Shottkey reverse leakage through limiting resistor

Thread Starter

Willard007

Joined Jan 22, 2020
11
Hi

I am a brand new member and hoping for a positive first foray into the forum.

I have what on the face of it is probably straightforward to most electrical engineers but the solution to this problem evades me.

I have a +5V supply to a 750 KOhm resistor connected to a cathode of a Shottkey diode which is grounded on the anode.

The shottkey has a specified 10uA reverse leakage at 5V.

Initially no current flows and the diode feels then the 5V and the reverse leakage current of 10uA flows this causes the resistor to drop the voltage at the cathode, as the voltage reduces the shottkey feels the drop in voltage and the current flow reduces.

What happens next, intuitively, the voltage should rise again at the cathode causing the leakage to increase again and repeat but I do not think this is how it would work.

What is the steady state result to this small arrangement.
 

ronsimpson

Joined Oct 7, 2019
400
We need a graph of leakage current verses reverse voltage. We know 10uA@5V and probably 0uA@0V, but we do not know all other points.
Just the resistor across 5V will have 6uA.
Initially no current flows and the diode feels then the 5V
You can also think about the reverse of that. Start out with 0V on the diode and 6uA. This will pull up on the diode until it find a voltage and current that works. (diodes have capacitance, the 6uA will charge up the capacitor)
 

Thread Starter

Willard007

Joined Jan 22, 2020
11
Thanks for the rapid response Ron, I'm new to this as you probably gathered. I don't have a graph as its really the principle that I'm trying to grasp.
I know that the leakage will reduce as voltage drops on the cathode due to the 750K limiting resistor but that will then cause the voltage to increase and the current to increase in response. How does it finally balance is the bit I cannot see.

Once the current gets to 6uV the voltage at the Cathode will be 0V and so no current will flow which means the voltage will be 5V

This seems to me like a paradox.

Best Regards

Will
 

MrChips

Joined Oct 2, 2009
19,915
Without knowing the I-V characteristics of the diode we have to resort to first order approximations.
Let us assume that the reverse current is proportional to the applied reverse voltage.
In other words, the diode looks like a resistor Rd, where

Rd = V / I = 5V / 10μA = 0.5MΩ = 500kΩ

Now you have a voltage divider consisting of 750kΩ + 500kΩ.

Diode voltage Vd = 5 * 500/1250 = 2V
I = 2V / 500kΩ = 4μA
 

Thread Starter

Willard007

Joined Jan 22, 2020
11
Thanks Dana, but where is the reverse leakage current featured. At 5 Volts there would be 10uA through the diode. Any help would be appreciated.
 

Thread Starter

Willard007

Joined Jan 22, 2020
11
Without knowing the I-V characteristics of the diode we have to resort to first order approximations.
Let us assume that the reverse current is proportional to the applied reverse voltage.
In other words, the diode looks like a resistor Rd, where

Rd = V / I = 5V / 10μA = 0.5MΩ = 500kΩ

Now you have a voltage divider consisting of 750kΩ + 500kΩ.

Diode voltage Vd = 5 * 500/1250 = 2V
I = 2V / 500kΩ = 4μA
 

Thread Starter

Willard007

Joined Jan 22, 2020
11
Thanks MrChips , however, if the voltage is now 2 Volts at the cathode the current will no longer be 10uA it will be less so the 750K will drop less Voltage and the 2V will increase.

I think I'm not quite getting this, and you are probably right but the logic escapes me.

Regards

Will
 

MrChips

Joined Oct 2, 2009
19,915
Thanks MrChips , however, if the voltage is now 2 Volts at the cathode the current will no longer be 10uA it will be less so the 750K will drop less Voltage and the 2V will increase.

I think I'm not quite getting this, and you are probably right but the logic escapes me.

Regards

Will
Review my calculations.
The diode voltage is now 2V. The diode current is 4μA.
The voltage across the 750kΩ resistor is 3V. The voltage adds up to 5V.
 

danadak

Joined Mar 10, 2018
3,785
Unfortunetaly the spice model may not be completye.

We also have the defining basic curves for diodes -

1579702682717.png


As you can see there is a zero leakage solution when Vdiode = 0, otherwise you would have a generator
of current.

Normally one uses load line analysis on actual curves to obtain solution.


Regards, Dana.
 

Thread Starter

Willard007

Joined Jan 22, 2020
11
Unfortunetaly the spice model may not be completye.

We also have the defining basic curves for diodes -

View attachment 197413


As you can see there is a zero leakage solution when Vdiode = 0, otherwise you would have a generator
of current.

Normally one uses load line analysis on actual curves to obtain solution.


Regards, Dana.
But the Voltage cannot get to zero volts because when current is Zero there is no voltage drop across the 750K resistor and the voltage will again be 5 volts
 

MrAl

Joined Jun 17, 2014
6,949
The reverse leakage should start at 0v reverse bias increasing as reverse voltage is increased and then at some point start to level off a little increasing less as voltage is raised, until the breakdown point is reached and then the current starts to increase much more.

Usually though the more important point is reverse leakage with temperature.
As the temperature is increased from 25C to 100C the leakage could go up by a factor of 100 and from 10C to 125C it could go up by another factor of 10. So from 25C to 125C it could go up by a factor of 1000.
 

danadak

Joined Mar 10, 2018
3,785
But the Voltage cannot get to zero volts because when current is Zero there is no voltage drop across the 750K resistor and the voltage will again be 5 volts
If there is no current thru junction then effectively its is infinite Z, therefore drop across R
and diode Z is 5 V which is placed across ~ the diode infinite Z = 0 current ?

Regards, Dana.
 
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