Diode characteristics ... how to calculate compared to resistors?

[Steben]

I was wondering ... can we use a diode in a calculation model as a "non linear" resistor? It kind of is in fact, but have to sum or parallel diodes to resistors?
It is all about output curves of clippers.
If we look at I/V graphs of diodes we could translate that into resistance at a given point, just as a FET.
At U 1.5V LED1 has 5 ohms ....At U 1.4V LED1 has 10 ohms .... etc....
If diodes are in series, one can add the needed voltage to get to a current (horizontal sum of the graph). This actually means a sum of resistances.
But what if resistors are added? This a complex model I guess....

----[>[--------[>[-----
              I               I
              ---^^^^---

Look at the next spreadsheet:



It is a rough approach I know.
In this the resistance is added and paralleled. But.... this is flawed as the second diode does not follow the same treshold....

[Steben]

[Rob Strand]

For the parallel resistor case you can do a rough estimate by assuming the there is a region where the diodes in parallel with R2 are off.   That would be when the voltage across R2 is less than a diode drop.    The you can treat the output as the diode voltage + R2 voltage.    Since the current down the series components, the voltage across  R1 and R2 are related   I = VR2 / R2 = VR1 / R1.   So that lets you work out the the range of input voltages where the lower diodes aren't active,

We know,

VR2  <    Vd

So,
Vin = VR1 + Vd + VR2
       = Vd  + (1 + R1/R2) VR2
       <  Vd  + (1 + R1/R2)Vd
or,
Vin  < (2  + R1 / R2) Vd

If R2 is a lot larger than R1, it ends up at as two diode drops.  As you would expect as the large R2 doesn't do much.
If R2 is a lot smaller than R1, the input needs to be very high before R2 does anything.  As expected since a small R2 shorts out the second of diodes.   In this case the clip point is only one diode.

If you want to calculate the behaviour for this type of circuit it's best to start with an assumed voltage across R2 and work backwards.     
- Assume voltage V2 across R2
- Calculate ID2, from diode equation with VD2=V2, and IR2 = V2/R2. 
- Since ID1 = ID2 + IR2 we can calculate VD1 from the diode equation.
- Then Vout = VD1 + V2 and Vin = Vout + R1 * ID1

So for different V2 voltages we can get a relationship between Vin and Vout.

[ElectricDruid]

If we look at I/V graphs of diodes we could translate that into resistance at a given point, just as a FET.

Absolutely could. This was the basic operating principle of the Diode Filter, based on the Moog Transistor Ladder filter.

https://electronicmusic.fandom.com/wiki/Diode_ladder

[Steben]

For the parallel resistor case you can do a rough estimate by assuming the there is a region where the diodes in parallel with R2 are off.   That would be when the voltage across R2 is less than a diode drop.    The you can treat the output as the diode voltage + R2 voltage.    Since the current down the series components, the voltage across  R1 and R2 are related   I = VR2 / R2 = VR1 / R1.   So that lets you work out the the range of input voltages where the lower diodes aren't active,

We know,

VR2  <    Vd

So,
Vin = VR1 + Vd + VR2
       = Vd  + (1 + R1/R2) VR2
       <  Vd  + (1 + R1/R2)Vd
or,
Vin  < (2  + R1 / R2) Vd

If R2 is a lot larger than R1, it ends up at as two diode drops.  As you would expect as the large R2 doesn't do much.
If R2 is a lot smaller than R1, the input needs to be very high before R2 does anything.  As expected since a small R2 shorts out the second of diodes.   In this case the clip point is only one diode.

If you want to calculate the behaviour for this type of circuit it's best to start with an assumed voltage across R2 and work backwards.     
- Assume voltage V2 across R2
- Calculate ID2, from diode equation with VD2=V2, and IR2 = V2/R2. 
- Since ID1 = ID2 + IR2 we can calculate VD1 from the diode equation.
- Then Vout = VD1 + V2 and Vin = Vout + R1 * ID1

So for different V2 voltages we can get a relationship between Vin and Vout.

Nice post and adds light on a very sound approach.
But there is one problem... It treats Vd1 and Vd2 as fixed or in other words Zd1 and Zd2 as 0 or infinite.
In that case we can draw straight lines according to the R1/R2 ratios.

The diode response however is logarithmic... every clipper (assuming R2 = 0, no Vd2 in play) will have a different output amplitude based on the R1.
One pair is fairly easy to assume, given the graph one can follow. I think it means a lot of iterative calculations.... Of course we "need" a load as well. But the combination with parallel components is plain rough. With Vin low, the diode is an almost infinite high resistor. But when Vin gets higher, the diode starts conducting.
Question is: is a diode a true voltage driven component? Even then, the slip of voltage will have influence on the total current flowing. It becomes a voltage divider. Which means the voltage rise will be compensated once the diode starts conducting. etc etc. Hence the iterative process.
On top of that the D2 subcircuit will be just as iterative. Once it starts conducting the second voltage divider opens up and will influence the first voltage divider.

[Rob Strand]

Nice post and adds light on a very sound approach.
But there is one problem... It treats Vd1 and Vd2 as fixed or in other words Zd1 and Zd2 as 0 or infinite.
In that case we can draw straight lines according to the R1/R2 ratios.

The first part is only ball-parking zones so that is assuming the diodes have constant Vd's.

The diode response however is logarithmic... every clipper (assuming R2 = 0, no Vd2 in play) will have a different output amplitude based on the R1.
One pair is fairly easy to assume, given the graph one can follow. I think it means a lot of iterative calculations.... Of course we "need" a load as well. But the combination with parallel components is plain rough. With Vin low, the diode is an almost infinite high resistor. But when Vin gets higher, the diode starts conducting.

