Schottky diode circuit mystery ...

Thread Starter

cmartinez

Joined Jan 17, 2007
7,880
This is very, very embarrassing for me. I'm supposed to be past this sort of basic level of knowledge. But I just can't wrap my head around understanding how this circuit works.

At first sight, I would've said that it shouldn't really work at all, and that current should be flowing through D2 back into V1, which is an undesirable condition. But the sims says otherwise:

upload_2019-7-10_8-48-7.png


If I were to change D1 and D2 to ordinary 1N914 diodes, the sim behaves more or less as I expected it would. Current would flow through D2 into V1.

But not if they're schottky diodes, in this case, only small current peaks through D2 are happening during rise and fall of the input square wave. But current through D1 goes from -25 uA to -90 uA between pulses.

There's obviously something I'm not understanding about schottky diodes here...

What gives? Why is current flowing the way it is? Why is the 12V pulse not feeding itself into V1 through D2?
 

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Thread Starter

cmartinez

Joined Jan 17, 2007
7,880
I've just read that schottky's can be used as voltage clampers ... but I still can't fully understand their behavior.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,880
There you go .... :oops:

upload_2019-7-10_9-15-53.png

The sim is now doing what I expected.... Now the question is which would be a better clamping technique ...
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,880
Depends upon how accurate you need the clamp voltage to be.
Not much actually... I need to drive an input pin of an MCU working at 3.3V. Specifically the PIC10LF322. That means that any voltage between 2V and 3.3V is acceptable, according to its datasheet.

Here's the whole problem:
  • A power source is generating pulses ranging between 5V and 24V at a frequency of about 300 Hz
  • These pulses are being used to charge a supercap, which in turn works as a steady power source for the MCU. A circuit limiting the supercap's charge is already in place, and I'm quite happy with it.
  • Using said MCU, I need to monitor the pulses, but using as little power as possible.
  • For that purpose, I've been using a TPS709B (which is a highly efficient linear regulator) to limit the pulses' voltage down to 3.3V, and then feeding its output to the MCU.

I'd like to get rid of the TPS709B being used to limit the pulses being fed into the MCU's input pin, and use a cheaper alternative if possible.
 

OBW0549

Joined Mar 2, 2015
3,566
Using said MCU, I need to monitor the pulses, but using as little power as possible.
When you say, "I need to monitor the pulses," do you need to measure their amplitude? Or do you want to measure their duration or their frequency, or just want to confirm that they're occurring?

Unless you're interested in the pulse amplitude, I'd say just feed the pulses to the MCU pin through a suitable resistor (say, 10 kΩ or thereabouts) and limit the pin voltage with a Schottky diode (anode to the pin, and cathode to Vdd).
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,880
Something like this?

upload_2019-7-10_12-18-56.png
Here's what I'm not sure I like: there's current flowing through D3 back into V2, and V2 is a TPS709B regulator, as I explained earlier. Wouldn't this condition affect it negatively in some way?
 

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OBW0549

Joined Mar 2, 2015
3,566
Here's what I'm not sure I like: there's current flowing through D3 back into V2, and V2 is a TPS709B regulator, as I explained earlier. Wouldn't this condition affect it negatively in some way?
Why did you keep the TPS709B? I thought the whole point of this exercise was to get rid of it?

Or is the TPS709B also supplying 3.3V to your MCU? In which case don't worry about any current flowing back into the regulator; it'll go to the MCU instead.

In the latter case ditch the 470k resistor and just go with what you had in post #5, except change the 100k resistor to 10k to reduce susceptibility to any noise coupling.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,880
Why did you keep the TPS709B? I thought the whole point of this exercise was to get rid of it?

Or is the TPS709B also supplying 3.3V to your MCU? In which case don't worry about any current flowing back into the regulator; it'll go to the MCU instead.

In the latter case ditch the 470k resistor and just go with what you had in post #5, except change the 100k resistor to 10k to reduce susceptibility to any noise coupling.
Yes, in my previous post V2 is the TPS709B that is supplying 3.3V to the MCU.

