Good morning-) any one to answer my Question?...can ι put amperage on this circuit?

 

Your question is not clear. Maybe the answer is: The 9 volt battery will provide the number of milliamps that is can push through the resistors. In Gopher T's circuit (or is it #12's circuit?) the current will be about 9 milliamps.

 

Your question is not clear. Maybe the answer is: The 9 volt battery will provide the number of milliamps that is can push through the resistors. In Gopher T's circuit (or is it #12's circuit?) the current will be about 9 milliamps.

Ηello-) I can raise that milliamps? it is possible?

 

In my first circuit, I told you to use a 10k resistor and a 60 cm piece of bare, 20-gauge wire (every 3cm from ground is a out one microVolt when using a 9V battery to power it. 60 cm from ground will give you 20 microvolts.

If you want a potentiometer instead of moving a clip somewhere along the 60cm wire...

9V +
|
1k resistor
|
10k pot linear taper (wiper connected to 1k resistor side of pot)
|
6 cm piece of wire. - tap here for various voltages (1 to 20 uV)
|
Ground (negative terminal of 9V battery)

Turn the pot to change the voltage from the tap used off of the bare copper wire. Arrange tap position to adjust voltage range when pot is turned far to one side. Then turn far to other direction to get (~10x voltage). You may find you get 10% of initial voltage if pot was connected revers of what I imagine but you will figure it out easily).

please can't understood very good the system for potentiometer, can you drow for me? please

 

Sorry styven, you seem to be requesting two different things that do not seem related to each other.

Somebody already asked what do you actually intend to do. Yo did not explain with enough detail.

As Gopher asked, are you trying to generate a signal varying from 2 to 20 uV? If yes, what the increase of current has to do with it?

 

please can't understood very good the system for potentiometer, can you drow for me? please

Try this.

What are you connecting it to? It may make a difference in accuracy.

Note that 20 gauge wire is about 3.3mOhms per 10 cm. Use ohms law to confirm my math if you wish.

Remember, you are monitoring a very small voltage drop across the 10cm of copper wire (20 gauge). Do not measure across the resistors.

 

Sorry styven, you seem to be requesting two different things that do not seem related to each other.

Somebody already asked what do you actually intend to do. Yo did not explain with enough detail.

As Gopher asked, are you trying to generate a signal varying from 2 to 20 uV? If yes, what the increase of current has to do with it?

Hello-)..i was wrote fiew days ago, i will use that for detect fich in sea..all fich emitting voltage, so i can detect that-)

 

Try this.

What are you connecting it to? It may make a difference in accuracy.

Note that 20 gauge wire is about 3.3mOhms per 10 cm. Use ohms law to confirm my math if you wish.

Remember, you are monitoring a very small voltage drop across the 10cm of copper wire (20 gauge). Do not measure across the resistors.

View attachment 109223

Hello GopherT,.i understood now-) thxs-).. but the first circuit is the best, working very well -)).i do vey nice job in the sea-))