# Why does voltage stay the same regardless of resistor connect to the source

I'm learning about clipping circuits, and I have hit a road block with my understanding of one particular problem. The circuit in question is a serial limiter. My issue is why does the output voltage equal exactly 10V when the diode is reverse biased, that is, why doesn't the resistor affect the voltage on the output (I know that when no current flows through the resistor there is no voltage drop on it, but why does it stay the same)? Also, why doesn't the DC voltage source affect the amplitude of the sine wave on the output when the diode is forward biased. Bellow are Multisim screenshots of the circuits and their instruments. The question that bothers me is illustrated with another circuit where the voltage is the same. I used it to try to understand why the resistor changes nothing, but haven't quite got it. The last screenshot is my try emphasized in hope you understand the problem at hand. Thanks in forward :)

simulate this circuit – Schematic created using CircuitLab

• Your schematic is hard to read. Please use the built-in schematic editor to make a more clear one. Apr 5 '18 at 18:10
• No problem. I'll make a new one. Apr 5 '18 at 18:12
• @billykalfus you can click on the image to open it, it has an ok resolution Apr 5 '18 at 18:13
• @PlasmaHH the problem is not so much the size, but the fact that the display of the circuit can be greatly simplified. Sometimes just by removing all the extraneous parts and rearranging the circuit I've found that I can answer my own question. Apr 5 '18 at 18:15
• @BillyKalfus I've added a simplified schematic. Apr 5 '18 at 20:00

Probably the easiest way to see this is like:

simulate this circuit – Schematic created using CircuitLab

Now, just think about that sine wave. It's oscillating between $V_\text{IN}=-12\:\text{V}$ and $V_\text{IN}=+12\:\text{V}$. When $V_\text{IN}\ge +10\:\text{V}$ (a short period) then the diode doesn't conduct because it is reverse-biased. So there is no current through it and therefore also no current in $R_1$, either. Since there is no current in $R_1$ there is no voltage drop across $R_1$. Therefore, $V_\text{OUT}=+10\:\text{V}$ for the period of time while $V_\text{IN}\ge +10\:\text{V}$.

When $V_\text{IN}\lt +10\:\text{V}$ (most of the period of time), then the diode is forward-biased. Now the best way to imagine is to realize that the diode's cathode is essentially nailed to $V_\text{IN}$, so its anode will be one diode drop above it. In short, the anode of $D_1$ will "follow" the sine wave source during this period, but will follow it biased upward by one diode drop away. The anode has to follow the cathode. It has no choice. So as a consequence, $V_\text{OUT}$ during this period will be a diode drop above $V_\text{IN}$.

That's it.

• Okay this sums up my questions. I was wondering about the voltage drop on the resistor when in reality the only drop in forward bias is the diode drop. Marked as an answer. Apr 5 '18 at 20:56

Your Scope Ch1 is AC coupled so do not be confused by peak Voltage readings centred around average DC voltage of 0.

Use DC couple or enable auto scale if possible.

THis is a webpage with a very useful javascript applet and many components and common component circuits.

This is my attempt (link) to simplify it for you. The scope window has a hidden edge to drag with mouse. Also Menu > Edit> centre circuit ( to fit)

More advanced variations of same theme with Diode or battery reversed

• Okay, I coupled the first oscilloscope channel 2 with DC, and it shows proper readings. Can you explain what happens when the diode is forward biased, when the source is in the negative half-period. Why is the sine on the output not a bit smaller, it's like there is no voltage drop on the resistor. Apr 5 '18 at 20:48
• THere is drop only when the Vbat+ Vf of diode begins to conduct. Then it has a very low current and Vf drop on diode and the voltage clips due to drop on R. Click on my photo to expand and the link to simulate. Explore tinyurl.com/y94xhpft Sorry I used the opposite diode orientation in my example.... above , IF YOU WANT to learn all the combinations.. see this and stretch DSO window tinyurl.com/ya8sthgn Apr 5 '18 at 22:50

Assume you have a DC voltage source of 12V instead of AC. Your diode prevents reverse voltage, so no current flows. No current in R1, means no drop. So you get 10V.

Now assume your source is 0V. Your diode conducts. And diode drop is Vf of your diode. In this case, ~2V. Your battery still is 10V. So resistor sees 8V voltage difference. Regardless of resistor value. All that changes is current through resistor.

Now when your AC is -12V, same thing applies. But resistors sees 10V-(-12+2) => 20V voltage difference. All that changes is current through resistor.

The answer to your question is current changes, not voltage. (V=R*I) This is because your battary will remain at 10V reagardless of resistor plugged on it (ok for smaller resistor, you would cap current capacity of your source at some point, but well for simulation it might not show up). If you want to see voltage change, you need a voltage divider, so that voltage redistributes between two resistors.

At much higher currents, a real diode would have a larger Vf, causing a small voltage difference. However, your resistor is way too large for this. If you were using a 100 ohm resistors and an accurate diode model you might seem something.

Edit

To answer your comment: resistors do not store energy. They convert energy to heat. Dissipated power is given by P=UxI, where P is power, U is voltage and I is current. Resistors follow Ohm's law: U=RxI, where R is resistance in Ohm.

If you plug a resistor directly to a battery like this:

simulate this circuit – Schematic created using CircuitLab

The voltage across the resistor will always be 9V, because the battery as a voltage difference of 9V. Since voltage is forced 9V and resistor is a "constant" of your component, the only variable is the current. If instead you do this:

simulate this circuit

The voltage at node will be U*R2/(R1+R2) = 9*100/(100+100) = 4.5V. This equation (voltage divider) holds true, because the total resistance across the circuit is R1+R2). The voltage is forced to U by the battery. Therefore there is I2=U/(R ampere passing through the resistors. The voltage at "NODE" is the voltage across R2. Since there is I2=U/(R1+R2) amperes running through R2, NODE=R2*I2 = U*R1/(R1+R2).

This is all application of Ohm's law. When you have branches (resistors in parallel), you have to apply Kirchhoff's law. That law states that the sum of what goes in is equal to the sum of what goes out. Here we are talking current. So if your battery feed I1 amperes into two branches I2 and I3, then I=I2+I3. When two branches I2 and I3 return to battery, the current I4 returning to battery is sum of I2 and I3.

Coils and capacitors are whole another beasts. Kirchhoff laws still applies, but I-V relation is different. The U=R*I rule doesn't apply anymore for those components (specific to resistor). Instead you have those rules:

$$U=RI\qquad\text{for resistor}$$ $$I=C\frac{\text{d}U}{\text{d}t}\qquad\text{for capacitor}$$ $$U=L\frac{\text{d}I}{\text{d}t}\qquad\text{for inductor}$$

As you can see, capacitor and inductors have a notion of time (because they "charge").

• Okay, I've understood the first part. The second part is a bit fuzzy but I've got it too. Now, why doesn't the resistor affect the potential, is it cause it acts as a 'dumb' wire or some other law, also, would a coil act the same if it didn't store any energy (during the first half-period)? Apr 5 '18 at 20:22
• Good explanation, although I've found what I was looking for in the answer above. Apr 5 '18 at 20:57