Diodes in Series question on simulation

Thread Starter

pll_check

Joined May 28, 2017
23
Hi,

I have two diodes back to back in series ( V -> N1 - P1 -> N2 - P2 -> GND).

If source have zero internal resistance, the voltage drop across Diode 1 and 2 is equal, V/2.
As soon as I add some internal resistance, the voltage drop across diodes is ~ 0.7 V.

Is this simulation artifact or some physics behind it?
 

Thread Starter

pll_check

Joined May 28, 2017
23
yes. its LT Spice.

Copying .asc file content.

Version 4
SHEET 1 880 680
WIRE 496 32 320 32
WIRE 496 144 496 96
WIRE 496 160 496 144
WIRE 304 176 240 176
WIRE 320 176 320 32
WIRE 320 176 304 176
WIRE 240 224 240 176
WIRE 496 288 496 224
FLAG 240 304 0
FLAG 304 176 Vin
FLAG 496 288 0
FLAG 496 144 VD2
SYMBOL voltage 240 208 R0
SYMATTR InstName V1
SYMATTR Value 3
SYMATTR SpiceLine Rser=0
SYMBOL diode 480 160 R0
SYMATTR InstName D1
SYMBOL diode 480 32 R0
SYMATTR InstName D2
TEXT 288 360 Left 2 !.dc V1 0 3 0.1
 

AlbertHall

Joined Jun 4, 2014
8,381
If the diodes are forward biased then the results are expected. With no source resistance the current would be huge and in real life the diodes would blow up but you can't blow up a simulated diode.
 

Thread Starter

pll_check

Joined May 28, 2017
23
If the diodes are forward biased then the results are expected. With no source resistance the current would be huge and in real life the diodes would blow up but you can't blow up a simulated diode.
That is what I thought, but I used a 30 mOhm source resistance for the behavior to change.
With 0 ohm V Source Resistance, there is V/2 across diodes.
With 30 mOhm V Source Resistance, there is 0.7 V across diodes.

Why such small resistance is important to trigger simulation results change?

but is it a good simulation practice to put some source resistance for a supply in general?
 

Thread Starter

pll_check

Joined May 28, 2017
23
Thanks Albert.

I edited my previous post and you replied before it was effective. Could you please respond to the new post?
 

WBahn

Joined Mar 31, 2012
24,700
Hi,

I have two diodes back to back in series ( V -> N1 - P1 -> N2 - P2 -> GND).

If source have zero internal resistance, the voltage drop across Diode 1 and 2 is equal, V/2.
As soon as I add some internal resistance, the voltage drop across diodes is ~ 0.7 V.

Is this simulation artifact or some physics behind it?
If your source has zero resistance, then there is no mechanism for the source to drop any voltage as it supplies and an arbitrarily large amount of current. Thus the voltage drop HAS to occur across the two diodes -- there's simply no other place for it to be dropped. Depending on the diode model used, you might get huge currents in the diodes. It looks like your source is 3 V (it would be nice if you posted a screenshot of your schematic), so 1.5 V across each diode. Assuming the diode model has no effective series resistance, then that is about 800 mV more than "normal", which would equate to 13 orders of magnitude more current than would result in a diode forward drop of 0.7 V since, at room temperature, the current increases by an order of magnitude (factor of ten) for each ~60 mV increase in junction voltage drop.

What does your simulation say that the current is?

Because of the exponential rise in current with voltage drop, it doesn't take much series resistance in the source to bring the diode drop back close to that 0.7 V mark.

Let's say that the diode mode is such that there is 600 mV at a current of 10 mA (I.m just shooting in the wind on that). What would the resistance need to be to result in 780 mV across each diode? Since that's an increase of 180 mV, the current has increased by three orders of magnitude, or 10 A. That would mean that the voltage across the internal resistance would be 3 V - 2(780 mV) or 1.44 V, which would be 144 mΩ.
 

WBahn

Joined Mar 31, 2012
24,700
That is what I thought, but I used a 30 mOhm source resistance for the behavior to change.
With 0 ohm V Source Resistance, there is V/2 across diodes.
With 30 mOhm V Source Resistance, there is 0.7 V across diodes.

Why such small resistance is important to trigger simulation results change?

but is it a good simulation practice to put some source resistance for a supply in general?
The more realistic your simulation models are, the more realistic your simulation results will be.

But there are the inevitable tradeoffs and you will need to learn what level of effort is reasonable to put into your simulation model and where that effort should be spent.

If you plan on shorting a 3 V source with two diodes like this, then you need much better models for both the source and the diodes. But in most (read "most", not "all") circuit simulations, you can use an ideal source and not compromise the results in any significant way. If you are drawing relatively small amounts of current from a source (relative to what it is indented to supply) and the circuit is not overly sensitive to the consequent voltage drop, then it is probably not worth the effort. But if you have a reasonable idea of what a realistic source resistance value is, then there isn't much downside to using it.
 
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