The second part of the post where I say,

Calculate ID2, from diode equation with VD2=V2

and

we can calculate VD1 from the diode equation.

is where the exponential diode equation comes in.

The reason behind starting with an assumed V2 is all the equations unravel in a way where ID2 and VD1 can be calculated *directly* from the diode equation.   You don't need to do any iteration or solve any equations and the solution should be more accurate.   Each step is a simple calculation and the answer is built up.

The only thing you have to do is choose representative steps in V2.    If you know the range of Vin you want to cover you can use Goal-Seek in the spreadsheet to find the upper and lower values of V2.   After that just fill in a hundred or so V2 points between those limits.

With very small R2 values you will end-up very small V2's corresponding to a large range of Vin.

I've plugged a lot for diode and transistor circuits through spreadsheets over the years so I'm always looking for ways to avoid iteration in spreadsheets.   For other circuits it's hard to avoid and I've got few techniques to get good starts values and good convergence.

FWIW, you can remove the effect of a resistive load RL by replacing Vin R1 and RL with the Thevenin equivalent  Vin' = Vin * RL/(R1 + RL) and R1' = R1 // RL.    Once you solve for Vin' you can calculate the real Vin using Vin = Vin' (R1 + RL) / RL. 

Question is: is a diode a true voltage driven component?

Both Vd and Id are linked by the diode equation so it's not really possible to say whether voltage or current is the primary variable.  When you do the physics on the diode you generally have an assumed current and a resulting voltage.

Even then, the slip of voltage will have influence on the total current flowing. It becomes a voltage divider. Which means the voltage rise will be compensated once the diode starts conducting. etc etc. Hence the iterative process.
On top of that the D2 subcircuit will be just as iterative. Once it starts conducting the second voltage divider opens up and will influence the first voltage divider.

I do get what you are trying to do.     The idea of finding the break points is fine and it does give an approximation to the characteristic.    The only loose end is exactly what the diode voltage is.   We assume silicon diodes are 0.6V but we know that's not constant.   If it's 0.6V at 1mA it might be 0.5V to 0.54V at 100uA.

Your idea of treating the diode as a small resistor also works but it still only produces an approximation.

The voltage + resistor model only works over a narrow range of diode voltage.    The root of the problem is the diode voltage is not constant with current.    If we assume a diode of 0.6V at 1mA the slope of the diode curve *around 0.6V* is the (dynamic) diode resistance.    You can calculate the slope from,

         rd  = (n*26mV / Id)              ;  n = 1 to 2 for diodes

Notice the slope depends on Id so treating the diode as a resistor is only an approximation.

Assume n = 1, just for the argument

         rd  = 26mV / Id

For 1mA we get,   rd = 26 ohms.

The resistance appears in series with diode voltage drop the 0.6V.   The 1mA current produces a 26mV drop across the slope resistor,  so the diode looks like 0.6 - 26mV = 0.574V in series with 26 ohms.

        Vd(Id)   = 0.574  + 26 * Id

If we use the exponential diode model we end up with a different result,

        Vd(Id)   = 0.6V  +  26mV * ln(Id / 1mA)    ;  ln() = natural log

          Linear   Exponential   
Id [uA]    Vd [V]      Vd [V]
10          0.574       0.480          <--- as you get further from 0.6V error gets larger
100        0.577       0.540
200        0.579       0.558
500        0.587       0.582
700        0.592       0.591
1000      0.600       0.600            <--- both pass through 0.6V 1mA
2000      0.626       0.618
5000      0.704       0.642
10000   0.834       0.660         <--- as you get further from 0.6V error gets larger


As Id get small the voltage drop + resistance model can't get below 0.574V but the exponential model can.   
As far as iterative solutions go it is actually possible to use the exponential equation in the spreadsheet.

[Steben]

I actually found based on "a" red LED graph a factor of +/- 1.9 with each step of 100mV which comes very close.

V   R / LED (ohms)
0,1   5992550
0,2   3153974
0,3   1659986
0,4   873677
0,5   459830
0,6   242016
0,7   127377
0,8   67040
0,9   35284
1,0   18571
1,1   9774
1,2   5144
1,3   2708
1,4   1425
1,5   750
1,6   400
1,7   213
1,8   113
1,9   59
2,0   31
2,1   16
2,2   8
2,3   4
2,4   2
2,5   1

This actually means the R1 can be a "gain" control or at least character. Low R1 gives lower ratio = higher output.

[Rob Strand]

This actually means the R1 can be a "gain" control or at least character. Low R1 gives lower ratio = higher output.

In your method you have a table of *large signal* effective resistances for the diode   Ra = Vd1 / Id1.    You can get VLED vs ILED from measurement or by calculating using a diode model.

If you have a second diode Rb = Vd2/Id2.

For diodes in series you find the (large signal) resistance   Rtotal = Ra + Rb.

In order for that to work the resistance Ra and Rb need to be at the same current on both diode characteristics.   The reason for that is Rtotal must be Vtotal / Id = (Vd1 + Vd2) / Id  and Id = Id1 = Id2.     If the diode currents are not equal, for example when you add a parallel resistance, the sum does work.

I'm not 100% sure how you are computing the last column in your table.       In your tables  my understanding is the first column is Vin and the columns "A/LED" and "DIODE' are the effective (large signal) resistance of the diodes.   To get the diode resistances you are driving Vin into R1 and a single diode clipper, then measuring Vd across the single diode, then calculating the diode resistance from,
Rdiode = Vd / Id   where Id  = (Vin - Vd) / R1.
In other words  Vd / Vin  = Rdiode / (Rdiode + R1) like a divider.