Would this diode clamp work in real application?

Question

I am reading the book The Art of Electronics and came across the diode clamping section. It has this circuit using a diode to clamp the voltage to specific power supply voltage (my example is 5.6 V.)

How would it work in real life? What happens when Vin gets above 5.6 V? Where does the excess voltage and current now flowing through the diode go? Straight to V1 power supply? How would this affect it? I have seen examples using a voltage divider, so excess voltage goes through it to ground, but this example, although working on LTspice, raises some questions.

Answer 1

That's a very common protection scheme on the inputs of logic devices example here, and the reason the input voltage range is specced to between (ground - a few hundred mV) and (Vdd + a few hundred mV); any more and you'd cause the input clamp diodes to start conducting.

Depending on what the application is and what the expected transients are, you might leave them as-is for ESD protection, add a resistor to the line connected to the input to prevent burning the diodes out (they can't handle very much current), or supplement them with additional external components.

Answer 2

Basic idea

The OP's question is fundamental and deserves a thorough answer. So I will first consider the circuit idea and then answer the questions.

Circuit name

I would call the OP's circuit a parallel diode positive voltage limiter because a diode element is connected in parallel to the load when a certain positive input voltage is reached; thus the voltage across the load is fixed to this value. Here is a simple intuitive explanation of the trick on which it is based.

Circuit phenomenon

Imagine that you have two voltage sources (e.g., batteries) with different voltages and internal resistances. One of them is a "weak" voltage source with a voltage of 9 V and a high internal resistance of 1 kΩ. However, when we connect a voltmeter to it, it shows exactly 9 V because no current flows in the circuit and there is no voltage drop across the internal resistance.

^{simulate this circuit – Schematic created using CircuitLab}

The other is a "strong" voltage source with a voltage of 6 V and zero internal resistance. Here there is no doubt that if we connect a voltmeter to it, it will show 6 V.

^{simulate this circuit}

Now, if we connect the two sources in parallel, the voltage of the second source is established across them, and current flows from the first source to the second source.

^{simulate this circuit}

In this way, something like an "automatic switching" of the voltage sources takes place.

Useful application

It allows us to switch two voltage sources by using a simple 2-terminal SPST switch instead of the more complex 3-terminal SPDT switch. This is especially useful in electronic circuits because semiconductor devices (diodes and transistors) are simple 2-terminal switches.

SPDT switch: Both voltage sources can be perfect.

^{simulate this circuit}

SPST switch: BAT1 is imperfect, BAT2 is perfect.

^{simulate this circuit}

The clever idea

This is the OP's idea with the diode limiter where a diode switches two voltage sources - "weak" and "strong" - to a common load (voltmeter).

Implementation

Above, I have used batteries because their internal resistance can be set in the CircuitLab parameters field.

Conceptual circuit

If we only have an "ideal" voltage source Vin, we can artificially "worsen" it (increase its internal resistance) by connecting a resistor R in series.

Vin < Vref, SW is OFF -> Vout = Vin

^{simulate this circuit}

Vin > Vref, SW is ON -> Vout = Vref

^{simulate this circuit}

Practical circuit

The conceptual switch above is electrically controlled by the voltage applied across it. In the OP's circuit of a diode limiter, it is implemented by a diode. I have used an "ideal" diode (with VF = 0 V) from the CircuitLab library in order not to deviate into details that are not essential for revealing the main idea.

Vin < Vref, D is OFF -> Vout = Vin

^{simulate this circuit}

Vin > Vref, D is ON-> Vout = Vref

^{simulate this circuit}

What is inside the "circle"?

So, we have finally arrived at the real OP's questions, which actually concern the internal structure of a voltage source:

How would it work in real life? What happens when Vin gets above 5.6 V ? Where does the excess voltage and current now flowing through diode goes? Straight to V1 power supply? How would this affect it?

I will answer them by considering several possible circuit solutions. For the purpose of this explanation, I have removed the diode.

Natural voltage sources

The first thing that can be represented by the voltage source symbol (especially in humble electric circuits) is a unregulated natural source, e.g. battery, rechargeable battery, charged capacitor, etc. They are bilateral sources that can both source and sink current. The first two examples below are like that.

Artificial voltage sources

In more sophisticated electronic circuits, there is a need for adjustable sources with a precisely defined voltage. For this purpose, the voltage of an unregulated voltage source is reduced and its value is kept constant. This is made by the help of various electronic devices that usually are unilateral, and this can be a problem in the OP's circuit.

Voltage divider

I have seen examples using voltage divider, so excess voltage goes through it to ground...

Generally speaking, any voltage output device is some kind of voltage divider that is made up of two elements in series stretched between the supply rails. In the simplest case, they are resistors (R2 and R3 below).

So, since the resistor R3 is grounded, the input voltage source Vin can pass its current through R3 to ground. Since Vout is a "reduced Vcc voltage", the current "prefers" to flow through it, and not through Vcc.

^{simulate this circuit}

Zener diode stabilizer

Similarly, in a Zener diode regulator, the input voltage source Vin can pass its current through the Zener diode (not through Vcc) to ground.

^{simulate this circuit}

As a conclusion, in the voltage sources above (implemented as shunt regulators), there is a connection between the output and ground; so the input voltage source Vin can pass its current through them.

"Push" voltage follower

However, if the voltage source is implemented as a series regulator, there is no such connection. Let's consider, as an example, an NPN emitter follower that can only source current.

"Pushing" current: When Vin < Vref (or Vin = 0 V), the current exits the emitter and passes through R1 to ground.

^{simulate this circuit}

No current: But when Vin > Vref, the transistor base-emitter junction is backward biased. The transistor is off and it impedes the Vin's current.

^{simulate this circuit}

"Pull" voltage follower

If the emitter follower is implemented by a PNP transistor, it can only sink current.

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"Push-pull" voltage follower

Pushing current: The NPN part sources current.

^{simulate this circuit}

Pulling current: The PNP part sinks current.

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Answer 3

The maximum current through the diode is (15V - 5.6V)/15kΩ = 630uA.

Most regulators are not designed to sink much current (many will sink a little but often enough they want the opposite- a minimum load), so if you add a resistor from 5V to ground of, say, 4.7kΩ, which conducts a bit over 1mA the regulator won't have to sink current until the input exceeds 21V. If there is no chance that will happen, then you're done.

If you are trying to clamp unknown input voltages that could be very high or low then it's better to have a Zener diode or some other kind of clamp that conducts current directly to ground.

Here's an example that clamps at +5.6/-0.6. The diode is necessary so that the Veb breakdown is not violated. The 5V supply only has to sink the base current, most of the current is conducted through the transistor to ground.

^{simulate this circuit – Schematic created using CircuitLab}