So I have two batteries of 9V (so 18V) and a 1MegaOhm resistor.

What I want is a 1 mA current through the resistor. How do I accomplish this? What other components do I need in order to force 1 mA through this resistor?

If I just connect the 18V to the resistor, by Ohm's law, I would not accomplish anything.

Obs: Of course, all this without increasing the voltage so I get 1mA.

Is it possible?

 

nope.

 

So I have two batteries of 9V (so 18V) and a 1MegaOhm resistor.

What I want is a 1 mA current through the resistor. How do I accomplish this? What other components do I need in order to force 1 mA through this resistor?

If I just connect the 18V to the resistor, by Ohm's law, I would not accomplish anything.

Obs: Of course, all this without increasing the voltage so I get 1mA.

Is it possible?

Why would you think it is possible?

By Ohm's Law, how much voltage MUST appear across the resistor to get 1 mA through it?

 

1000 volts will "push" 1 mA through a resistance of 1 Meg.

A flyback transformer circuit could possibly give you short duration pulses that approach 1000 volts from 18 volts DC.

 

1000 volts will "push" 1 mA through a resistance of 1 Meg.

A flyback transformer circuit could possibly give you short duration pulses that approach 1000 volts from 18 volts DC.

Though that would violate his spec of doing this without increasing the voltage.

@RdAdr : Could you clarify what you meant by not wanting to increase the voltage? Are you saying you truly don't want to increase the voltage but still somehow get 1 mA through a 1 MΩ resistor? Or are you just saying that you want to keep the 18 V as your power source but are willing to include components that increase that to the necessary 1000 V?

If the latter, keep in mind that even with 100% efficiency in the voltage conversion you would need to draw 56 mA from the batteries. This is not an unreasonable current for a 9V alkaline battery, though your continuous-duty service life would only be about 8 hours. Realistically, you would probably be pulling upwards of 100 mA and thus only seeing about four hours of life.

 

1000 volts will "push" 1 mA through a resistance of 1 Meg.

A flyback transformer circuit could possibly give you short duration pulses that approach 1000 volts from 18 volts DC.

Years ago I was developing an SMPSU for a pocket xenon strobe. The transformer came from an 80s plastic computer ( Atari/Amiga - something like that).

Basically just a blocking oscillator driving the chopper transformer backwards for step-up - just hooked up to see if I could get away with it, I didn't include any regulation loop - the flyback pulses produced a steady arc between the (originally) primary pins. It blew a 100W/240V bulb - my best guess was somewhere in the general direction of 900V or so.

 

So THAT'S how you strobe pocket xenons!

 

Stop and think. You have E = 18 volts. You have R = 1M-Ohm. Ohm's Law says I cannot be 1mA. So, you're not going to get it done with those two. In order to change that, you need just 1 more resistor. But how do you determine it?

First, you have to determine what Resistance _is_ necessary to achieve what you want, so you have a target. Ohm's law again comes to the rescue:

R = E/I
R = 18/0.001
R = 18000 Ohms

So.... how do you add just 1 resistor in parallel with the 1M-Ohm resistor to achieve what you need? There's an equation for that-- and if you know Ohm's Law, you should also know the equations for series and parallel resistances.

1/R = (1/R1 + 1/R2)

 

Stop and think. You have E = 18 volts. You have R = 1M-Ohm. Ohm's Law says I cannot be 1mA. So, you're not going to get it done with those two. In order to change that, you need just 1 more resistor. But how do you determine it?

First, you have to determine what Resistance _is_ necessary to achieve what you want, so you have a target. Ohm's law again comes to the rescue:

R = E/I
R = 18/0.001
R = 18000 Ohms

So.... how do you add just 1 resistor in parallel with the 1M-Ohm resistor to achieve what you need? There's an equation for that-- and if you know Ohm's Law, you should also know the equations for series and parallel resistances.

1/R = (1/R1 + 1/R2)

And that is somehow magically going to result in 1 mA through the 1 MΩ resistor ?!?!?!

 

And that is somehow magically going to result in 1 mA through the 1 MΩ resistor ?!?!?!

Not exactly, but... 18V across 2 parallel resistors of the right values will result in a 1mA current in the circuit. As close as they are going to get. You know your electronics fundamentals. In which case you know that parallel resistance between two resistors results in a resistance being equivalent to being closer to the lower of the two resistor values (less resistance).

