# Puzzled with parallel forward biased diodes with common anode and different cathode voltages

If two identical diodes are in parallel and forward biased such that their anodes are at the same potential but their cathodes are at different potential how can we find the anode potential?

As an example I found the following circuit where diodes are identical and their forward voltage drop is 0.6V:

I can simulate this and figure out what V6 is but I tried to find myself and puzzled.

First of all the diodes are identical and the diode on the left is reverse biased so it is out of the picture.

Now if only the diode with -6V cathode voltage existed V6 would be -5.4V. And if only the diode with -12V cathode voltage existed V6 would be -11.4V. My logic here is that the voltage drop must be 0.6V so one can find the anode terminal voltage by adding 0.6V to the cathode voltage. And both diodes are forward biased.

But when their anodes are tied together there must be only one voltage at that node V6. I have been thinking about it but probably missing a point. How should we approach this topology to figure out the voltage at V6?

• Hint: they're not in parallel if they're not connected at both ends. May 18, 2019 at 2:05
• Oh thats true I should have named not in parallel but naming it different doesn’t help me. May 18, 2019 at 2:07
• The diodes themselves are not connected in parallel, but rather two networks (each containing a diode and a voltage source in series) are in parallel. We can call these networks "one-way voltage sources". Feb 16, 2023 at 16:53

if only the diode with -12V cathode voltage existed V6 would be -11.4V.

And if $$\V_6\$$ is at -11.4, then what operating regime is the other diode in?

But maybe it will be how can we know?

The general approach to these problems is,

1. Guess which diodes are forward biased and which are reverse biased.
3. If it does, go back to 1. and make a different guess.

It's possible (in a more complicated circuit) that more than one guess will produce no contradiction. In that case, the circuit may have multiple stable states, and the actual state will depend on the history of the input voltages over time, or may even develop randomly (through metastability, for example).

• It would be reverse biased May 18, 2019 at 2:08
• But maybe it will be how can we know? Or it cant? May 18, 2019 at 2:09
• Is there any non-physical result if it's reverse biased at -11.4 V? What if $V_6=-5.4\ V$? Is there any non-physical result in that case? May 18, 2019 at 2:14
• -5.4V yields contradiction. Because in that case both are forward biased. So my question arises. But if somehow -11.4V it doesn’t yield weirdness. One becomes reverse biased. But is there a way to analytically reach that point? Or this is the way? May 18, 2019 at 2:16

This circuit is a bit different, but I have always liked to think of this topology as a voltage selector.

If I’m not mistaken, this circuit is used in your phones and laptops, because if you connect a charger to your device it introduces a bit of a problem. With the charger connected, how does the device decide to take power from your battery or the charger? You would definitely want to use the higher of the two voltages, but how do you go about doing this?This circuit handles this by taking whatever voltage is higher, and allowing it to pass to the rest of the circuit.

In your case, it is simply the opposite, where instead of picking the highest voltage, the diode with the lowest voltage on its cathode will conduct.

If we draw the two voltage sources, we will easily see that they are connected (through the diodes) in parallel to the common resistor. D1 is backward-biased and turns V1 off; D2 is forward-biased and turns V2 on. So the voltage across and current through the resistor are determined by V2; the output voltage V6 is -11.4 V.

simulate this circuit – Schematic created using CircuitLab

In the following lines, I will use the chance that this question gives me to reveal the philosophy of this kind of diode switching circuits.

# Why are device outputs bidirectional?

To control the output voltage, device's outputs need a network of two elements in series - pull up and pull down, connected between supply rails (voltage divider configuration). In simple output stages, only one element is a transistor; the other is the load or a resistor.

## Only-sourcing outputs

For example, the simple 1-transistor output stage made with an NPN transistor (emitter follower or open emmiter) can only source current to a pull-down (grounded) load.

simulate this circuit

It cannot drive pull-up loads because both the transistor and load would be pull-up elements (there would be no voltage divider formed).

simulate this circuit

## Only-sinking outputs

Conversely, the emitter follower made with a PNP transistor can only sink current from a pull-up (connected to Vcc) load.

simulate this circuit

It cannot drive pull-down loads because both the transistor and load would be pull-down elements (there would be no voltage divider again).

simulate this circuit

## Both sourcing and sinking outputs

Figuratively speaking, the output stages above behave as "unidirectional voltage sources". But what do we do if we need to both source and sink current in order to drive all sorts of loads? We can simply combine the two 1-transistor stages above into one complementary stage (the figure below). It can both source and sink current to pull-down and from pull-up loads behaving as a bidirectional voltage source (here the RL1-RL2 voltage divider serves as a load).

simulate this circuit

# Why do we sometimes need one-way outputs?

Complementary stages are perfect but in some cases a problem appears. It sounds a bit paradoxical but in electronic circuits it is sometimes necessary to connect several voltage outputs to each other (usually we connect input to output). Typical examples are computer buses, diode logic gates, backup power supplies, etc.

simulate this circuit

But imagine we only have devices with perfect complementary outputs. If we directly join them, a conflict (short connection) between them can appear. For example, in the figure above, the left voltage follower supplies roughly 15 V to the load (RL1-RL2 voltage divider) and the right follower only 5 V. This is equivalent to a 10 V voltage source shorted. How do we resolve such a conflict?

