Hi,

I can figure out the voltage divider easy enough if -0- current is leaking through the diode using:

12*200/(200+800)=2.4V

But, what if the diode was leaking 500 uA ? What would the formula be to calculate its affect on voltage between X1 and GND?

Thanks,

Alan

 

If the 12V source is ideal then there will be no difference on the calculation.

 

But, what if the diode was leaking 500 uA ? What would the formula be to calculate its affect on voltage between X1 and GND?

The leakage has no effect for the ideal circuit as you have shown it. However, for a real 12 volt voltage source, there would be a source resistance (Rs), and the leakage current (Is)would change the voltage. The presence of the source resistance will change the voltage anyway, even if there is no leakage current through the diode.

I quickly calculate the formula as follows, but double check me.

\( V_{x1}={{R_2(R_s I_s+12V)}\over{R_1+R_2+R_s}}\)

 

Hi,

Yes, I should have stated both the 12 and 20 are not ideal. The 12 is a battery. The 20 is likely a regulated voltage adapter.

What is R3 and I3 in the formula? - edit - I see it is Rs not R3...

Thanks,

Alan

 

Hi,

Yes, I should have stated both the 12 and 20 are not ideal. The 12 is a battery. The 20 is likely a regulated voltage adapter.

What is R3 and I3 in the formula? - edit - I see it is Rs not R3...

Thanks,

Alan

Oh, those are "s" not "3". The Rs is the 12V source resistance and the Is is the leakage current in the diode.

 

Hi,

Where I am going with this is that I will know the value (20V), and value of X1, and I will know the 200K/800K because they are fixed. I believe I can figure out a formula to estimate the current the reverse current through D1 based on the 20V and X1. What I am trying to figure out is the actual voltage at "12V" without the influence of the current leaking through the diode...

Is this possible without calculating the source resistance, or can I calculate it based on the above known values?

Thanks,

Alan

 

Hi,

I'm still trying to figure this. Is the leakage current through the diode similar to if you had an appropriately sized resistor where the diode is? Is the leakage current driven by the voltage difference between the 12V and 20V? Why does the internal resistance of the 12V matter? I understand it matters in the overall formula, but how does it affect the amount of leakage current?

Thanks,

Alan

 

Hi,

I'm still trying to figure this. Is the leakage current through the diode similar to if you had an appropriately sized resistor where the diode is? Is the leakage current driven by the voltage difference between the 12V and 20V? Why does the internal resistance of the 12V matter? I understand it matters in the overall formula, but how does it affect the amount of leakage current?

Thanks,

Alan

Personally, I'm confused and am unsure what you are trying to do. So, I don't know how to help you further. Can you explain more about what the goal is. Is this an academic problem or a practical one. If the latter, what is the problem you are trying to solve.

Basically a good battery is going to have a small source resistance compared to the R1 and R2 values you specified. So it's going to be difficult to back out any data from simple measurements. The effects of the leakage should be small. Are you seeing the leakage affect your circuit? Keep in mind that leakage current not only depends on voltage, but also on temperature and the particular device you are using.

 

Hi Steveb,

I appreciate the help. I *thought* it was leakage from the diode that was causing the issue, but it was something internal to the ADC that was causing it. It was designed for 10K impedance, but that allows too much current to be wasted (1ma) from my battery. I ended up finding a place around 60K where it works well and cut my current waste down quite a bit.

Thanks,

Alan