Very Wide input voltage circuit for frequency counter?

Thread Starter

Dyslexicbloke

Joined Sep 4, 2010
559
Hi folks,
I do quit a bit of work with induction motors configured as SEIG machines (Capacitivly excited) an occasionally just as motor/generators, DOL, that need to be spun up prior to going online. (Small hydroelectric for those of you wondering)

Anyhow...
There is almost always sufficient residual magnetism to get a few volts that will drive a counter but I would like to design something that is opto-isolated and capable of accepting a signal of as little as 500mv and as much as 500V.

The rational is to allow the frequency to be counted with a PLC but to avoid having to have any form of interlocked switch.
Partially because a switch is expensive and cumbersome but mainly because I would like to keep counting the SEIG as its excitation builds.

FYI I use encoder wheels and proximity switches right now and don't even attempt to connect a counting instrument to the output of the machine.

I have in the past used simple transformer, which is fine once the voltage comes up but no us at all when the machine is offline.

I would want two versions of the circuit, one operating at 24VDC, on the control side, and one battery operated to use in conjunction with a multi-meter or similar test equipment. Obviously with appropriately fused/class probes.

I suspect that there is some clever way to do this, and do it safely, and many more ways to blow stuff up and be very unsafe, hence the question.

I am confident that a simple restive divider feeding a clamp circuit, possibly just a zenner, would work but I am equally sure its a very unsafe way to go and have already discounted the idea.
Unless of course such an arrangement could be used to drive an opto-isolator directly... Whats the minimum forward voltage I am likely to find?

Thoughts?
Is there anything commercially available that I am unaware of? Ideally I would want a square wave output with a short pulse every half cycle.

Cheers,
Al
 

Thread Starter

Dyslexicbloke

Joined Sep 4, 2010
559
Available current...
The signal source is a generator so many tens of amps, often hundreds or even thousands if you consider fault current when grid connected.

Without knowing the circuits it would be silly to say current isn't an issue but the source is probably such a low impedance that it can effectively be ignored as a limiting factor in the design.

In addition, designing for the lowest input current that is practical would just make the 'thing' more useful wouldn't it.

Brododynov...
Wow, thanks,
that didn't take 5 mins...

I have a handle on the comparators and the astable but I cant get my head around the 'active'? input divider.
I can see you have used FET's but I am struggling to understand how that is working and what exactly it is doing.
Probably just being vague but it looks like a learning opportunity to me, if you wouldn't mind.

Many thanks,
Al

Al
 

Thread Starter

Dyslexicbloke

Joined Sep 4, 2010
559
Current limiter... OK so that makes more sense now but, not thinking clearly it would appear, but I am struggling to understand how that is working below the minimum VGS. Of course I may be reading the spec incorrectly or simply not understanding what I am reading in the first place.

It looks like the body diodes have a minimum Vf of about 0.42, which is fine but why would the other FET, in whichever direction we are talking, turn on at all with say 500mV as an input?

What am I missing?
 

Bordodynov

Joined May 20, 2015
2,469
I used transistors with an operating voltage of 1000 V in the current limiter! Together with the diodes it is a quality voltage limiter. The power dissipated by the transistors is quite acceptable.
Limiter2.png
 

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Bordodynov

Joined May 20, 2015
2,469
Current limiter... OK so that makes more sense now but, not thinking clearly it would appear, but I am struggling to understand how that is working below the minimum VGS. Of course I may be reading the spec incorrectly or simply not understanding what I am reading in the first place.

It looks like the body diodes have a minimum Vf of about 0.42, which is fine but why would the other FET, in whichever direction we are talking, turn on at all with say 500mV as an input?

What am I missing?
The threshold of these n-channel transistors is negative (~ -2.5V). And for small signals (~1mV) the transistor is a resistor. Note the transistor symbol.
 

Thread Starter

Dyslexicbloke

Joined Sep 4, 2010
559
Depletion... Completely missed that bit... I wonder sometimes if I relay get this stuff at all... Sort of explains why I couldn't find VGS (on)

I still don't understand the deal with the 2 BEV99's? Is R7 actually supposed to be connected to the node between them as opposed to the inp_comp node? That would offer a force voltage on that node, irrespective of current through R7.
As is they don't appear to be doing anything... At least I cant see what, which isn't the same thing at all :)
 

Bordodynov

Joined May 20, 2015
2,469
The R7 resistor ensures that there is no signal (low level) in the event of an open input. This is to eliminate the influence of the comparator input currents. However, this resistor may not be used for normal operation. Also for noise immunity, a capacitor should be placed in parallel to this resistor (to reduce short-term voltage spikes). And diodes - BAV99.
 

Thread Starter

Dyslexicbloke

Joined Sep 4, 2010
559
OK so the left pointing arrow, below the bav99 label, is a ground... Sorry being silly, thinking it was the low side of a floating input.
I see now its a pull down and that there is a 100n to ground from the same node.

However the two diodes are in series, with the anode of the first and cathode of the second connected to the same node, inp_comp.
Sorry just not getting that bit... Am I misreading drawing?

Node inp_comp... M2(D), U1(In-), U4(In+), R7, C2, bav99(Top - Anode), bav99(Bottom - Cathode) - That is as it is drawn, right?
 

Thread Starter

Dyslexicbloke

Joined Sep 4, 2010
559
Thanks that makes sense now...
I don't know if it was just the rendering or an issue with the node placement but the ground between D1 and D2 wasn't showing up as a connection.
It is clear now that the forward voltage of D1or D2, dependent on the signal polarity, is working as a clamp, which make sense.

I was beginning to wonder what I was missing.

I like this circuit and will be trying it soon...

In a related matter could you recomend a way of limiting current to 25mA without applying any significant impedance increase until the current exceeds 20mA?
The application is input card protection so the signal is a current loop which is driven by a loop powered instrument.

Typically 24V is fed to the sensor, which drops about 8v to run and draws less than 4mA internally.
if the measured parameter is 0 than the instrument will drive the overall loop current to 4mA into any impedance that there is suficiant voltage to run. PLC card typically, but not always, present 50 Ohms and measure the voltage across this to determine the current.
If the instrument is measuring full scale then the loop is driven to 20mA

Adding impedance isn't a problem provided you don't drop more than the supply in total. With a typical instrument and PLC card there is about 5V of headroom. Any current to ground will generate an offset in the measurement which would be hard to compensate for if it wasn't constant or negligible.

Thanks for the help, I am learning loads.
 
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