Solving simple diode resistor circuit

[Chai]

How do I go about solving this circuit's voltage drops and current draw? Just a step by step on how it's done for each section. The diode on the left has 2.2Vf at 1A, the right has 0.8Vf.

http://tinyurl.com/jnklyel

[rstofer]

You know the voltage drop across the top resistor (415) so work out the current.
It better equal the given current in the two legs.
All of the above is redundant, you already know the voltage and current for the top resistor

You don't have Vf of the diodes but the Vf is simply the 2.7V rail - the voltage drop across the leg resistor

For the left leg, Vf = 2.7 - (500 * 0.0019) or 1.75V => 2.7 - (500 Ohm resistor times given current of 1.9 mA)
For the right leg, Vf = 2.7 - (100 * 0.0204) or 0.66V => 2.7 - (100 Ohm resistor times given current of 20.4 mA)

[Chai]

Lets say you don't know drops or currents from the start. I put those in just to see. I want to solve all that myself. Sorry. The diode on the left has 2.2Vf at 1A, the right has 0.8Vf.

[Ian.M]

Assuming you aren't using a SPICE package, with good models for both diode, you don't solve it!

The problem is you have insufficient data for the diodes Vf at low currents.  You can make a rough stab at it by assuming that the diodes follow the simplified ideal Shockley diode equation, then doing mesh analysis, or by iteration, treating the diodes as voltage sources, calculating the current in each arm, then repeating for the diodes new Vf calculated from the previous step's currents, but its a total PITA and fairly inaccurate without more Vf vs If values for each diode to let you fit the ideality factor and internal resistance to improve your diode models

If it was a single diode, you could solve it graphically by plotting a load line on the diode's Vf vs If curve . . . . if you had that curve in the first place!

[Chai]

nm

[Chai]

So there is no way to kinda intuit the voltage level where the line spits after R415?

[rstofer]

Sure!  Consider the diodes to be 0.7V batteries of polarity opposite the battery and slide them up toward the source.  Instead of 12V, work the problem for 11.3V

I'd have to work through the solution and see if the transformation is valid but I think it is.  That horizontal rail puts the same voltage on each diode (draw the diode as the first thing after the rail).  If that's the case, just put 1 diode in series with the top leg.  Then slide it all the way to the power supply and subtract 0.7V to get 11.3V.


Note:  This is making the gross assumption that both diodes originally dropped 0.7V.  This might not be valid with the differences in current through the legs.  But for a first approximation, 0.7V works for me!

[Chai]

I guess I need to understand how diodes with different forward voltages behave when put in parallel. I'll look into it.

[Ian.M]

Take V_t as 25.85 mV. 
Plug your specified Vf and the current If it was specified at into the simplified Shockley equation

I_f=I_se^(V_f/V_t)

Solve for Is for both diodes.

Then estimate the possible current through the diodes.  The one on the right looks like it dominates so solve that first.  12V/(415R+100R)=23.3mA as a first estimate.  Plug that back into the diode equation and you get 0.7V (0.703) for Vf if it got all the current.
That would give you 2.33 + 0.7V at the juntion if there was no current through the other diode.   
Now assume the other diode is a dead short - the 500R and the 415R form a potential divider with a Thevenin equivalent voltage of 6.56V and equivalent resistance of 227R
Recalculate: 6.56V/(237R+100R)=20.1mA, and plug that back into the diode equation for the right diode to get 0.699V for Vf - a negligible change.

Now treat the 415R and 100R resistors as a potential divider between two sources, 12V and 0.7V and find the Thevenin equivalent voltage of 2.89V and equivalent resistance of 80.6R.  The max current through the left diode cant be more than 2.89V/(80.6R+500R)= 5.0mA.  Plug that into the diode equation, and you get a Vf of 2.06V.  Re-solve for If: (2.89V-2.06V)/(80.6R+500R)=1.43mA, and back into the diode equation for a revised Vf of 2.03V, again a negligable change. 

At this point, you have a good estimate for If and Vf of the left diode, but If for the right diode is still wrong, although you know its Vf must be 0.7V

Now you have Vf for both diodes at close to their actual If, you can treat them as pure voltage sources and finish solving the circuit by superposition.

Then throw the result out of the window if you need better than a ballpark estimate as you *ASS*U*ME*ed the diodes were ideal with no evidence to support that assumption, and also have made absolutely no allowance for the dissipation in each diode increasing the junction temperature and decreasing Vf.

Due to the need to repeatedly solve the diode equation it goes a lot easier if you use a spreadsheet rather than doing it the old-skool way with pencil and paper and a slipstick.

 

[Chai]

Yeeesssssss! That's what I wanted. This kind of stuff is so hard to search for on google. You're the man.

[Ian.M]

As a very very rough rule of thumb, decrease diode Vf by 10% of its specified value for each decade that the current is lower than the specified If by.   That's good enough for ball-park back of the envelope calculations for about 3 decades of If.  Its not much good for higher currents as the internal resistance term takes over.  The percentage varies with the general type of diode, but if you are getting into that sort of detail you need to throw out rules of thumb and model it properly.

[jh15]

Gads! I used to know this stuff. Using hybrids, (learned from Chinese teachers I revere) etc. To ramble on, today playing with my 6" slide rule model the Apollo astronauts took with them. Used to use it during the first fuel crisis to figure gas mileage.

I used to use and understand all the scales. Wonder if Dave does.

Hybrid models would be a good Dave thing.

[Chai]

So thinking through this assuming the diode on the left leg is a an LED with Vf of 2.2 and the diode on the right is a boring 0.8Vf diode...

1) Current will first find it's way through the right leg since it will push through the 0.8 before 2.2.

2) This sets the first voltage level at the fork of about 2V which opens up the left leg a bit.

3) Current now flows and drop across the left leg 500 resistor raising the voltage at the fork a bit more.

4) More current now flows through the right leg and 100 resistor raising the voltage at the fork more.

5) Bounce back and fourth until a balance is found

.. all happening instantaneously?