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Topic Title: Convert 9 V battery into two separate 1.5 mv currents?
Topic Summary: Convert 9 V battery into two separate 1.5 mv currents
Created On: 12 December 2013 08:48 PM
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 12 December 2013 08:48 PM
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jeffanderson

Posts: 1
Joined: 12 December 2013

I am looking to design a small box that houses a single 9-V battery (as power supply), and has two anodes (separate wires) and two cathodes (also separate wires) as outputs.

I need for the current to get 1.5 Mv coming out of each anode (to flow across a surface to get to the respective cathode) , and it is important that not go above this value.

Can anybody help me design the appropriate circuit to try out on my breadboard. Not sure what I would need or how to do it.

Thanks,

Jeff
 12 December 2013 08:57 PM
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kengreen

Posts: 400
Joined: 15 April 2013

hi Jeff,

I am a bit rusty on this sort of puzzle but I think that if you go through the catalogues you will find an i/c that will perform the trick of providing two isolated outputs.

Alternatively, if you want to build it yourself, take two separate feeds each via an emitter-follower type of circuit.

Ken Green
 12 December 2013 11:16 PM
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ANFierman

Posts: 136
Joined: 25 July 2008

Assuming you mean that you want 1.5mA (and not 1.5x10^6 Volts!) then the very simple little circuit I have posted on EasyEDA should do what you want:

http://easyeda.com/file_view_D...battery_5zVPWQKqW.htm

If you follow the instructions in the schematic then you can run a DC Sweep of the output currents from each source vs. the battery voltage.



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Andy Fierman

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http://signality.co.uk
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 13 December 2013 08:36 AM
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ectophile

Posts: 546
Joined: 17 September 2001

I think you need to be a bit clearer as to what you want. Is it 1.5 millivolts or 1.5 megavolts? The circuits for the two would be rather different!

Also, 1.5 mV (or 1.5 MV) is a voltage, not a current. When designing such a circuit, the required current would matter.

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S P Barker BSc PhD MIET
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