- 10th September 2005, 21:03 #1

- Join Date
- Sep 2005
- Posts
- 11
- Helped
- 0 / 0
- Points
- 1,515
- Level
- 8

## How to convert mV to db?

anyone know the equation converts mV to db?

- 10th September 2005, 21:03

- 10th September 2005, 23:46 #2

- Join Date
- Aug 2005
- Posts
- 15
- Helped
- 6 / 6
- Points
- 1,592
- Level
- 9

## 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.

- 11th September 2005, 07:53 #3

- Join Date
- Aug 2004
- Location
- Egypt
- Posts
- 1,185
- Helped
- 145 / 145
- Points
- 9,782
- Level
- 23

## 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.

- 11th September 2005, 09:20 #4

- Join Date
- Dec 1999
- Location
- on third flor
- Posts
- 1,426
- Helped
- 189 / 189
- Points
- 16,909
- Level
- 31

## db to mv

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

- 11th September 2005, 09:23 #5
## db mv

macro82 is right.

- 11th September 2005, 09:24 #6

- Join Date
- Aug 2004
- Location
- Egypt
- Posts
- 1,185
- Helped
- 145 / 145
- Points
- 9,782
- Level
- 23

## convert mv to db

This last post is wrong.

Regards,

Amr.

- 11th September 2005, 09:52 #7

- Join Date
- Dec 1999
- Location
- on third flor
- Posts
- 1,426
- Helped
- 189 / 189
- Points
- 16,909
- Level
- 31

## convert mv to dbm

Wrong.

dBm == power compared to power of 1 mW.

- 12th September 2005, 11:06 #8

- Join Date
- Sep 2004
- Posts
- 53
- Helped
- 1 / 1
- Points
- 1,703
- Level
- 9

## mv db

I think mario82 is right.

- 12th September 2005, 11:06

- 12th September 2005, 16:05 #9

- Join Date
- Jul 2005
- Posts
- 452
- Helped
- 78 / 78
- Points
- 4,907
- Level
- 16

## 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.

- 12th September 2005, 16:05

- 14th September 2005, 09:36 #10
## mv to dbm conversion

What's meaning of these equations, jallem?

- 14th September 2005, 13:35 #11

- Join Date
- Sep 2005
- Location
- Romania
- Posts
- 38
- Helped
- 4 / 4
- Points
- 1,551
- Level
- 9

## convert dbm to mv

p = 20 log V

- 14th September 2005, 16:26 #12

- Join Date
- Jul 2005
- Posts
- 452
- Helped
- 78 / 78
- Points
- 4,907
- Level
- 16

## 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.

Added after 27 minutes:

I meant to say dBm becomes dB

- 16th September 2005, 07:51 #13

- Join Date
- Sep 2005
- Posts
- 5
- Helped
- 0 / 0
- Points
- 1,316
- Level
- 8

## db to mv converter

Originally Posted by**jallem**

Ok,the last to do is change the format of mw to dBm,with this formula,

10log(P(mw)).

- 16th September 2005, 08:18 #14

- Join Date
- Dec 1999
- Location
- on third flor
- Posts
- 1,426
- Helped
- 189 / 189
- Points
- 16,909
- Level
- 31

## db to mv conversion

Jallem's only two first formulas are right.

- 16th September 2005, 18:30 #15

- Join Date
- Jul 2005
- Posts
- 452
- Helped
- 78 / 78
- Points
- 4,907
- Level
- 16

## mv dbm

Jallem's only two first formulas are right.

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)

- 16th September 2005, 19:07 #16

- Join Date
- Oct 2004
- Posts
- 143
- Helped
- 14 / 14
- Points
- 2,543
- Level
- 11

## convert db to mv

so simple

there should be an r

and

dbm=10lg(vmv^2/r/1mw)

- 17th September 2005, 05:02 #17

- Join Date
- Dec 1999
- Location
- on third flor
- Posts
- 1,426
- Helped
- 189 / 189
- Points
- 16,909
- Level
- 31

## 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.

- 17th September 2005, 13:50 #18

- Join Date
- Jul 2005
- Posts
- 452
- Helped
- 78 / 78
- Points
- 4,907
- Level
- 16

## 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

- 2nd December 2005, 10:54 #19

- Join Date
- Jan 2003
- Location
- Dubai
- Posts
- 1,230
- Helped
- 53 / 53
- Points
- 10,345
- Level
- 24

## 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 ..

- 2nd December 2005, 11:31 #20

- Join Date
- Sep 2005
- Posts
- 129
- Helped
- 4 / 4
- Points
- 1,731
- Level
- 9

## convert dbm to vrms

I often use db not use mv