentioned in
US Patent: US 7923710B2 "Digital isolation with communication across an isolation barrier". I already implemented capacitive coupling isolation on IC with a single-ended signal by using the oscillator schematic mentioned in my question. To implement capacitive isolation with a differential signal I found this US patent.

If I understood correctly. In the patent, they use NOT gate to invert/differentiate control signal and then modulate inverted signals with modulators. The coupling capacitor coupled modulated signal by providing isolation on a silicon substrate.

The reason for using NAND gate in single-ended signal setup: Due to the NAND gate, it only modulates when the control signal is high. This is because the control signal is held low when no communications are occurring. The modulator is idle when no communication is occurring, making this modulator design power efficient.
 

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Harald Kapp

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US Patent: US 7923710B2 "Digital isolation with communication across an isolation barrier".
Do you have a link to at least the abstract of that patent for us?

then modulate inverted signals with modulators.
I think what you want to express is that the control signal and the inverted control signal modulate the oscillator/carrier signal(s).
Due to the NAND gate, it only modulates when the control signal is high.
You seem to confuse modulation with oscillation. Modulation here is the action of turning either of the two oscillators on or off. The control signal is not modulated, it is the signal modulating the oscillators.
The modulator is idle when no communication is occurring, making this modulator design power efficient.
Since you have the inverter at the front end, at any time one of the two oscillators (not modulators) will be active, even with the control signal being inactive. So no power savings here.
If you wanted to shut down both oscillators when the control signal is idle, you'd have to implement an additional idle detection circuit, e.g. by a monostable timer which is re-triggered by a change in the control signal but resets if there hasn't been a change in the control signal for some time. You'd also have to implement a similar scheme on the receiving end to generate a fixed output when the oscillators are stopped.
That'S quite a bit more logic than you currently have.
 

bertus

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Hello,

Have a look at the ISO122 isolation amplifier datasheet.
That will explain the working of the isolation barrier.

Bertus
 

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Alex_Bam

Sep 28, 2020
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Do you have a link to at least the abstract of that patent for us?


Link of a patent for your reference: https://patents.google.com/patent/US20080218258

I think what you want to express is that the control signal and the inverted control signal modulate the oscillator/carrier signal(s).

Yes, you are right, Control and inverted control signal modulate the oscillator.


You seem to confuse modulation with oscillation. Modulation here is the action of turning either of the two oscillators on or off. The control signal is not modulated, it is the signal modulating the oscillators.

Thanks for the correction.

Since you have the inverter at the front end, at any time one of the two oscillators (not modulators) will be active, even with the control signal being inactive. So no power savings here.
If you wanted to shut down both oscillators when the control signal is idle, you'd have to implement an additional idle detection circuit, e.g. by a monostable timer which is re-triggered by a change in the control signal but resets if there hasn't been a change in the control signal for some time. You'd also have to implement a similar scheme on the receiving end to generate a fixed output when the oscillators are stopped.
That'S quite a bit more logic than you currently have.
 
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