Thanks for pointing out the *contextual difference* of the decibel (voltage, amplitude, … vs power, watt, …). Mathematically speaking, the *equal sign is formally used this way though:* *Power:* A/B = 10W/1W = 10log(10W/1W)dB = 10log(10)dB = 10dB = 1B *Field:* A/B = 10V/1V = 20log(10V/1V)dB = 20log(10)dB = 20dB = 2B
I used to think prefix deci in dB stood for the log to the base '10'. But it looks it stood for metric prefix, ie represent multiplier 10 scale, not the base. But if deci is removed how do you say B; is it 'bels'
Time stamps for the different topics covered in the video: 0:56 Decibel (dB) in electronics 2:15 Representation of Power Gain in dB 6:26 Representation of Voltage Gain in dB 8:40 What is dBW, dBm, and dBV? 11:40 dBm to dBW and dBW to dBm conversion
Super helpful. Now I know what exactly decibel is. For my class, teacher just start using decibel in the graph without explain what it actually is. Like only want us to memorize.
Yes, that is true. Direct addition nahi hoga. First you need to convert each power into mW or watt. And then you can add them. So, P1 in mW will be equal to 10 mW and P2 will be equal to 3.2 mW. So, P1 + P2= 13.2 mW And if we again convert it back into dBm then it would be equal to 11.2 dBm. I hope its clear to you.
convert 5dBm into mW. In mW, it will be equal to 10^(5/10). If you calculate, it will come out around 3.16 mW. Or roughly 3.2 mW. Rest of the procedure I have already mentioned in the earlier comment. I hope it will clear your doubt.
@@MRPerfect-zr6mv Because P is proportinal to U^2 (voltage raised to 2). This "2" goes multiplying 10 (that comes from conversion of bells to decibels). 2.10 log (V(out)) (for 1V reference voltage).
Hello. I need explanation of what is dBm in RF communication. For example I have two RF Module. The Transmit Power is 14dBm.What dose this mean? After connection RSSI is -9 dBm.What dose this mean? Can you help me please?
dBm is used to represent the power in milliwatt. 14dBm means the power is around 25mW. When the power is less than 1mW then in dBm it will be negative. -9dBm means power is approximately equal to 0.125 mW or 125 uW. For the calculation, you can use the equation given in the video. Direct conversion calculators are also available on the web. You can also use them. I hope it will help you.
if the input power is -3dBm which suffers a gain of -4dB on insertion . What will be the output power? And if the outpower is equally splitted into 4 branches. What will be the output power per branch in dBm? Please help.
Because at that frequency, the output voltage reduces to 0.707 of the maximum value. If you convert it into the dB then Vo / Vin is -3dB. That's why it's called -3dB frequency. I hope it will clear your doubt.
respectable sir please make video, that why are they using 3dB in many place. and how derive that 3db, as much as possible. really i couldn't understand that. thank you
Please check my RC low pass filter. There I have explained. 3 dB frequency is where the power becomes half or the voltage is 0.707 times the input voltage.
The table is for voltage. The voltage becomes 0.707 times the actual value at -3dB. For power, it becomes half at -3dB. The equation for power is Power (in dB) = 10 log (Power in Watt/ 1W). For voltage its Voltage (in dB) = 20 log (Voltage in Volt/ 1V)
@@vinaykumarkv5830 I also think it is the right answer. 10dBm ~ 10mW 5dBm (half of prior potence) = 10dBm - 3dBm = 7dBm Total potence in dBm is 10dBm + 7dBm = 17dBm Also ask you to say if this is correct. Thank you in advance!
Power P = V^2 / R. So, when we replace P with V^2/ R then in the log, 2 will get multiplied with 10. And hence for voltage, there is a 20 in dB equation.
The thing is it's converted into dB so that the quantity can be added directly without involving any further complexity. If two quantities are in dB then they can be added directly.