Here's a more complete explanation of what I'm doing:

upload_2019-7-10_14-8-5.png

The main source is a AC sine wave that can vary in amplitude anything in between 10 and 18 VDC, but I've represented it as a constant 18 VAC for simplicity. This source has an impedance of 270 ohms, and is responsible of charging C1. As you can see, it takes a while. D1 is a 6.6V zener that limits the charge voltage reached by C1. All excess current exceeding the zener voltage is shunted to ground. A TPS709B regulator is then used to power the PIC10LF322 at 3.3V, although the MCU starts running as soon as the voltage reaches 1.8V.

What I want is to measure the frequency being produced at the 18VAC source. To that effect, a diode (D3) rectifies the source, and its cathode is connected to a second TPS709B regulator whose output is then connected to an MCU's input pin. The MCU's pin is currently configured as an input with an internal weak pull-up activated. That is why I'm using a 470K resistor (not shown) to pull it down when that second TPS709B is inactive.

What I'm trying to do is to get rid of that second TPS709B while at the same time maintaining current draw to a bare minimum. But that regulator is extremely efficient and I have no idea as to how to supplant it with cheaper discrete parts.
 

OBW0549

Joined Mar 2, 2015
3,566
What I want is to measure the frequency being produced at the 18VAC source. To that effect, a diode (D3) rectifies the source, and its cathode is connected to a second TPS709B regulator whose output is then connected to an MCU's input pin. The MCU's pin is currently configured as an input with an internal weak pull-up activated. That is why I'm using a 470K resistor (not shown) to pull it down when that second TPS709B is inactive.

What I'm trying to do is to get rid of that second TPS709B while at the same time maintaining current draw to a bare minimum. But that regulator is extremely efficient and I have no idea as to how to supplant it with cheaper discrete parts.
Then get rid of D3 and the second TPS709B, and go directly from the top end of your 18 VAC source to the left end of R1 in the diagram of your post #5. You don't need anything else.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,880
Then get rid of D3 and the second TPS709B, and go directly from the top end of your 18 VAC source to the left end of R1 in the diagram of your post #5. You don't need anything else.
You mean like this? The problem I see with that arrangement is that negative current going through R3 back into the TPS709B peaks at about -8.3 uA, and we're dealing with a circuit that (so far) draws with 100 uA tops. That's an 8% loss there, I think. Also, why is it I'm not seeing a negative voltage at the MCU_pin? Is it because current is being sank at the TPS709B?

upload_2019-7-10_19-20-38.png

 

OBW0549

Joined Mar 2, 2015
3,566
You mean like this?
Yes.

The problem I see with that arrangement is that negative current going through R3 back into the TPS709B peaks at about -8.3 uA, and we're dealing with a circuit that (so far) draws with 100 uA tops. That's an 8% loss there, I think.
Umm... the way I see it, that 8.3 uA simply lightens the load on the TPS709B during periods when D3 conducts. It does not go "into" the TPS709B; rather it goes to your load (whatever that is), replacing current that would otherwise have to have been supplied by the TPS709B.

Besides, your Zener diode D1 has nearly 70 milliamps going through it during peaks of the 18 VAC source; why are you worried about a handful of microamps?

Also, why is it I'm not seeing a negative voltage at the MCU_pin? Is it because current is being sank at the TPS709B?
Probably because of the clamping action of Schottky diode D2.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,880
rather it goes to your load (whatever that is)
I'm thinking it rather draws more current from the MCU's pin, risking damaging it ... but maybe I'm wrong about that

Besides, your Zener diode D1 has nearly 70 milliamps going through it during peaks of the 18 VAC source; why are you worried about a handful of microamps?
Right ... maybe I'm splitting hairs (or "counting chiles", was we say down here :D) ... it's the perfectionist in me

I guess the only way to really know what will happen is to build the thing, test it, and perform a few thorough measurements.

Many thanks for your help, my friend.
 

Sensacell

Joined Jun 19, 2012
3,098
Surprisingly, Microchip shows this crazy zero-crossing circuit with no external clamping diodes.

As long as the peak current is low, all is ok.

You can eliminate the diodes.
 

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ebeowulf17

Joined Aug 12, 2014
3,307
Surprisingly, Microchip shows this crazy zero-crossing circuit with no external clamping diodes.

As long as the peak current is low, all is ok.

You can eliminate the diodes.
It's not too surprising - @OBW0549 has described the logic behind this simple one-resistor protection scene numerous times here (at least I think it was him,) although this may be the first time I've seen an application note describing the same thing.
 
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