## From two-way back to one-way

We know the one-way outputs do not have this problem. So we start looking for a way to turn the two-way output back into a one-way output. The funny thing here is that we assembled it with two transistors and now, since we cannot take it apart into its component parts, we are looking for another way.

And here diodes help us to do that. By connecting a diode in series with a voltage source we make it a "unidirectional voltage source". Now we can join such outputs to solve different tasks.

# Applications

## Only sourcing two-way output

simulate this circuit

## Two only sourcing two-way outputs joined

simulate this circuit

## Only sinking two-way output

simulate this circuit

## Two only sinking two-way outputs joined

simulate this circuit

## Diode gates

In diode logic gates, diodes prevent conflict between the input sources (having equal voltages).

In OR gates, diodes act as switches that connect the respective input source (X1) producing logic "1" to the input of the logic gate and disconnect it when it produces logic "0". It is as if these are "double-acting" sources - they supply high voltage and close the switch in series.

simulate this circuit

So if, for example, one of the input sources produces high voltage, the others producing low voltage will be switched off and no short circuit will occur. Note: I have used ideal diodes (VF = 0 V) and real voltmeter (Rv = 10 k) as a load.

In AND gates, the diodes act as switches that connect the respective input source (X1) producing logic "0" (ground) to the input of the logic gate and disconnect it when it produces logic "1". Now the "double-acting" sources supply negative voltage and close the switch in series.

simulate this circuit

So if, for example, one of the input sources produces low voltage, the others producing high voltage will be switched off and no short circuit will occur. Note: I have used ideal diodes (VF = 0 V).

## Maximum voltage selector

In other cases, the input voltages have different values. The unidirectional sources (equipped with diodes in series) are connected in parallel. Only the source with the highest voltage will be connected via its diode to the common point; the other sources will be disconnected by their diodes.

For example, let's consider a simple power supply with a backup battery. Figuratively speaking, the two sources BAT and Vs are made "unidirectional" by means of diodes D1 and D2. Since Vs is selected slightly higher than VBAT, it dominates and imposes its voltage on the load.

simulate this circuit

## Minimum voltage selector

If we reverse the polarity of the sources and diodes, we get a "minimum voltage selector"; such is the OP's circuit.

simulate this circuit

# Generalization

This arrangement of two voltage sources (V1 and V2) connected in parallel through two devices (D1 and D2) can be seen in many other circuit solutions. Actually the two sources are connected in series (through the ground) so their voltages are summed/subtracted and the resulting voltage is applied across the network of the two devices in series. Each of the sources tries to set the voltage of the midpoint equal to its voltage. Let's see some typical applications of this arrangement.

## Diode limiters

• Parallel diode limiter

simulate this circuit

• Series diode limiter

simulate this circuit

## Inverting configuration

simulate this circuit

## The role of the resistor

In order to operate in the vertical part of their IV curve, diodes need a minimum current. It can be provided by the load resistance. But when the load resistance is very high or open circuit we have to connect an additional resistor in parallel with the "load". We choose its resistance in order to ensure the minimum necessary current through the diode.

The diode connected to -12v is forward biased since its the lower of the two voltages (-6 and -12). Hence the -12v diode anode (V6)will be forced to -11.4v, and this will make the -6v diode reverse biased by (-11.4 minus -6) 5.4volts. There is no ambiguity.

The value strongly depends on the resistor. It is anywhere between 0 V and around -11.95 V, considering all three diodes are equal.

If the resistance is low, then both of the bottom two diodes will be forward biased and the node voltage will be close to 0V.

The other answers assume a large resistance. In that case, current through R will drop enough voltage to bring the node sufficiently negative to reverse-bias the bottom diode.

Why doesn't the voltage reach -12 V ? The saturation current of the two reverse diodes must flow through the forward diode and drops about twice the thermal voltage of ~25 mV over it. You can see in the plot below, that the voltage never reaches -12 V.

## Plot of the voltage over resistance

• I'm trying to imagine how a resistor can set a voltage in a circuit of diodes and voltage sources... In my view, it can only set the current... Feb 19, 2023 at 16:40
• @Circuitfantasist Just simulate it and step the resistor from say 1e-12 Ohm to 1e12 Ohm. Feb 19, 2023 at 17:34
• That is, short circuited and open circuited? To be honest, the resistor was the last thing I would think of in this circuit... but I'll give it a shot... Feb 19, 2023 at 18:36
• I have added a text in the end of my answer about the role of the resistor. Feb 19, 2023 at 19:01