@@ALLABOUTELECTRONICS If two quantities are in dB then they can be added directly. Then why can't we simply add P1 and P2 to get P i.e. why is P=10+5=15 not true? Please explain.
@@tariqislam9388 hi, 40db+50db=90db is correct, because this is a system of two amplifiers in series and amplification can be represented as Pin * A1 * A2. P=10dBm + 5dBm =15dBm not true because we just want to make addition powers (not multiplication) (this is the key) to sum up, multiplication in normal scale is equivalent to addition in dB (LOGARİTHMİC) scale.
@@ekremfidan6311 Hi. Thanks for your explanation. But I still do not understand this thing well. Apparently it seems easy but I think I have problem some where in the basics. Can you please tell what background knowledge should I have in order to understand this db, dbm etc thing. Thanks.
I have always been kind of confused about decibels. You have made it crystal clear. Thank you.
🔮
Thanks for pointing out the *contextual difference* of the decibel (voltage, amplitude, … vs power, watt, …).
Mathematically speaking, the *equal sign is formally used this way though:*
*Power:* A/B = 10W/1W = 10log(10W/1W)dB = 10log(10)dB = 10dB = 1B
*Field:* A/B = 10V/1V = 20log(10V/1V)dB = 20log(10)dB = 20dB = 2B
I used to think prefix deci in dB stood for the log to the base '10'. But it looks it stood for metric prefix, ie represent multiplier 10 scale, not the base. But if deci is removed how do you say B; is it 'bels'
@@lolvivo8783 Correct 1 B = "one bel"
Time stamps for the different topics covered in the video:
0:56 Decibel (dB) in electronics
2:15 Representation of Power Gain in dB
6:26 Representation of Voltage Gain in dB
8:40 What is dBW, dBm, and dBV?
11:40 dBm to dBW and dBW to dBm conversion
ALL ABOUT ELECTRONICS what.is watts power in music amplifier
It is the unit of power measurement. Generally speaking, the amplifier with higher wattage produces a louder sound. (more amplification)
Спс бро .Я начинающий радиолюбитель и теперь я глубже буду изучать технику .
excellent explanation of difficult looking easy concept. Thank you.
btw... P1 + P2= 10 + 3.162 = 13.162 mW
In terms of decibel it is 11.193310
In terms of decibel it is 11.193310
@@He.Who.Remains. power will remain the same, it will not depend on reference
@@He.Who.Remains.
P= -18.806 dB
P1=0.01W = 10mW
P2=0.00316W = 3.16mW
P1 = P2 = 13.16mW
how did 3.16 came ? log 5 is 0.698 right, So 10*0.698 is 6.98
@@karthikeyaprasadnapa8847 you should add that
Super helpful. Now I know what exactly decibel is. For my class, teacher just start using decibel in the graph without explain what it actually is. Like only want us to memorize.
Your video clear my concept about 20bd thanks a lot
You Made it very simple for me
too much helpful video, highly recomended
Let me know your answer here:
If P1= 10 dBm and P2=5dBm
Then P1+P2 = ??
Is it 15 dBm or something else?
Directly addition to nahi hoga.
I have tried in 2 ways but don't know it is correct or not .
Pls give solution.
Yes, that is true. Direct addition nahi hoga. First you need to convert each power into mW or watt. And then you can add them.
So, P1 in mW will be equal to 10 mW and P2 will be equal to 3.2 mW.
So, P1 + P2= 13.2 mW
And if we again convert it back into dBm then it would be equal to 11.2 dBm.
I hope its clear to you.
sir plzz
explain..i did'nt get it.
how u get 3.2..plzz rpy
convert 5dBm into mW. In mW, it will be equal to 10^(5/10). If you calculate, it will come out around 3.16 mW.
Or roughly 3.2 mW.
Rest of the procedure I have already mentioned in the earlier comment.
