# How to convert mV to db?

1. ## How to convert mV to db?

anyone know the equation converts mV to db?

2. ## dbm to mv

You can't convert mV to dB because dBs are an adimensional magnitude, like the quotient of two powers or two tensions. Thus, L(dB)=k.log(x2÷x1) where k is 10 for the powers and 20 for voltages.

I think you can use this equation L(dBm)=10log(p(mW)/1mW), like a mW is as to power developed by an effective voltage of 0,775 V (rms) on a resistance of 600Ω. You can say that L(dBm)=10.log(v²÷R×600÷(0.775V)²)=20.log(v/0.775)+10.log(600/R)=L(dBV)+10log(600/R) and obtain the relative power of the signal.

Regards.

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3. ## mv to dbm

dB is used for ratio.
You can convert mV directly to dBm as follows:
P(dBm) = 20 log V(mV)

Regards,
Amr.

4. ## db to mv

This last post is wrong. You can not convert mV to dBm this way. How to calculate dBm explained mario82.

5. ## db mv

macro82 is right.

6. ## convert mv to db

This last post is wrong.
dBm == power compared to power of 1 mV.
Regards,
Amr.

7. ## convert mv to dbm

Wrong.
dBm == power compared to power of 1 mW.

8. ## mv db

I think mario82 is right.

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9. ## dbm mv

P=(Vrms)^2/R

P(mW) = ([Vrms(V)]^2)10^3/R = {[Vrms(mV)1000]^2}1000/R

10log[P(mW)]=20log[Vrms(mV)]+90-10log(R) = P(dBm)

P(dBm)=20log[Vrms(mV)]+90-27.77(R=600 ohm)
P(dBm)=20log[Vrms(mV)]+90-17(R=50 ohm)

P(dBm)=20log[Vrms(mV)]+62.21dB ......(R=600 ohm)
P(dBm)=20log[Vrms(mV)]+77dB ..........(R=50 ohm)

1 members found this post helpful.

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10. ## mv to dbm conversion

What's meaning of these equations, jallem?

p = 20 log V

12. ## dbm to mv conversion

dB as someone said before is a ratio.
However, when the reference is understood as in
mW(dBm) becomes an absolute quantity.
In other words dB is not a phisycal quantity, dBm
is. When refering to dBm, most of the people omit
the "m", so dBm becomes dBm. That is incorrect.
dB is not the same as dBm.
So, I thought muaythai2006 was asking for
the formula to convert from mV to dBm.
These two unit are different, one is a voltage and
the other one is power. The quantity that relates
them is the Resistance as in P=V^2/R, the formula
then depends on the impedance.
However, if the formula asked for is to convert from
mV to dBmV(or dBV) then the formula is like other said.

I meant to say dBm becomes dB

13. ## db to mv converter

Originally Posted by jallem
P=(Vrms)^2/R

P(mW) = ([Vrms(V)]^2)10^3/R = {[Vrms(mV)1000]^2}1000/R

10log[P(mW)]=20log[Vrms(mV)]+90-10log(R) = P(dBm)

P(dBm)=20log[Vrms(mV)]+90-27.77(R=600 ohm)
P(dBm)=20log[Vrms(mV)]+90-17(R=50 ohm)

P(dBm)=20log[Vrms(mV)]+62.21dB ......(R=600 ohm)
P(dBm)=20log[Vrms(mV)]+77dB ..........(R=50 ohm)
jallem's formulas are right.Of course ,ypu can calculate more easily.First, you need computate the power with the voltage,Vrms(V)]^2/R. Attention here ,the impedance 's value is 50 Ω or the charateristic impedance of your system. Then you can convert the power from Watt to mW.
Ok,the last to do is change the format of mw to dBm,with this formula,
10log(P(mw)).

14. ## db to mv conversion

Jallem's only two first formulas are right.

15. ## mv dbm

Jallem's only two first formulas are right.
Then what is the solution, know-it-all?

The only error is:
Error:
P(dBm)=20log[Vrms(mV)]+77dB ..........(R=50 ohm)
must be
P(dBm)=20log[Vrms(mV)]+73dB ..........(R=50 ohm)

16. ## convert db to mv

so simple
there should be an r
and
dbm=10lg(vmv^2/r/1mw)

17. ## db/mv

Jallem's only first formula is right but the calculation is wrong.
Power of 1mW has on 50ohms a voltage of 223.6mVrms. This power is 0dBm. His formula gives 120dBm after last try.

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18. ## db a mv

Correction:

P(dBm) = 20log[Vrms(mV)]-30-10log(R).

Example(1):

P=1mW(0 dBm) ---> V=223.6 mV
using the formula(for R=50ohm):

P(dBm)=20log(223.6)-30-10log(50)=
=46.98-30-16.99=-0.01 dBm~0dBm

Example(2):

Vrms=0.34 =340mV, R=50 ohm
P(W)=(0.34)^2/50=0.00231W=2.31mW
P(dBm)=10log(2.31mW)=3.63 dBm
using the formula,
P(dBm)= 20log(340)-30-10log(50)
= 50.63-30-16.99=3.64 dBm

19. ## dbm to mv converter

what I see is completely CHAOS .. and too many different answers ..
the issue is very primitive guys .. here is a simple definition to "dB" :

The dB is a logarithmic unit used to describe a ratio. The ratio may be :
1-POWER.
2-SOUND PRESSURE
3- VOLTAGE
4- or intensity or several other things

Again .. VOLTAGE ratio .. same as power ratio .. so what's all that about ! .. if you can describe the ration between 2 power levels measured in "mW", then you can still also use the "dB" to describe the ratio between 2 voltage levels measured in "mV" ..
Notice .. if you want to express power in "mW", then the formula will be like this :
10 Log(P2/P1)
if you express POWER in terms of the originating voltage, then the formula will be like this :
10 log(P2/P1) = 10 log([V2/V1]^2) = 20 log(V2/V1)

that's it ..

20. ## convert dbm to vrms

I often use db not use mv

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