May I know some facts here?

a. The 1N5819 diode with a white band denotes negative (-) terminal, so as to capacitors, right?
b. The -3V connection can be made by making the -3V connecting to the +ve terminal and the +3V to ground?
c. Is 0V consider as "-" terminal? As shown in red, how can i actually connect the negative source to ground providing I have only one source of voltage?

Thanks for the silly questions.

 

a. The band on a diode denotes the cathode. Not all capacitors are so marked - a radial tantalum cap will have a stripe on the anode side.

b & c. Voltages are relative to the measuring point. The lower rail, marked 0 volts, is going to be the common point for all voltage measurements. Attaching the battery positive terminal to it means that a voltage from that battery to a load will be negative. Swapping the terminals simply makes the applied voltage positive.

 

a. Most diodes have a band on their cathode (-) end. The positive end is called the anode.

b. I don't know where you got that schematic, but it's terrible along with being nonsensical. You wouldn't get -3v at the point where it is shown.

c. The upside-down "T" is one symbol used for "ground", which is considered to be a 0v reference point. Every circuit must have some point of reference.
Other common symbols used for ground are an inverted triangle, and a stylized inverted triangle made of several horizontal lines.

You would connect your battery negative (-) terminal to the "Source-" connection point.
You would connect your battery positive (+) terminal to the "Source+" connection point.

 

a. The band on a diode denotes the cathode. Not all capacitors are so marked - a radial tantalum cap will have a stripe on the anode side.

b & c. Voltages are relative to the measuring point. The lower rail, marked 0 volts, is going to be the common point for all voltage measurements. Attaching the battery positive terminal to it means that a voltage from that battery to a load will be negative. Swapping the terminals simply makes the applied voltage positive.

Thanks for your reply. Well,

.. Attaching the battery positive terminal to it means that a voltage from that battery to a load will be negative ...

Does this means that I may get the -3V connection by doing so? (Connecting the positive terminal to the ground while load to the positive terminal?) Does this capacitor the right connection?

In a circuit board, normally the two upper line will be used for positive and negative from the source. How to provide a common reference point (0V) to it?

Thanks again.

 

What are you trying to do with this circuit?

 

a. Most diodes have a band on their cathode (-) end. The positive end is called the anode.

b. I don't know where you got that schematic, but it's terrible along with being nonsensical. You wouldn't get -3v at the point where it is shown.

c. The upside-down "T" is one symbol used for "ground", which is considered to be a 0v reference point. Every circuit must have some point of reference.
Other common symbols used for ground are an inverted triangle, and a stylized inverted triangle made of several horizontal lines.

You would connect your battery negative (-) terminal to the "Source-" connection point.
You would connect your battery positive (+) terminal to the "Source+" connection point.

Thanks for your reply.

..b. I don't know where you got that schematic, but it's terrible along with being nonsensical. You wouldn't get -3v at the point where it is shown...

Well that is part of the schematic. The -3V basically connected to a 100nF capacitor. Can I reverse the terminal of the source/voltage supply in order to get the -3V?

Thanks

 

Your schematic makes no sense. Please post a better one or tell us what the circuit is going to do.

 

What are you trying to do with this circuit?

Basically it is just part of the schematic. There are still components for the dotted line. That circuit is basically for watt meter (schematic here). I just want to know some basic connection on the source, -3V and the ground.

Thanks.

 

Well, if you need a unipolar supply (just positive voltage) connect the negative terminal of the battery to zero volts rail and the positive terminal to the positive voltage rail on the schematic. If you need a bipolar supply (both positive and negative voltages) connect another battery with its positive terminal connected to the same zero volts rail and use the negative terminal for the negative rail on the schematic.

 

Well, if you need a unipolar supply (just positive voltage) connect the negative terminal of the battery to zero volts rail and the positive terminal to the positive voltage rail on the schematic. If you need a bipolar supply (both positive and negative voltages) connect another battery with its positive terminal connected to the same zero volts rail and use the negative terminal for the negative rail on the schematic.

