Here's the schematic:
I circled each 9-18VDC in red, +V in orange, 1/2V in green, and +5V in blue.
Any help/lessons on what these exactly mean would be greatly appreciated!
Schematic notes are confusing me a bit...
Hello, this is my first post on the forums, and I've learned a lot so far just from reading all of the information posted here. I've done multiple projects involving soldering, but I'm rather inept when it comes to reading even slightly complex schematics, so I could use some help... I came across a schematic from a BYOC Reverb unit that I'd like to put on veroboard (I consider myself a beginner at schematic-veroboard transfer too, but let's save that for another time) that left me a bit confused, particularly with the voltage divider, power supply, and a couple of passive components. I see the notes 9-18VDC, +V, 1/2V, and +5V, and I don't quite understand where to put them or how to separate them, or even what the last 3 on that list mean. That may seem like a trivial question, but I am in the Absolute Beginners area, so hopefully this won't be seen as silly...
Here's the schematic:
I circled each 9-18VDC in red, +V in orange, 1/2V in green, and +5V in blue.
Any help/lessons on what these exactly mean would be greatly appreciated!
Here's the schematic:
I circled each 9-18VDC in red, +V in orange, 1/2V in green, and +5V in blue.
Any help/lessons on what these exactly mean would be greatly appreciated!
-
PokeyPete
- Resistor Ronker
The 9-18VDC is simply your power source. The circuit is designed for any voltage in that range. Generally speaking, the higher the voltage used, the more "headroom" you will have.
The power source can be a 9 volt battery, or two 9 volt batteries in series (18V), or an external power supply.
The +V is whatever you decide the power source to be. In other words 9-18VDC = +V
The 1/2V is taken from the voltage divider. Notice that +V is connected to two 10K resistors in series, then goes to ground. These resistors are equal, so each drops half of
the supply voltage. For example.....if using a 9V source, with a volt meter, you would measure 4.5V across each resistor. So, with one probe at ground, and the other probe at
the point between the two resistors you would read 1/2 of the supply voltage. This 1/2V is used as a reference voltage at various points in the circuit.
Some IC's, such as TTL devices, use 5V as the supply voltage. But what is taking place here is that your power source is going through a 5V voltage regulator (78L05). Because the
9V supply would be too much to power that IC, switch, or etc., the power source (9V or whatever) is reduced by the regulator to the 5V level that the device needs for power.
Hope this makes a little sense.
The power source can be a 9 volt battery, or two 9 volt batteries in series (18V), or an external power supply.
The +V is whatever you decide the power source to be. In other words 9-18VDC = +V
The 1/2V is taken from the voltage divider. Notice that +V is connected to two 10K resistors in series, then goes to ground. These resistors are equal, so each drops half of
the supply voltage. For example.....if using a 9V source, with a volt meter, you would measure 4.5V across each resistor. So, with one probe at ground, and the other probe at
the point between the two resistors you would read 1/2 of the supply voltage. This 1/2V is used as a reference voltage at various points in the circuit.
Some IC's, such as TTL devices, use 5V as the supply voltage. But what is taking place here is that your power source is going through a 5V voltage regulator (78L05). Because the
9V supply would be too much to power that IC, switch, or etc., the power source (9V or whatever) is reduced by the regulator to the 5V level that the device needs for power.
Hope this makes a little sense.
“No man is so foolish but he may sometimes give another
good counsel, and no man so wise that he may not easily err
if he takes no other counsel than his own. He that is taught
only by himself has a fool for a master.”
–Hunter S. Thompson
good counsel, and no man so wise that he may not easily err
if he takes no other counsel than his own. He that is taught
only by himself has a fool for a master.”
–Hunter S. Thompson
Thanks for the quick and informative reply! So just to make sure I have this right... In terms of veroboard rows, I'd connect the 9-18VDC and V+ to the +9V row, connect the point between the two resistors to another row and connect all 1/2v points to that row, and connect the left end of the voltage regulator on the schematic to the +9V row and connect the right end to another row for the 5V connection on the IC?
-
PokeyPete
- Resistor Ronker
That is essentially correct. But please be advised of one more thing.
The schematic shows your supply voltage going to three places:
1) The LED indicator
2) The Voltage divider
3) And, as input supply to the voltage regulator
But, what is not shown (just implicitly implied) is the +V supply, and ground, of the TL072 op amps.
These need their power source as well.
The schematic shows your supply voltage going to three places:
1) The LED indicator
2) The Voltage divider
3) And, as input supply to the voltage regulator
But, what is not shown (just implicitly implied) is the +V supply, and ground, of the TL072 op amps.
These need their power source as well.
“No man is so foolish but he may sometimes give another
good counsel, and no man so wise that he may not easily err
if he takes no other counsel than his own. He that is taught
only by himself has a fool for a master.”
–Hunter S. Thompson
good counsel, and no man so wise that he may not easily err
if he takes no other counsel than his own. He that is taught
only by himself has a fool for a master.”
–Hunter S. Thompson