I hope it will clear your doubt.
oh..thanks a lot sir
Well explained. I have a question:
What happens if I add dB value of a transmitter and a dbW value? Can they be added?
If the reference is same, then they can be added. I mean if both represents the Power in dB then they can be added.
@@ALLABOUTELECTRONICS ok thanks. I appreciate your work and the time you spend to help other people
Why in voltage gain log of power gain is multiplied by 10? From where this formula is derived?
"d" in dB stands for "deci", i.e. 1/10th. Log(A/B) gives you 1bel. For the decibel you need to multiply by 10.
@@RimantasLiubertas ohk. But what about 20. Why 20 is used for voltage gain??
@@MRPerfect-zr6mv Because P is proportinal to U^2 (voltage raised to 2). This "2" goes multiplying 10 (that comes from conversion of bells to decibels).
2.10 log (V(out))
(for 1V reference voltage).
Thank you so much sir....very easy and understandable explanation.
Great delivery!
Hello.
I need explanation of what is dBm in RF communication.
For example I have two RF Module.
The Transmit Power is 14dBm.What dose this mean?
After connection RSSI is -9 dBm.What dose this mean?
Can you help me please?
dBm is used to represent the power in milliwatt.
14dBm means the power is around 25mW.
When the power is less than 1mW then in dBm it will be negative.
-9dBm means power is approximately equal to 0.125 mW or 125 uW.
For the calculation, you can use the equation given in the video. Direct conversion calculators are also available on the web. You can also use them.
I hope it will help you.
Thanks. I didnt out find how to convert dB to dBV? Can you please explain?
it is very simple way to understand the concept.
excellent
Thanks for clearing the concept.
tq u sir for this vedio now i have cleared my doubt
Thank you for this very clear explanation.
Great explanation
thank you so much for the explanation
interesting....very informative.
sir i want to ask one question
why zero watt can not be chosen as zero decibel level?
Good and nice explain
Very helpful video!
Thank you so much sir ❣️❣️😍😍😍
if the input power is -3dBm which suffers a gain of -4dB on insertion . What will be the output power? And if the outpower is equally splitted into 4 branches. What will be the output power per branch in dBm? Please help.
Very nice video.... Thank you please keep it up
You cleared my doubt why gain in synthesiser always show in negative
excellent way u explain.....
Perfectly clear, thanks!
conversion of negative dbm to milliwatt?
Why are lower cut-off frequency and upper cut-off frequency called -3dB frequencies???
Sir please explain.
Because at that frequency, the output voltage reduces to 0.707 of the maximum value. If you convert it into the dB then Vo / Vin is -3dB. That's why it's called -3dB frequency. I hope it will clear your doubt.
Thank you. This is very helpful.
Very Well explained
Can anyone explain about in this video 11.18
1000mw to dbw is 0dbw ?
Also
dBm to mw?
1000 mW is 1W.
1W in dBW is 10 log (1W/ 1W) = 0
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS
Thank you so much sir
That convertion of dBm to mw?
I'm in love with this channel!
excellent explanation
sir how to measure unknown power of microwave x band in dbm
thanks soo much, good explanation
are we get direct mw from dbm?
mw = 10 ^ ( dBm / 10 )
Talk slower please good video
thanks and awesome work
Nice explained.
Thanks alot.
Bro will you share the video of total harmonic distortion concept
excellent explanation...
Please explain BJT, transistors...
Thanks, that was helpful.
respectable sir please make video, that why are they using 3dB in many place. and how derive that 3db, as much as possible. really i couldn't understand that. thank you
Please check my RC low pass filter. There I have explained. 3 dB frequency is where the power becomes half or the voltage is 0.707 times the input voltage.
@@ALLABOUTELECTRONICS thank you sir
Thanks this is really useful
It's regarding half power it should be -3db but according u r table it showing -6db , correct me if i m wrong....
The table is for voltage. The voltage becomes 0.707 times the actual value at -3dB.
