Question on special measuring of Diodes voltage?

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russwr

Joined Aug 29, 2017
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Question - How do I measure on multimeter the output voltage from circuit seen from 2007 Georgia project, that was secondary side of Inverter of 120VAC connected on each output line single diodes that are opposite polarity each facing, going to resistor load? Is this still a portion of 120V AC output ,or approx 40 volt half wave DC plus and minus DC? The writer may have been thinking about back flow protection. This is not print error, or about clipping circuit. Please advise.
 

Papabravo

Joined Feb 24, 2006
18,422
Question - How do I measure on multimeter the output voltage from circuit seen from 2007 Georgia project, that was secondary side of Inverter of 120VAC connected on each output line single diodes that are opposite polarity each facing, going to resistor load? Is this still a portion of 120V AC output ,or approx 40 volt half wave DC plus and minus DC? The writer may have been thinking about back flow protection. This is not print error, or about clipping circuit. Please advise.
The short answer is that you can measure the voltage between any two points in a circuit with a multimeter. Some meters will protect themselves from overloads and other won't.
 

MisterBill2

Joined Jan 23, 2018
11,880
At least a detailed description of the portion of the circuit you are asking about. A two-diode parallel arrangement is often used to detect current flowing in an AC circuit. It works but it is not isolated and so all precautions must be for the voltage of the rest of the circuit, even though the voltage across the diodes is small.
 

Ramussons

Joined May 3, 2013
1,264
The short answer is that you can measure the voltage between any two points in a circuit with a multimeter. Some meters will protect themselves from overloads and other won't.
.... and some meters will change the parameters being measured by "loading" the measurement points.
 

MisterBill2

Joined Jan 23, 2018
11,880
If you use an auto-ranging digital multimeter it should not affect the voltage readings, if this is any sort of power circuit.
And I still hhave no information at all about any "2007 Georgia project", and at least an adequate description of the portion of the circuit under discussion, using accurate terminology, is mandatory, not an option. Otherwise no more assistance, at least not from me. Blind guesswork is a waste of time for many of us.
 

Thread Starter

russwr

Joined Aug 29, 2017
5
Question - How do I measure on multi meter the output voltage from circuit seen from 2007, that was secondary side of DC to 120V AC 60 cycles Inverter with one diode connected on each line of secondary, that has the diodes reversed, as cathode out and anode out. (opposite polarity) This went to resistor and inductor as load. This is not clipper circuit or misprint of diagram. Originator may have been thinking about reverse higher voltage protection . Is this reading on meter as AC voltage or DC voltage output, as each position on meter has different reading.
 

ThePanMan

Joined Mar 13, 2020
367
Wondering the same thing. "2007 Georgia project".
If you're not answering this question OR posting a schematic of what you're talking about - - - . Exactly what is the 2007 Georgia project? Until we know what YOU are asking WE can't help. We can be general about how to measure the voltage across a diode but to answer YOUR question WE need to know more.
 

Thread Starter

russwr

Joined Aug 29, 2017
5
If you're not answering this question OR posting a schematic of what you're talking about - - - . Exactly what is the 2007 Georgia project? Until we know what YOU are asking WE can't help. We can be general about how to measure the voltage across a diode but to answer YOUR question WE need to know more.
Main circuit is a blend of 2 power sources together, high voltage and low voltage inverter DC used during experimental engine testing in Georgia, 2007. Analogous to car battery jumper Lithium battery unit blends higher and lower voltage together same time so as starter motor receives proper voltage to operate. DC current - eventually, through final diode blends inverter power exactly when ignition fires spark plug. The relay was type used in Georgia traffic signal control pole box before the change over to LED. This is not clicking method of latching dual coil relay, it uses only the inductance iself in series circuit. Just the Inverter with the opposite diodes reads both on dc and ac on multimeter to resistor load. Negative flows through Cathode. Positive flows through Anode. Which setting is correct?
S1R relays.jpgS1R relays.jpg
 
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MisterBill2

Joined Jan 23, 2018
11,880
I was eventually able to see that picture of some wiring, which is not a circuit drawing by any definition, and because it is only a tiny fragment of some circuit it makes no sense at all. Worse yet, I see nothing about any inverters, nor diodes in anti-parallel connections.
In addition, voltages do not "blend" at any time under any conditions. Voltages may add or subtract, but never ever will they blend.
What I make of that drawing is that it is some arrangement of switching to make something happen.
The mechanics of some electro-magnet operated switching scheme is not obvious, and I am not willing to make guesses.

What I can deduce is that there are three connections leaving the circuit, two with series diodes. and that tells me no useful information at all.
And referencing the #1 post, I am wondering if in reality the "inverter" is actually a transformer fed by 120 volts AC from an external mains connection.

As for " Question - How do I measure on multimeter the output voltage from circuit seen from 2007 Georgia project, that was secondary side of Inverter of 120VAC connected on each output line single diodes that are opposite polarity each facing, going to resistor load? " That answer is simple and correct: "To measure the voltage at the output of any circuit, first set the multimeter to a voltage range greater than the anticipated magnitude of the voltage to be measured. If the measured voltage is anticipated to be alternating current, set the multimeter function to AC Volts, while if the measured voltage is anticipated to be a DC voltage, set the multimeter function selector to DC Volts. The connect the multimeter inputs to the output points of the circuit and energize the circuit, and observe the meter reading. You may need to reverse the polarity connections if the meter shows that the voltage is reversed."
These instructions should be adequate for an average second year journalism student to take reasonably accurate voltage readings without injuring themselves or the multimeter.
 

