Where did the - 0.7 V from the graph came from? (Diode Clampers Circuit)

[Jayce]

I am currently studying how Diode Clampers works. (Sorry for the newbie question.)

Regarding the picture on the attachment or in this link : https://drive.google.com/file/d/1KqsmmzlFRwVzQ8dE6zRtJlrQ_vh2kQD3/view?usp=sharing. On the result waveform after applying clampling / shifting techniques, I am wondering where did the -0.7 V on the graph came from?

On the negative alternation / half-cycle, the Vp(out) is equal to 0 V as there is no current flowing through the load resistor.
On the positive alternation, the Vp(out) is equal to 2 Vp(in) - 0.7 V through applying KVL : V(in) + V(in) - 0.7 - Vp(out).

As the result, the AC waveform shifted upwards. The Vp(in) - 0.7 V on the graph signifies the new / assumed / considered origin to show that it shifted upwards. I don't know why the -0.7 V is labeled on the peak of the negative alternation, wherein the V(out) is 0 V, hence why it is on the 0 / origin line.

Again, sorry for the newbie question but I will appreciate any help. Thank you.

 

[duke37]

The -0.7V is the voltage necessary to turn the diode on when it is driven negatively.

 

[Cannonball]

Welcome to EP.

On the negative half cycle the diode becomes forward biased with ground being the most + point in the - part of the input signal. The voltage drop across the diode will be .7 volts.

I hope this helps.

 

[Jayce]

The -0.7V is the voltage necessary to turn the diode on when it is driven negatively.

Welcome to EP.

On the negative half cycle the diode becomes forward biased with ground being the most + point in the - part of the input signal. The voltage drop across the diode will be .7 volts.

I hope this helps.

I think I know now why I didn't get that. I got confused, as on the videos I watched regarding the topic, when the diode is in forward bias, they set the diode as a short component, so current won't flow through the load resistor but that's when we assume the circuit to be ideal (Voltage drop is ignored). So for this case, when we acknowledge the voltage drop, there is a current flowing through the load resistor. So the voltage across the load resistor is equal to the voltage across the diode due to the 2 components (Diode and load resistor) connected in parallel with each other (where voltage is constant) and the diode in that alternation is in forward bias.

Tell me if I'm correct.

 

[duke37]

In the second diagram, consider removing the diode. The output will then be a sine wave centred on 0V. Adding a diode limits the negative voltage to about 0.7V and the sine wave is therefore shifted positively. It is still driven with a sine wave but the capacitor allows the average voltage to shift.

 

[(*steve*)]

The simple answer is that when forward biased, the anode if a diode is always a little bit (0.7V for a silicon PN junction) more positive than the cathode.

In the diagram you showed, the reference point is the anode (it is ground) and so, with respect to that, the measured point (the cathode) will be at -0.7V.

This is a little confusing for newcomers because in their first exposure to diodes they often see some AC source being rectified and the voltage at the cathode being shown as positive. When you then see a circuit like this (and it's often as you're learning theory) the first thought is "WTF, this is backwards!?!".

However, even in a simple rectifier, the voltage at the anode must be a little higher than that at the anode for the device to conduct current in the forward direction.

So even though "positive comes from the cathode" (in some sense), you need something more positive at the anode

 

[Ratch]

I am currently studying how Diode Clampers works. (Sorry for the newbie question.)

Regarding the picture on the attachment or in this link : https://drive.google.com/file/d/1KqsmmzlFRwVzQ8dE6zRtJlrQ_vh2kQD3/view?usp=sharing. On the result waveform after applying clampling / shifting techniques, I am wondering where did the -0.7 V on the graph came from?

On the negative alternation / half-cycle, the Vp(out) is equal to 0 V as there is no current flowing through the load resistor.
On the positive alternation, the Vp(out) is equal to 2 Vp(in) - 0.7 V through applying KVL : V(in) + V(in) - 0.7 - Vp(out).

As the result, the AC waveform shifted upwards. The Vp(in) - 0.7 V on the graph signifies the new / assumed / considered origin to show that it shifted upwards. I don't know why the -0.7 V is labeled on the peak of the negative alternation, wherein the V(out) is 0 V, hence why it is on the 0 / origin line.

Again, sorry for the newbie question but I will appreciate any help. Thank you.

A junction always conducts a minute amount of current in the forward direction when a very small voltage is applied. The current becomes significant at around 0.7 volts. It also conducts current in the reverse direction, but that current is limited to the small saturation current plus the surface currents of the PN slabs.


Ratch

 

[Jayce]

So basically, at the negative half cycle, the diode is in forward biased and will conduct current. I get that part. So since it is in forward bias, there will be a current through the load resistor. I mixed the analysis on ideal and practical model, hence the confusion.

V_(R-L) + 0.7 V = 0 ; So, V_o = - 0.7 hence the -0.7 V on the graph.

Thanks guys, i appreciate the answers.

 

[AnalogKid]

So basically, at the negative half cycle, the diode is in forward biased and will conduct current. I get that part. So since it is in forward bias, there will be a current through the load resistor.

Not quite. There will be current through the load resistor (and the capacitor) whether the diode is forward biased or not, only the mount of current through the resistor changes between the forward and reverse biased conditions.

 

[Jayce]

Not quite. There will be current through the load resistor (and the capacitor) whether the diode is forward biased or not, only the mount of current through the resistor changes between the forward and reverse biased conditions.

You are referring to reverse current which occurs in the diode when it is in reversed bias condition? Thank you.

Again, thank you guys.

 

[AnalogKid]

You are referring to reverse current which occurs in the diode when it is in reversed bias condition?

No, I'm referring to forward current when the diode is forward biased. When the diode is reverse-biased, the current through it is essentially zero, and there is a current path through the capacitor and the load resistor. When the diode is forward biased, the voltage across the resistor is clamped at the Vf rating of the diode. The current through the resistor is less because of the lower voltage across it, and the current through the capacitor is more because of the larger voltage across it.

ak