How do I convert db into volts?

[reddish75]

I've searched but there doesn't seem to be an easy for dummies explanation of how to convert db into volts, how do you do it? I need it to adjust an old bbc amp back into spec ( please see attachment)

Can anyone help?

[radiolistener]

dB represents logarithmic ratio, relative to some value. It looks that for your case it talks about dBm - dB relative to 1 mW.

Assuming that dBm is measured as RMS power, if you want to get RMS voltage you can calculate it as follows:

Urms = sqrt((10^(dBm/10) / 1000) * Z0),

where Z0 is impedance, in your example Z0=8 Ω

For Z0=8 Ω:
-26 dBm = 4.483 mVrms
-25.6 dBm = 4.694 mVrms
0 dBm = 89.443 mVrms
+13.3 dBm = 413.566 mVrms

Also note that this is voltage for a loaded output.

For peak sine voltage you can replace Z0 in equation with 2*Z0, but for other waveform it depends on exact waveform.

[shabaz]

dB represents logarithmic ratio, relative to some value. It looks that for your case it talks about dBm - dB relative to 1 mW.

Assuming that dBm is measured as RMS power, if you want to get RMS voltage you can calculate it as follows:

Urms = sqrt((10^(dBm/10) / 1000) * Z0),

where Z0 is impedance, in your example Z0=8 Ω

For Z0=8 Ω:
-26 dBm = 4.483 mVrms
-25.6 dBm = 4.694 mVrms
0 dBm = 89.443 mVrms
+13.3 dBm = 413.566 mVrms

Also note that this is voltage for a loaded output.

For peak sine voltage you can replace Z0 in equation with 2*Z0, but for other waveform it depends on exact waveform.

The above is incorrect.

To the OP:

Context is key, and the above calculations didn't consider that. You can see that by just visual inspection of those calculated results, such differences in value clearly make no sense for an audio amp.

There's a typo in those instructions in your attachment, and the author cannot have meant -26 dBm when referring to an audio amplifier.

It's most likely a reference to dBu, since there is most likely a 600 ohm audio input.
-26 dBu would be 39 mV RMS, which is the same as 55 mV peak or 110 mV peak-to-peak sinewave.

In other words, the instructions are advising you to apply a 600 Hz sine wave of 39 mV RMS at the input (measure that with a 'scope or an AC voltmeter (some DMMs can do it) that works at audio frequencies, because it's likely that the signal generator equipment may display a different output level, so it needs to be measured).

At the output it really doesn't matter what the actual output is, because it's dependent on the volume setting, i.e. it is up to the user to set it. All that matters, is that relative values across different frequencies are correct.

So, the procedure would be to set the volume to some nominal value (you can connect a speaker, or a dummy speaker load, or even leave the output disconnected, although for best accuracy you will want to connect a speaker or dummy speaker load), and then measure the output, and call it 0 dB.

Next, you'd change your input frequency (but keep it at the 39 mV RMS level!) and then see if the output varies to the amount in those instructions.

Here's an example.

Let's say with a 600 Hz sine wave input at 39 mV RMS, you measure an output of A volts, where A could be (say) 0.5V or 0.6V or 1V or whatever (it can be RMS or peak or peak-to-peak, no difference), it doesn't matter. Let's say A is 0.123 V RMS. That's going to be your 0 dB reference.

Now, if the instructions advise the output at 6 kHz to be D dB where D is +9.1 dB on the HF output, then you'll need to look for a voltage of A * 10^(D/20) which is 0.123 * 10^(+9.1/20) which is 0.123 * 10^0.455 which is 0.123 * 2.85 and that's 0.35 V RMS.

So, if you'd measured 0.123 V at the LF output at 600 Hz, then you'd want to see 0.35 V output at the HF output at 6 kHz.

I may have made a mistake in the above info, apologies if I did, but I think it's mostly correct, and is broadly along the lines of what's actually intended by the instructions that you showed.

