# Calculate Voltage Across 4k Resistor in Simple Diode Circuit

• snowJT
In summary, the conversation was about calculating voltage from point X to ground in a circuit with a 4 kilo-ohm resistor and a diode. The conclusion was that the voltage at point X is 5.3V, which is still more than 2V even if the diode is reversed. This is because the 8V battery is charging the 2V battery through the diode when it is forward-biased. The basic function of a diode is to allow current to flow in one direction and block it in the opposite direction. When a diode is reverse biased, no current flows through it.

#### snowJT

I have an interesting question about this circuit... if I were to calculate the voltage from point X to ground... essentially all it is, is voltage across the 4 kilo-ohm resistor because if it went the route of the diode, it would lose an additional 2.7 volts.. so its going to take the better path.. right? Thats how it works doesn't it?

so... we have... $$\frac{4k}{2k+4k} * 8 Volts = 5.7 V$$

Now... let's say we reversed the diode... wouldn't then.. we push 1.7V towards point X right? Now... does that 1.7 volts add up with the 5.7V, does the 5.7 volts push against the 1.7V. The voltage I know is still 5.7 volts even if the diode was reversed... but is that because the voltage won't go that way due to there being more across the 4 kilo-ohm resistor?

Picture of circuit:

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I got 5.3333333333 Volts using your method, and if the diode was forward biased, why wouldn't point X be 2.7 Volts? I think this will cause some current to oppose the 2V source

For the original circuit the Diode/Battery is the functional equivilent of a zener diode.

For the case where the diode is reversed.
What is the basic function of a diode?
What happens when a diode is reverse biased?

oops I ment to write 5.3 not 5.7 for everything

When a diode is reversed current doesn't go through it.

But what I'm wondering is why you don't add the resultant voltage with the 5.3V

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Yes, the voltage at X is 5.3V. Subtract 0.7V from 5.3V and it's still more than 2V. So the 8V battery is charging the 2V battery through the diode, which if forward-biased.

1)

## What is the formula for calculating voltage across a 4k resistor in a simple diode circuit?

The voltage across a 4k resistor in a simple diode circuit can be calculated using Ohm's Law, which states that voltage (V) is equal to current (I) multiplied by resistance (R). So the formula would be V = I x R, where V is the voltage across the resistor and R is the resistance of the resistor (4k).

2)

## Can the voltage across a 4k resistor vary in a simple diode circuit?

Yes, the voltage across a 4k resistor can vary in a simple diode circuit. This is because the voltage is dependent on the current flowing through the resistor, and the current can be affected by various factors such as the input voltage and the characteristics of the diode.

3)

## How do you determine the current flowing through a 4k resistor in a simple diode circuit?

The current flowing through a 4k resistor in a simple diode circuit can be determined using Ohm's Law. The formula is I = V/R, where I is the current, V is the voltage across the resistor, and R is the resistance of the resistor (4k). Alternatively, you can measure the current using a multimeter.

4)

## What is the ideal voltage across a 4k resistor in a simple diode circuit?

The ideal voltage across a 4k resistor in a simple diode circuit is dependent on the input voltage and the characteristics of the diode. Generally, the voltage across the resistor should be lower than the input voltage to ensure that the diode is operating in the forward bias region, where it conducts electricity.

5)

## Can the voltage across a 4k resistor affect the performance of a simple diode circuit?

Yes, the voltage across a 4k resistor can affect the performance of a simple diode circuit. If the voltage is too high, it can cause the diode to enter the reverse bias region, where it does not conduct electricity. This can result in the circuit not functioning as intended. Therefore, it is important to choose an appropriate voltage and resistor value to ensure proper performance.