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[SOLVED] reading schematics issue

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this

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Hi
I know it's probably very simple question, but I wasn't sure how to ask google :)
So, I got this schematic:

and I'm not sure what do I do with the part in left-bottom corner as it's separated from the rest of the circuit.

TIA
 

srizbf

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the left part of the circuit shows the power supply connection.
The +9v in left part is to be connected to all the points where it is marked as +9v.

and the gnd symbol, is to connected to all points where it is shown in the schematic.

it is a convenient way of drawing power supply.
so that you need not show the individual connection to supply by wire line.

in general , the rule is connect all points that have the same node symbols.

hope it is clear.
 
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jpanhalt

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That is the +9V supply for the circuit. Just connect the +9V output from the battery switch to the three points that are labeled +9V. Supplies are often drawn like that to avoid a lot of confusing lines. Similarly, wires connecting all of the grounds are not shown for clarity.

This schematic also doesn't show a "decoupling" capacitor across the power pins (4,8 ) of the TLV3702. I suspect there is one (0.1 uF ceramic) attached very close to the pins. There have been extensive discussions here on decoupling capacitors.

Here is a link to that thread: https://www.edaboard.com/thread212775.html

John
 

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Thanks for your answers guys.
From what I understand it should be something like this:(sorry for really bad drawing:) )

And all ground connections should just go to negative battery terminal. Is this correct?

Thanks for your patience :)
 

jpanhalt

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No! You have the battery connected backward. The triangle is "Ground." The plus side of the battery has the switch. That goes to th +9V symbols.

The battery symbol is the thing with the long line/short line/long line/short line,etc. The short line is ground ("-"). You do not need another battery.

Sorry, I misunderstood your original question. That "thing" is the battery.

John
 
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alexan_e

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All the triangles represent the ground and should be connected between them.
All the +9 symbols are the positive supply and should be connected between them

81860888.gif

Alex
 
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I know it's a battery symbol, sorry, I mixed it all up, brain fart.
Disregard my drawing please, let's start again.

"Supply circuit" - the separated part of the original drawing(the one I moved to the right-top corrner and added some rubbish)
"Main circuit" - rest of the original drawing.

1)All "+9" parts of the main circuit should be connected to "+9" lead of supply circuit.
2)All gnd parts of main circuit goes to negative battery terminal.
Is this correct now?

--edit
OK, thank you all, got this now. I knew it was simple :)
 

andre_teprom

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this,

the same way you can adopt GND symbol ( wich implicitly connects all them ), also can represents VCC separatelly ( here, denoted by +9v ), in order to clarify design.

+++
 

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I don't know if I should create new thread for this, so I'll ask here:
I don't have TLV3702 but I have LM393N, may I use it as a replacement?
Quick googleing suggest that I could, but I want just want to be sure.
 

jpanhalt

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I don't think so. The LM393 is open collector output. The TLV3702 is push-pull output -- it can actually source current. I need to try to find the block schematic for the TLV3702 to confirm that. If you look at your circuit, effectively grounding pin 7 (open collector) in the top comparator won' do much.

Unfortunately, I am at dinner time, so I need to take a break for an hour or so. That probably won't work well for your time zone. I will look at it later. Maybe someone else will have a definitive answer before then. Your circuit might be modified to work with the LM393, though. A simple pull-up resistor on each out put may work.

John
 

jpanhalt

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Here is a comparison of the two types of comparator outputs:

[/url][/IMG]

Everything inside the gray rectangles is inside the comparator. The push-pull type has an internal connection to V+. It can source or sink current. The open-collector (also called, open-drain) type can only sink current to ground when the output transistor is turned on. An external resistor ("pull-up" resistor), makes the output voltage high when the output transistor is off. Current is sourced through that resistor. For more detail see: Comparator - Wikipedia, the free encyclopedia. Scroll to Output type.

Can your LM393 replace the TVL3702? I don't know for sure, but you can give it a try. The circuit does not appear to be too demanding on the comparators, except they need to source current. So, you could try a pull-up resistor on each output.

What size resistor? For the greatest voltage swing, you would use a large resistor, such as 10K; however, that won't be able to provide much current. On the other hand, a much smaller resistor will use quite a bit of current when the output transistor turns on, because it will be connected directly from V+ to ground minus the voltage drop of that output transistor. You want both good current and a large voltage swing to turn on the 2N7000 mosfet. I would try a 1K resistor (the LM393 is limited to 18 mA current sink on its output). If that doesn't turn on the mosfet, then I would try 4.7K or whatever resistor you have that is larger.

Please give us updates. There are numerous push-pull comparators made. You may be able to get one of them instead of the TLV3702, if these modifications do not work.

John
 

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