I left it as an exercise for the OP to rework the parallel resistance equation to solve for that 2nd resistor value.

 

Not exactly, but... 18V across 2 parallel resistors of the right values will result in a 1mA current in the circuit. As close as they are going to get. You know your electronics fundamentals. In which case you know that parallel resistance between two resistors results in a resistance being equivalent to being closer to the lower of the two resistor values (less resistance).

I left it as an exercise for the OP to rework the parallel resistance equation to solve for that 2nd resistor value.

This is like telling someone that asks how to get 4 gpm of water to flow though a tiny pinhole without increasing the water pressure that all they need to do is open a faucet somewhere else in the house until the total flow rate in the entire house is 4 gpm. It solves a problem that doesn't exist and does nothing to solve the problem that does.

The TS doesn't care about how much current is flowing in the circuit -- he very specifically is asking about how to get 1 mA to flow through a 1 MΩ resistor.

 

That's true, but with Boba's solution, we can pretend.

 

This is like telling someone that asks how to get 4 gpm of water to flow though a tiny pinhole without increasing the water pressure that all they need to do is open a faucet somewhere else in the house until the total flow rate in the entire house is 4 gpm. It solves a problem that doesn't exist and does nothing to solve the problem that does.

The TS doesn't care about how much current is flowing in the circuit -- he very specifically is asking about how to get 1 mA to flow through a 1 MΩ resistor.

I read what the original poster (OP) asked for, and it sounded more like they didn't understand what they were doing. My example gives them something to think about that does provide a solution that may actually be what they needed. If nothing else it gives them more understanding or to think about.

Given the scenario stated by the OP, what I provided is as close as he is going to get by simply adding another component.

 

Hello,

To have 1 mA in an 1 MegOhm resistor a voltage of 1000 Volts is required.
(just look at Ohms law : http://www.allaboutcircuits.com/textbook/direct-current/chpt-2/voltage-current-resistance-relate/ )

Bertus

 

So I have two batteries of 9V (so 18V) and a 1MegaOhm resistor.

What I want is a 1 mA current through the resistor. How do I accomplish this? What other components do I need in order to force 1 mA through this resistor?

If I just connect the 18V to the resistor, by Ohm's law, I would not accomplish anything.

Obs: Of course, all this without increasing the voltage so I get 1mA.

Is it possible?

In light of other respondents on this thread, could you please elaborate as to why you want precisely 1mA throught that specific resistor at 18V or is this simply for argument's sake, as it is physically impossible, not ever going to happen.

 

Since you guys are working on solutions for impossible problems. Could you get my 401k to generate more interest with the stability of a bond fund?

 

Can we pretend?

 

So I have two batteries of 9V (so 18V) and a 1MegaOhm resistor.

What I want is a 1 mA current through the resistor. How do I accomplish this? What other components do I need in order to force 1 mA through this resistor?

If I just connect the 18V to the resistor, by Ohm's law, I would not accomplish anything.

Obs: Of course, all this without increasing the voltage so I get 1mA.

Is it possible?

Current is a result of voltage across a resistance. You do not "generate a current". You generatev a voltage. How much current will flow is a matter of resistance.

If your objective is to het 1 mA out of 18 V with a 1 M ohm load on the18 V you need to chop up the 18 V DC then run it through a step down transformer to about 300 mV.

 

Though that would violate his spec of doing this without increasing the voltage.

@RdAdr : Could you clarify what you meant by not wanting to increase the voltage? Are you saying you truly don't want to increase the voltage but still somehow get 1 mA through a 1 MΩ resistor? Or are you just saying that you want to keep the 18 V as your power source but are willing to include components that increase that to the necessary 1000 V?

If the latter, keep in mind that even with 100% efficiency in the voltage conversion you would need to draw 56 mA from the batteries. This is not an unreasonable current for a 9V alkaline battery, though your continuous-duty service life would only be about 8 hours. Realistically, you would probably be pulling upwards of 100 mA and thus only seeing about four hours of life.

I could be wrong but I read that as a need to keep theb1 M ohm load on the18 V. If the objective is to get1 mA regardless of voltage step down the voltage.
In the final word I think his answer is simply just "no".

 

This must be where that imaginary "i" variable comes into play..