[radiolistener]

The above is incorrect.

There's a typo in those instructions in your attachment, and the author cannot have meant -26 dBm when referring to an audio amplifier.

It's most likely a reference to dBu, since there is most likely a 600 ohm audio input.

Are you sure that he talks about dBu? There is no mention about 600 Ω, he talks about 600 Hz...

And I see that he clearly mentioned dBm:

Look for 0 dBm on the LF side

And since he mentioned that the load is 8 Ω and mentioned about dBm units, I think that he is talking about dBm

[shabaz]

And I see that he clearly mentioned dBm:

See my earlier comment that that's likely a typo by the original author.

And since he mentioned that the load is 8 Ω and mentioned about dBm units, I think that he is talking about dBm

That logic makes no sense (and you likely know it since you arbitrarily used an 8-ohm impedance on the _input_ side when you referred to -26 dBm, and you're not a beginner, so you don't have that excuse).

I wasn't going to go there, but since you're digging your heels in, I'll say you're frequently incorrect. And you don't care enough to try to use common sense because, as mentioned, you _know_ you're using a silly input impedance to come up with the values that you did.

[radiolistener]

That logic makes no sense (and you likely know it since you arbitrarily used an 8-ohm impedance on the _input_ side when you referred to -26 dBm, and you're not a beginner, so you don't have that excuse).

I agree that the input impedance is unlikely to be 8 Ω, as there is no specific information provided about it. However, the text explicitly refers to output power values in dBm, mentioning an 8 Ω load and a 0 dBm reference.

While the units for the input signal remain unclear, it seems unlikely that dBu would be used to measure output power on an 8 Ω load, given its typical association with 600 Ω systems and potential for confusion with other impedance.

However, my response concerns the output signal level in the table. As for the input level, I cannot confidently say in which units it is specified. It is entirely possible that it is in dBu, but I would not rush to conclusions and would review the other pages of the manual for clarification.

I wasn't going to go there, but since you're digging your heels in, I'll say you're frequently incorrect. And you don't care enough to try to use common sense because, as mentioned, you _know_ you're using a silly input impedance to come up with the values that you did.

I acknowledge that I may be incorrect at times. But in this case, I believe your assertion regarding the use of dBu is mistaken. The context explicitly involves output power measurements on an 8 Ω load, which is incompatible with the typical application of dBu, as dBu is primarily associated with 600 Ω impedance. Hence, it seems unlikely that dBu is the unit being referenced here.

Moreover, the text explicitly specifies 0 dBm as the reference level, clearly indicating the use of dBm units. This is why I concluded that the instruction pertains to dBm. Additionally, I find it unlikely to be a typographical error, as using dBu for an 8 Ω impedance would be unconventional and could lead to confusion.

Correct me if I'm wrong.

[daisizhou]

I have a firmware here that corresponds to a load of 50 ohms,Internally, it contains corresponding DB values converted into voltage values, and also calculates power, but it needs to be reversed to display the source code

If you need it, I can send it to you

[shabaz]

Correct me if I'm wrong.

I did...
You're a timewaster, I'm not going to repeat the reasoning.

[radiolistener]

I have a firmware here that corresponds to a load of 50 ohms,Internally, it contains corresponding DB values

What exactly dB units it references? dBm? dBV? dBW? dBu? ...

As I said before, dB is just a ratio, but you need to specify reference level, otherwise it don't have sense.
And you need to carefully take into account impedance.
For example, the same dBm power value will give you different voltage for 50 Ω load and for 8 Ω load.

[Andy Chee]

I agree that the input impedance is unlikely to be 8 Ω, as there is no specific information provided about it. However, the text explicitly refers to output power values in dBm, mentioning an 8 Ω load and a 0 dBm reference.

Failing to define input impedance will result in an answer in the rather inconvenient units of dBm/ohm, rather than dBm.

I agree, make an assumption about input impedance (e.g. 600 ohms) and use dBu.

