small voltage supply

[the cracken]

i was wondering if anyone had any good ideas on making a voltage supply using 2 aa or aaa batteries (3V) that could go down to about 0.001V???
i have know i idea how to go about this!!
thanks for any help!

 

[MikeMl]

Is this for calibrating a thermocouple, for example?

I have done something similar where I needed a calibrated signal of 15.000mV. I started with a fresh Alkaline cell (I think I used a D-cell). I measured the voltage at 1.540V. A voltage divider will work for this:

Say that the upper resistor is R1 and the lower one is R2. Vout = Vin*R2/(R1+R2)

or, in my case, 0.015 = 1.54*R2/(R1+R2).

or R2/(R1+R2) = 0.015/1.54 = 0.00974

Now, for my example, I knew that the input impedance of where the output was going to be connected was very high (LCD Voltmeter), so I could pretty much pick any value of resistors without regard to loading the voltage divider. If you know what impedance the output voltage is fed to, you may have to consider that.

Back to my example. To not appreciably load the battery, I choose R1 to be 100K.

That allows me to solve for R2. Namely:

R2/(100000+R2) = 0.00974

R2 = 0.00974(100000 + R2)

R2 = 974 + 0.00974R2

R2 = 974/(1-0.00974) = 983Ω

Now 100K is a standard value, but 983 is not. However, if you break 983 into two 1% resistors, you can buy a 909Ω and a 75Ω. So three 1% 1/4W metal film resistors will do it...

Do the math for your specific example.

Note that you could just simplify the math thusly:

The current through the voltage divider is 99.99% determined (ignoring R2) by R1.

To get 1mV from 3V, and say you are willing to have 1mA flowing in the divider, then:

R1 = E/I = 3/0.001 = 3000.

Now to get 1mV with 1mA, R2 = E/I = 0.001/0.001 = 1Ω

So R1=3KΩ and R2=1Ω will work; so would 30KΩ and 10Ω, or 300KΩ and 100Ω, but that brings back the input impedance question?

 

[the cracken]

Is this for calibrating a thermocouple, for example?

I have done something similar where I needed a calibrated signal of 15.000mV. I started with a fresh Alkaline cell (I think I used a D-cell). I measured the voltage at 1.540V. A voltage divider will work for this:

Say that the upper resistor is R1 and the lower one is R2. Vout = Vin*R2/(R1+R2)

or, in my case, 0.015 = 1.54*R2/(R1+R2).

or R2/(R1+R2) = 0.015/1.54 = 0.00974

Now, for my example, I knew that the input impedance of where the output was going to be connected was very high (LCD Voltmeter), so I could pretty much pick any value of resistors without regard to loading the voltage divider. If you know what impedance the output voltage is fed to, you may have to consider that.

Back to my example. To not appreciably load the battery, I choose R1 to be 100K.

That allows me to solve for R2. Namely:

R2/(100000+R2) = 0.00974

R2 = 0.00974(100000 + R2)

R2 = 974 + 0.00974R2

R2 = 974/(1-0.00974) = 983Ω

Now 100K is a standard value, but 983 is not. However, if you break 983 into two 1% resistors, you can buy a 909Ω and a 75Ω. So three 1% 1/4W metal film resistors will do it...

Do the math for your specific example.

Note that you could just simplify the math thusly:

The current through the voltage divider is 99.99% determined (ignoring R2) by R1.

To get 1mV from 3V, and say you are willing to have 1mA flowing in the divider, then:

R1 = E/I = 3/0.001 = 3000.

Now to get 1mV with 1mA, R2 = E/I = 0.001/0.001 = 1Ω

So R1=3KΩ and R2=1Ω will work; so would 30KΩ and 10Ω, or 300KΩ and 100Ω, but that brings back the input impedance question?

i like the simplicity but this all depends on the voltage of the battery! i was thinking more of a small voltage regulated supply if they exist?

 

[MikeMl]

Then start with a band-gap voltage reference like an LM431 or TL431, and a minimum 3V battery. The voltage divider is unchanged. A fresh Alkaline cell, very lightly loaded, has a VERY predictable and stable voltage.

 

[Vizier87]

What kind of device is powered up with 0.0001V?

 

[Hero999]

How much current do you require?

Use a reference such as the TL431 as mentioned above.

If you need to drive a low impedance load (<10k) then you'll need to add a buffer amplifier, maybe a small op-amp.

 

[the cracken]

im powering a diode!
i want to have an adjustable hard clip. from no signal to no clipping. the only way i could think of doing that is to have a 3 diodes and a small variable voltage. when i turn up the voltage it clips less!

 

[ericgibbs]

im powering a diode!
i want to have an adjustable hard clip. from no signal to no clipping. the only way i could think of doing that is to have a 3 diodes and a small variable voltage. when i turn up the voltage it clips less!

hi,
The problem with the clipping diode biasing that you are setting to 1mV resolution, is the fact that the diode is effected by temperature changes in the order of 2mV/Cdeg.

 

[the cracken]

hi,
The problem with the clipping diode biasing that you are setting to 1mV resolution, is the fact that the diode is effected by temperature changes in the order of 2mV/Cdeg.

do you know of another way of doing it? i cant think of any other way!