For power, it becomes half at -3dB. The equation for power is Power (in dB) = 10 log (Power in Watt/ 1W).
For voltage its Voltage (in dB) = 20 log (Voltage in Volt/ 1V)
@@ALLABOUTELECTRONICS got it thanks..
well explained dude...
Needs correction at 3:09
cool explanation sir
sir, Why we are taking -3dB in LPF and HPF?
Because at these points on the curve, the amplitude is 3dB less than the maximum value.
thank you sir
Thanks a lot sir 🙏
I don't understand why 20 for voltage and 10 for power
P prop to v square
So if v =10
Then log means p = 20 ?
Please explain
P(out) = (Vo^2/Ro).
P(in) = (Vi^2/Ri2),
dB = 10log(Po/Pi) = 10log((Vo^2/Ro)/(Vi^2/Ri)) -> (if Ro = Ri) ->
-> 10log(Vo/Vi)^2 => 20log(Vo/Vi)
My answer to question is 11.2 dBm
Thank you very much for Clearing concepts
Could you please explain about dbi
Intro music madhosh kar deta hain😂
good one.....
Nice
Good
Thank you bro
thanks a ton, buddy!!
What about current
p1,+p2= 10mw+3.162mw=13.162mw
bro, how did 3.162 came ?
@@karthikeyaprasadnapa8847
I forgot all these things.
I don't know 😁
P1=10 dBm(gain)
P2=7dbm (gain)
Total=17 dBm sir plzz tell me it's correct or not
I think its
@@vinaykumarkv5830 I also think it is the right answer.
10dBm ~ 10mW
5dBm (half of prior potence) = 10dBm - 3dBm = 7dBm
Total potence in dBm is
10dBm + 7dBm = 17dBm
Also ask you to say if this is correct. Thank you in advance!
Thank you very much!
I'm stupid I didn't understand nothing.
thank you so much!
Nice. Thank.
13.16 mw
Why 20 is used for voltage gain????
Power P = V^2 / R. So, when we replace P with V^2/ R then in the log, 2 will get multiplied with 10. And hence for voltage, there is a 20 in dB equation.
i think so is that correct
P = 15 dBm
P = 15 - 30 = -15 dBW
Anyone got the answer for the last question?
B for? Begetables
Ans : 15 dbm
13.162 is the answer
Thanks
8:46
Bai Jan urdu ya hindi mein btatay to behter hota
How to write 100dB =_______kw
dB = 10 log (P), for the power.
For 100 dB the power is 10^10 W. Or 10^7 kW.
P=11.16dB
P(dBm)= P1+P2= 11.1933 dBm
not 10+5 = 15dBm
15 dBm
Please explain why cats are so lazy.
40db+50db=90db is not correct. In db u cant add directly..
The thing is it's converted into dB so that the quantity can be added directly without involving any further complexity. If two quantities are in dB then they can be added directly.
@@ALLABOUTELECTRONICS If two quantities are in dB then they can be added directly. Then why can't we simply add P1 and P2 to get P i.e. why is P=10+5=15 not true? Please explain.
@@tariqislam9388 hi, 40db+50db=90db is correct, because this is a system of two amplifiers in series and amplification can be represented as Pin * A1 * A2.
P=10dBm + 5dBm =15dBm not true because we just want to make addition powers (not multiplication) (this is the key)
to sum up, multiplication in normal scale is equivalent to addition in dB (LOGARİTHMİC) scale.
@@ekremfidan6311 Hi. Thanks for your explanation. But I still do not understand this thing well. Apparently it seems easy but I think I have problem some where in the basics. Can you please tell what background knowledge should I have in order to understand this db, dbm etc thing. Thanks.
3.16mw
45dBm
Remember to pronounce Watt and not Volt when you write Watt.
(10+10^0.5)milliwatts
Sir i want to join ur whatsapp group
p1+p2=16.989
15 dbm
50dbm