Thread Starter

russwr

Joined Aug 29, 2017
5
I understand about anticipation of polarity on test meter. It is obvious you never encountered use of 2 diodes in this UNUSUAL opposite polarity configuration that was explained 3 ways. The further exact circuit attatched to power supply was not neccessary. I can see an AC reading on meter on circuit with 2 diodes just after power supply Inverter across resistor load. I have same Inverter as used back then, as 400 Watt Power On Board Inverter 12VDC -110VDC , 60 cycles, model VEC031P0B. An AC voltage can be present in 1st part of load circuit, and then change to DC later with final output diode. I am assuming a proper answer will be needed elsewhere. Most likely the reverse protection of 600v diode would be under much higher stress and still fail with reverse KV. Most diodes fail with heat, not with overvoltage.
 

MisterBill2

Joined Jan 23, 2018
11,880
I have both encountered unusual diode arrangements and included them in original designs. I have even used a similar arrangement with diodes on opposite ends of a transformer secondary winding. If that winding is not tapped, then the effect is the same as two diodes in series pointing the same way. And I have used multiple diodes in series.
I am guessing now that it is the output winding of the inverter transformer that connects to the two diodes, one at each end of the transformer 120 volt winding.
If that is the case, then the portion that we do not need to see is the transformer secondary, just below where the picture ends. And if that is the case, then the diodes are in series pointing in the same direction and having the same effect as if they were immediately adjacent to one another. So then below the diodes you have an AC voltage while above the diodes you have half-wave rectified DC. Been there and done that. It just looks funny.
 

Thread Starter

russwr

Joined Aug 29, 2017
5
I have both encountered unusual diode arrangements and included them in original designs. I have even used a similar arrangement with diodes on opposite ends of a transformer secondary winding. If that winding is not tapped, then the effect is the same as two diodes in series pointing the same way. And I have used multiple diodes in series.
I am guessing now that it is the output winding of the inverter transformer that connects to the two diodes, one at each end of the transformer 120 volt winding.
If that is the case, then the portion that we do not need to see is the transformer secondary, just below where the picture ends. And if that is the case, then the diodes are in series pointing in the same direction and having the same effect as if they were immediately adjacent to one another. So then below the diodes you have an AC voltage while above the diodes you have half-wave rectified DC. Been there and done that. It just looks funny.
Thanks for clarifing this matter, as I had wondered about that for quite some time. Many others following the same subject matter over long period had used the expression "110v AC" Inverter output use as full final circuit voltage, while the relay was only .6ohm coils - not enough resistance for that level power supply. The Relay was also called 110v, which is impossible . Relay was originally intended for 6V DC automotive bulbs in the upper traffic lights. Timer would activate relay as per red/green lights and click it's separate contacts in S1R dual coils relay drawing.jpglatching mode. There is no relay clicking, only voltages passing through inductors, diodes, and onto spark plug. The engine plugs probably provided the high resistance necessary so as relays, one per cylinder, did not burn out windings. Ignition volts and low inverter DC volts both go through inductors, for purpose of delay line for extending spark time of firing. I only wished the original forum mentioned whether Average or Peak current of 6-7 Amps that was required.
 
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MisterBill2

Joined Jan 23, 2018
11,880
OK, now I recall that there was a time when there was experimenting with longer spark times to better ignite the leaner mixtures. I do recall that one researcher had asked for an ignition system delivering 10,000 volts at one amp. (This was in about 1980)
We pointed out that the person had just solved the fuel consumption problem by running the car on pure spark. Ten KW per stroke is a lot of power at 3000 RPM. There was created an oscillator that could deliver that kind of power, but it was also able to burn up distributors and spark plugs, but only after some much more serious ignition cables were provided. I think that the oscillator ran a couple hundred kilohertz. One monster RFI generator by any standard. And way to expensive for any car.

Using a mechanical relay does not seem like the best scheme for firing a spark plug, no matter what. back in 2007 there were all kinds of electronic devices that could have done the job very well.
In addition, there was the MARK 10 CD ignition system, and I had one, and that delivered a solid 300+ volt pulse into the ignition coil primary that normally got less than 12 volts because of the ballast resistor. That did fire even fouled plugs, and if ever there was an exposed conductor the spark would jump well over an inch, and make a loud "PoP noise.
 

MisterBill2

Joined Jan 23, 2018
11,880
The first problem that I see with electolyzing water to produce hydrogen and oxygen as fuel is that it consumes more energy than it delivers. The thermodynamic proof is simple: Electrolyzing water generates heat, and that heat is lost, not recovered. Thus the energy to produce the gas mix contains both the energy to split the hydrogen to oxygen bonds as well as the energy to produce the heat lost in the process. Thus even if the use of the fuel mix is 100% efficient, the production of the fuel consumes more energy than the fuel can deliver. How can that problem be avoided???
So certainly it can be done, but not economically, because of those losses.
 

MisterBill2

Joined Jan 23, 2018
11,880
The interesting fact within chemical reactions is that the energy released when the reaction goes in one direction is equal to the energy required to make the reaction go in the opposite direction. Some systems are able to come much closer to that point than others, but it still takes at least as much energy to break the hydrogen to oxygen bond as is released when they join. So at best it is a zero-sum game, at 100% effective conversion.
With the carbon arcs generating carbon monoxide as a fuel, yes, that is effective, BUT the combustion is no longer carbon-free. So there is a bit of compromise there.
 
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