[pienari]

How about this page?.
https://www.random-science-tools.com/electronics/dBm-Watts-volts.htm

[Bud]

Given the so poorly written instructions, their author may himself not understand dB units.

[radiolistener]

I did...
You're a timewaster, I'm not going to repeat the reasoning.

It’s unfortunate and unproductive to resort to rudeness or make accusations like calling someone a "timewaster" - especially in a technical discussion where mutual respect is key. Let’s keep the discussion professional and focused on technical accuracy.

Regarding the OP question, the context clearly indicates that the output is specified in dBm. The units of the input signal are less critical here, as the instruction specifies adjusting the input to achieve 0 dBm reference level on output at 600 Hz. Subsequent measurements are then conducted relative to this established reference 0 dBm.

From my opinion, what is the key here is that all values in the table are related to the output, not for input. So it is highly unlikely that there was a typo and these values ​​were given in dBu for 8 Ω load. It more looks like dBm, as it mentioned in instruction.

[Andy Chee]

From my opinion, what is the key here is that all values in the table are related to the output, not for input.

Given that dB is a ratio, then you CANNOT ignore input impedance.

Impedance units divided by impedance units should mathematically cancel out, leaving millivolt units. 

[radiolistener]

Given that dB is a ratio, then you CANNOT ignore input impedance.

In this case, the input impedance and units doesn't matters. The key is to maintain a consistent input level across all frequencies, which is straightforward with a lab signal generator.

The instruction emphasizes adjusting the input signal so that the output reads 0 dBm at 600 Hz. After this calibration, you vary the input frequency while maintaining the same input level, then measure the output in dBm and compare it with the provided table.

So, the OP can use the formula that I provided above to translate dBm values from table to Volts RMS:

Urms = sqrt((10^(dBm/10) / 1000) * Z0),

where Z0 is load impedance, in this context Z0 = 8 Ω


For example, according to the table, at 100 Hz it should be +11.2 dBm. Lets calculate:

Urms = sqrt( (10^(11.2/10) / 1000) * 8 ) = 0.325 Vrms

For 0 dBm:

Urms = sqrt( (10^(0/10) / 1000) * 8 ) = 0.089 Vrms

As you can see, there is no need to know what is input impedance, exact value and units, as it depends on the adjusting procedure.

From my point of view, my answer is pretty clear and easy to understand. Isn't it?

[iMo]

Afaik the audio people often mess with "0dB" as 1mW into 600ohm what is 0.775Vrms. They use "0dBxyz" rather freely (often do not care on impedance too), but when talking those preamps/equalizers and such things usually it is that meaning (0.775Vrms)..

[reddish75]

So here is another part of the manual which describes in more detail and does reference dbm

I think the equations will take a few re reads to understand and get right for me

[Tation]

Surely the doc attached by the OP is referring to dBu = dBm, which is defined over 600 Ohm, thus 0 dBu/dBm/dB_ is 0.775 VRMS. Old, needle, audio voltmeters use such 600 Ohm reference. Thinking that the reference is 8 Ohm may seem theoretically correct, but surely wrong.

To the OP:
In any case, the adjustment procedure only ensures that, at 6 kHz, the signal output at the HF output (over 8 Ohm) is 9.1 dB above the signal ouput at the LF output (over 8 Ohm) at 600 Hz, the absolute level of the signals is meaningless, only the 9.1 dB ratio matters, so use any output level of your liking and adjust the potentiometer to fit such 9.1 dB ratio. High enough level to be easy to measure, low enough to neither saturate the amplifiers nor toast the loads.

ADDED: or, if you prefer, the adjustment procedure target is to ensure that the gain of the HF amplifier at 6 kHz is 9.1 dB higher than the gain of the LF one at 600 Hz.

[radiolistener]

Afaik the audio people often mess with "0dB" as 1mW into 600ohm what is 0.775Vrms. They use "0dBxyz" rather freely (often do not care on impedance too), but when talking those preamps/equalizers and such things usually it is that meaning (0.775Vrms)..