Thanks for your reply. I think from the schematic, I might having bipolar connection as "+" and "-" terminals are needed. One connected to the + rail, other to - rail. So basically there is no 0V or reference point, am i right?

 

From the schematic you need a unipolar power supply, thus use only one battery with its negative side on the zero volts rail.

 

Well that is part of the schematic. The -3V basically connected to a 100nF capacitor. Can I reverse the terminal of the source/voltage supply in order to get the -3V?

I see. The schematic you linked to uses and ICL7660, which is a voltage inverter. It'll take an input up to 10v, and output up to around -9v. This "negative rail" is used by IC3A, an opamp.

Polarized capacitors must always have their negative lead connected to the side of the circuit with the lower (more negative) voltage. Some capacitors (Tantalums, for example) have only their positive lead marked - these must have the positive lead connected to the more positive side.

If you reverse the polarity across a polarized capacitor, it will go out with a "bang".

 

I see. The schematic you linked to uses and ICL7660, which is a voltage inverter. It'll take an input up to 10v, and output up to around -9v. This "negative rail" is used by IC3A, an opamp.

Polarized capacitors must always have their negative lead connected to the side of the circuit with the lower (more negative) voltage. Some capacitors (Tantalums, for example) have only their positive lead marked - these must have the positive lead connected to the more positive side.

If you reverse the polarity across a polarized capacitor, it will go out with a "bang".

I uses the normal capacitors with a white band indicates negative terminal, i guess. I connected exactly the same as shown by the schematic. So for a battery of 9V for example, the + terminal will the positive terminal while "-" having the ground? May I know the differences between "-" terminal and ground in terms of connection from a battery? Thanks a lot.

P/S: So the ICL uses 10V. But the LM2936 has output a low voltage to 3.3V if not mistaken. So the source should have more than 10V?

 

I uses the normal capacitors with a white band indicates negative terminal, i guess.

What's a "normal capacitor?"
There are many different types of capacitors.
Tantalum capacitors tend to have their positive terminals marked with a white band.
Electrolytic capacitors tend to have their negative terminals marked with a black band.

I connected exactly the same as shown by the schematic. So for a battery of 9V for example, the + terminal will the positive terminal while "-" having the ground?

The "ground" is the "0v" reference point. It really doesn't have anything to do with the battery; merely where is the ground considered to be in the circuit.

May I know the differences between "-" terminal and ground in terms of connection from a battery?

As I mentioned above, it all depends upon the schematic diagram.

P/S: So the ICL uses 10V. But the LM2936 has output a low voltage to 3.3V if not mistaken. So the source should have more than 10V?

The ICL7660 has an input voltage range of 1.5v to 10v. It's output will be nearly the same amplitude, but the opposite polarity. Since it's being powered from the output of the LM2936, it's output will be somewhere between -3v and -3.2V, depend upon load.

If it's input voltage is greater than 10v, it will be destroyed.

The ICL7660A can operate with input voltages from 1.5v to 12v.

 

What's a "normal capacitor?"
There are many different types of capacitors.
Tantalum capacitors tend to have their positive terminals marked with a white band.
Electrolytic capacitors tend to have their negative terminals marked with a black band.

Sorry, I am using the Electrolytic capacitors. There are two types. One with black band as you'd mentioned. Another one is with white band. A triangle-like symbol on the band is negative terminal.

The "ground" is the "0v" reference point. It really doesn't have anything to do with the battery; merely where is the ground considered to be in the circuit.

As I mentioned above, it all depends upon the schematic diagram.

Well, on a circuit board, it will have two rails to support the positive and negative source. Normally positive above and negative right below. If I am using just a battery of 9V, I am doubtful to get a ground connection.

The ICL7660 has an input voltage range of 1.5v to 10v. It's output will be nearly the same amplitude, but the opposite polarity. Since it's being powered from the output of the LM2936, it's output will be somewhere between -3v and -3.2V, depend upon load.

If it's input voltage is greater than 10v, it will be destroyed.

The ICL7660A can operate with input voltages from 1.5v to 12v.

Thanks for the info.