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I understand that decibels are used to express power and we can use 20log(V1/V2) and 10log(P1/P2) to find the voltage and power ratios. But what are we meant to do when it comes to non-linear components?

If I had a diode and connected it to a voltage. And then doubled the voltage and wanted to describe the voltage increase in dB.

Do I say that there was a 6dB voltage increase?

Or do I calculate the power consumed by the diode at each voltage, find the power ratio convert it to dB?

The second seems more correct.

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  • \$\begingroup\$ A forward-biased small 1N4148 diode with a low current (10mA) voltage of 0.7V will be extremely hot with a high current (500mA) voltage of 1.4V but its voltage drops when it gets hot. Why use db's for such a small voltage range? \$\endgroup\$
    – Audioguru
    Sep 26, 2023 at 12:37

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The definition of dB is 10log(P1/P2) where P1 and P2 are powers. The conversion to 20log(V1/V2) only applies when the voltages are across equal valued resistors. Often dB's are used in other situations such as when the voltages are not across the same value resistor. For example, if the voltage gain of an amplifier is 10, it is often expressed as 20 dB. This is not strictly true since the input and output voltages of the amplifier appear across different resistances. However is has become common usage in Electrical Engineering because it is convenient for doing calculations since it replaces multiplication with addition and, in many cases, avoids large numbers (it is easier to say that an amplifier has a gain of 120 dB rather than 1,000,000). In your case the power and the voltage are not related in the same way as for a resistor, so the 2 uses of dB cannot both be applied.

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Think of dB as something like a comparison tool to compare one value to another (reference). One important point here is that the things to be compared should be of same type and unit.

This can be used for virtually anything e.g. sound/audio (SPL), power (transmission), amplification, etc.

As you stated, it's \$10 \log(P/P_r)\$ where \$P_r\$ is the reference power and \$P\$ is the power that we are comparing to the reference. If these power values/measurements are for the same load (resistance) then, from \$P=V^2/R\$, dB for voltages becomes \$20 \log(V/V_r)\$ where \$V_r\$ is the reference voltage and \$V\$ is the voltage level that we are comparing to the reference.

For example, we can use dB for amplification i.e. to compare output voltage, V, to the input voltage, Vr (although they are not applied across the same resistances). You can go one step further and use it for your weight change in a year (sorry, I don't know how to represent weight in dB).

If I had a diode and connected it to a voltage. And then doubled the voltage and wanted to describe the voltage increase in dB.

So, what is your reference here? Input voltage applied to the diode's anode or output voltage that you measured at the diode's cathode? If the input voltage is what you mean then ...

Do I say that there was a 6dB voltage increase?

yes, it's going to be 6 dB. But if the output voltage is what you mean then, assuming the load resistance stays the same, it should be measured on both cases (for low power Si diodes you can assume 0.6~0.7V drop, otherwise it should be measured) then put into the dB equation.

Or do I calculate the power consumed by the diode at each voltage, find the power ratio convert it to dB?

Again, if your reference is diode's power dissipation then measure the dissipation for both cases and put into the dB equation.

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The first sentence of your question is already problematic. Decibels themselves cannot be used to express power, unless you also give what is the reference power, such as "power dissipated is 3 dB in reference to 1 milliwatt", sometimes abbreviated just as 3 dBm.

Decibels are only used to express a ratio of two things, it does not matter if they are voltages, currents, powers, sound pressure, digital numbers, or representing the amount of sugar when baking.

I could say I have 6dB more sugar than you, but without a specific reference, we only know I have double the amount what you have but we don't know how much either of us have.

So decibels apply to give ratio for any two numbers, no matter what the two numbers are.

So regarding your diode, feel free to compare two voltages or two powers in any way you wish what you think makes most sense in your scenario. It does not matter if it is a diode or resistor or whatever. However, do consider if it even makes much sense to compare diode voltages or dissipated power in decibels. Much like we can compare amount of sugar or eggs in decibels when baking, it may not be very convenient.

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