# Resistance of the resistor and of the diode

1. Sep 29, 2016

### moenste

1. The problem statement, all variables and given/known data
A semiconductor diode and a resistor of constant resistance are connected in some way inside a box having two external terminals, as shown in the diagram below. When potential difference of 1.0 V is applied across the terminals the ammeter reads 25 mA. If the same potential difference is applied in the reverse direction the ammeter reads 50 mA.

What is the most likely arrangement of the diode and the resistor? Explain your deduction. Calculate the resistance of the resistor and the forward resistance of the diode.

2. The attempt at a solution
I would say the arrangement should be as follows:

Since the diode consumes energy, therefore the current is smaller on its side. And current is higher on the side of the resistor.

But I am not sure about it. I am also not sure whether "when a PD of 1 V is applied across the terminals" it means that both the diode and the resistor have a PD of 1 V?

2. Sep 29, 2016

### Staff: Mentor

If 50 mA is flowing into one side of the diode and only 25 mA is flowing out, what happens to the "extra" 25 mA? Is charge accumulating on the diode? What effect would that have?

You should review the characteristics of series and parallel connected components in terms of potential difference and current. What must be the same for all parallel connected components? What must be the same for all series connected components?

3. Sep 29, 2016

### moenste

Voltage is the same for parallel components and current is the same for series. That means that diode and the resistor are in parallel and the diode is parallely connected and when 50 mA is shown it is when current does not go into the diode and 25 mA is when current goes into the diode (when it is forward-biased).

The extra 25 mA are accumulated by the resistor. I think charge is not accumulated on the diode.

Last edited: Sep 29, 2016
4. Sep 29, 2016

### Staff: Mentor

No. If they are connected in parallel then you'd expect the net resistance to be lower when the diode is conducting. So more current when the diode is conducting, less current when its cut off.

Do this: Since according to the problem description we are looking at a non-ideal diode with some forward resistance, model the real diode as an ideal diode in series with a resistor. Call it $R_d$. The ideal diode has zero forward resistance and no voltage drop when forward biased, and is an open circuit (infinite resistance) when reverse biased. Then draw the two cases for the applied voltage polarity.

5. Sep 29, 2016

### moenste

But isn't it what I did draw? A diode in series with the resistor?

6. Sep 29, 2016

### Staff: Mentor

Does that choice work for both polarities of applied voltage? Can any current at all flow when the diode is reverse biased?

7. Sep 29, 2016

### moenste

Hm, that's true -- current won't flow when the diode is reverse biased.

So I should change the current diode into an ideal diode? How does it look like? From here it looks like the variable voltage DC supply.

So it should be like this.

8. Sep 29, 2016

### Staff: Mentor

See the second paragraph of post #4. For an ideal diode current flows only in the forward direction.

9. Sep 29, 2016

### moenste

In an ideal diode current flows only in the forward direction. In a non-ideal diode current flows in both directions.

Ideal diode in a series with a resistor:

Open circuit means it is shut down? Turned off? So it doesn't affect the circuit?

10. Sep 29, 2016

### Staff: Mentor

No. The difference between a real diode (non-ideal) and an ideal diode is that the real diode has some resistance when it conducts current and also exhibits a threshold voltage (forward bias voltage) that must be surpassed before it conducts. The forward bias voltage is close to 0.7 V for a silicon diode. Neither the ideal nor the real diode will conduct in the reverse direction.

There are various levels of detail possible for modeling a real diode with ideal components. The "quick and dirty" way is to simply consider the diode to be an ideal diode (so it behaves as a simple one-way valve for current). The next level is to add in the forward voltage drop by placing a 0.7 V voltage source in series with an ideal diode. The next level up again is to add in the series resistance associated with the diode. You then have a series connection of three ideal components to represent a "real" diode.

Three levels of model for a simple diode

Here we are only concerned with modelling the diode's basic properties of conducting current in only one direction and having some forward resistance. That's just an ideal diode in series with a resistor. This combination replaces the diode symbol in your proposed circuit:

A simple diode model that exhibits resistance when conducting

So that "subcircuit" represents your real diode. You still need to connect this with another resistor in some fashion to create the circuit inside the box.

Right. Open means no current can flow. You have to be more specific about what you mean by "doesn't affect the circuit". Certainly cutting off current flow when it would otherwise flow can be considered an effect

11. Sep 29, 2016

### moenste

So we will have a resistor RD, a diode and a second resistor?

12. Sep 29, 2016

### Staff: Mentor

Yes. Rd is part of the diode model. The 'real' diode is being replaced by this model.

13. Sep 30, 2016

### moenste

So the graph should look like this?

14. Sep 30, 2016

### Staff: Mentor

Have you checked to see if this circuit can fulfill the observed behavior? For example, can it conduct current in both directions?

15. Sep 30, 2016

### moenste

I think the ideal diode with it's resistor RD should be placed parallel to the resistor.

If current goes anti-clockwise, the ammeter reads 25 mA, since both resistors consume energy and the diode allows the current flow. If the current goes clockwise, the ammeter reads 50 mA, since only resistor consumes energy and the diode does not allow the current flow.

16. Sep 30, 2016

### Staff: Mentor

Yes, that looks like a viable circuit. Well done.

17. Sep 30, 2016

### moenste

We can calculate the total resistance of the circuit. V = I R so R = V / I = 1 / 25 * 10-3 = 40 Ω. So in the circuit where the diode is forward-biased the total resistance of the parallel circuit is 40 Ω. The total resistance when the diode is reverse-biased is 1 / 50 * 10-3 = 20 Ω. That means that the resistor has a resistance of 20 Ω and the diode has a resistance of 20 Ω (half of 40 Ω).

18. Sep 30, 2016

### Staff: Mentor

I think you've muddled the logic a bit. When two resistors are put in parallel the net resistance of the combination is LESS than either of the two resistors. So when the diode is forward biased, placing its resistance in parallel with the external resistor, that is when the net resistance needs to be halved (to double the current). The lower current occurs when just the external resistor carries current.

19. Sep 30, 2016

### moenste

You mean that the 40 Ω that we got when the current is 25 mA and the current is flowing through two parallel resistors (i. e. the diode is forward-biased), this is a halved resistance from that of what they would have if they would be on their own? So 40 Ω is per resistor?

Why then we have R = V / I = 1 / 50 * 10-3 = 20 Ω when the resistor or on its own?

20. Sep 30, 2016

### Staff: Mentor

The smaller current occurs when the resistor is by itself. The current doubles when the diode conducts, providing another current path.

Current is greater when the net resistance is lower. Resistance is lowered when resistors are placed in parallel. When the diode conducts it places its own resistance in parallel with the external resistor, lowering the net resistance and raising the current. Two paths → more current. One path → less current.