This is why distinct units such as dBu and dBm exist:
- dBu is a unit of voltage and does not depend on impedance when converted to voltage. It is commonly used in audio applications.
- dBm, on the other hand, is a unit of power and always considers impedance during conversion to voltage.

The values of dBu and dBm are numerically equivalent only when the impedance is 600 Ω, otherwise, they differ.


The document references dBm, not dBu, and talks about output with an 8 Ω load.
If it is really intended to refer to dBu, then this is mistake.

[coppice]

Afaik the audio people often mess with "0dB" as 1mW into 600ohm what is 0.775Vrms. They use "0dBxyz" rather freely (often do not care on impedance too), but when talking those preamps/equalizers and such things usually it is that meaning (0.775Vrms)..

This is why distinct units such as dBu and dBm exist:
- dBu is a unit of voltage and does not depend on impedance when converted to voltage. It is commonly used in audio applications.
- dBm, on the other hand, is a unit of power and always considers impedance during conversion to voltage.

The values of dBu and dBm are numerically equivalent only when the impedance is 600 Ω, otherwise, they differ.


The document references dBm, not dBu, and talks about output with an 8 Ω load.
If it is really intended to refer to dBu, then this is mistake.

A lot of audio people are really sloppy about the difference between voltage ratios and power ratios. They will compare a voltage at a high impedance point early in an amp, and a voltage at the low impedance output of the amp, and use the ratio of those voltage to come up with a bogus dB gain figure that completely ignores the actual power ratio between those 2 points. Furthermore they usually get really rude when you point out the error.

[Andy Chee]

The instruction emphasizes adjusting the input signal so that the output reads 0 dBm at 600 Hz.

The instruction says:

Bung -26dB at 600Hz into the input.

This instruction is very different to yours.

[radiolistener]

The instruction says:

Bung -26dB at 600Hz into the input.

This instruction is very different to yours.

It seems you haven’t read the instruction in full. The purpose of applying a -26 dB signal at 600 Hz to the input is not the goal of the instruction, it's just prepare before doing the main task. The goal of applying this signal so that you can adjust the input signal level and volume control knob to achieve 0 dBm at the output. This is the main goal of this step.

The main goal of the instruction, however, is further down: it aims to verify the output level across different frequencies.

Input level doesn't matters here and can be different. It just mentioned as an example and needs to be adjusted in order to get 0 dBm on the output.

Do you actually understand the purpose of the instructions, or are you simply following the steps mindlessly without considering the reasoning behind them?

It should be obvious, isn't it? 

[Andy Chee]

Do you actually understand the purpose of the instructions, or are you simply following the steps mindlessly without considering the reasoning behind them?

Depends if one's reasoning has determined that the instructions are incorrect/incomplete/inaccurate.

[radiolistener]

Depends if one's reasoning has determined that the instructions are incorrect/incomplete/inaccurate.

I see no reason to believe there is an error in the instructions. The units are specified as dBm for the output signal table, and I don’t see any issues with that. It was @shabaz who claimed that dBm was mistakenly indicated and that it should actually refer to dBu.

If the author of the instructions did indeed mean dBu instead of dBm for the output signal on an 8 Ω load, then that would be a real mistake. However, even in that case, I don’t see any problem with conducting the measurements in dBm units according to the instruction - the instruction will still work as expected. This is due to property of logarithmic dB scale. The problem may happens if there is required absolute value, but this is not the case.

[magic]

I see no reason to believe there is an error in the instructions. The units are specified as dBm for the output signal table, and I don’t see any issues with that.

It's your problem that you don't see any issue with units of "dBm into 8Ω load"

No one uses that.

Besides, it doesn't matter. If the amplifier is remotely linear, the HF/LF ratio will be the same regardless of volume setting. Just stay away from clipping and from the noise floor.