G
George Gabriel
Guest
Hello to everyone,
it's my first time posting on this newsgroup, I'm not a mothertongue and
my english is very much rusty, so I hope to don't do much mistakes...
I have a little problem in understanding the meaning of some
specifications about murata piezotransducers, on their catalogs.
I tryed to figure out something but I'm not sure if the reasoning is ok.
I write several times to murata about this problem, but sadly I never
received any answer. I hope somebody could be so kind to confirm or
reject my interpretation.
The catalogue I refer to is the P19E7 accessible at the following URL:
http://www.murata.com/catalog/p19/p19_01e.pdf
the device(s) are the ultrasonics sensors (Piezotite), for example the
high frequency type shown on page 24.
The parameter I'm not able to figure out completely is the "Overall
Sensitivity" in the 5th column.
For the MA80A1, for example, they state that the overall sensitivity is
-47 dB, referred to 0dB=18Vp-p, @ 50cm.
I interpret these as:
1. we have a couple of transducers of the same kind, one used as
transmitter, the other used as receiver.
2. they are placed at a relative distance of 50cm.
3. the 0dB = 18Vp-p = V_rx_ref is the reference value for the
voltage at the receiver side. We apply a voltage value V_in at the
transmitter side.
4. the voltage we read at the receiver end, assuming V_in = V_tx_ref
could be calculated by (we assume V_in = V_tx_ref, but I was not
able to find it on the cited page on the catalog)
-47dB = 20*log_10(V_out/V_ref)
where V_out is the voltage I could expect at the receiver side and
V_ref is 18V, so
V_out = V_ref * 10^(-47/20) = 18*10^(-47/20) =
= 0.08043 V (p-p) = 80.43mV (p-p)
5. if the voltage applied to the transmitter side is different from
the reference value, we have to expect a reduction or an increase
in the overall sensitivity:
O.S.' = O.S. + 20 log(V_in/V_tx_ref)
6. if the distance from S and E is different from the stated 50cm,
we have to expect another alteration of the O.S.:
O.S.' = O.S. + 20 log(50cm/d)
where d = effective distance from tx-rx (in cm).
My biggest perplexity is: at which V_in is the value referred to? I read
carefully the page but I can't figure out... perhaps the max Input
voltage? I don't think so.
Perhaps the V_ref isn't referred to the receiver side, but to the
transmitter one? So how can we calculate the V_out?
Perhaps all the reasoning is wrong...
is there anybody experienced with these devices who can help me?
Thanks in advance, and sorry for the length of the message!
Pierpaolo
it's my first time posting on this newsgroup, I'm not a mothertongue and
my english is very much rusty, so I hope to don't do much mistakes...
I have a little problem in understanding the meaning of some
specifications about murata piezotransducers, on their catalogs.
I tryed to figure out something but I'm not sure if the reasoning is ok.
I write several times to murata about this problem, but sadly I never
received any answer. I hope somebody could be so kind to confirm or
reject my interpretation.
The catalogue I refer to is the P19E7 accessible at the following URL:
http://www.murata.com/catalog/p19/p19_01e.pdf
the device(s) are the ultrasonics sensors (Piezotite), for example the
high frequency type shown on page 24.
The parameter I'm not able to figure out completely is the "Overall
Sensitivity" in the 5th column.
For the MA80A1, for example, they state that the overall sensitivity is
-47 dB, referred to 0dB=18Vp-p, @ 50cm.
I interpret these as:
1. we have a couple of transducers of the same kind, one used as
transmitter, the other used as receiver.
2. they are placed at a relative distance of 50cm.
3. the 0dB = 18Vp-p = V_rx_ref is the reference value for the
voltage at the receiver side. We apply a voltage value V_in at the
transmitter side.
4. the voltage we read at the receiver end, assuming V_in = V_tx_ref
could be calculated by (we assume V_in = V_tx_ref, but I was not
able to find it on the cited page on the catalog)
-47dB = 20*log_10(V_out/V_ref)
where V_out is the voltage I could expect at the receiver side and
V_ref is 18V, so
V_out = V_ref * 10^(-47/20) = 18*10^(-47/20) =
= 0.08043 V (p-p) = 80.43mV (p-p)
5. if the voltage applied to the transmitter side is different from
the reference value, we have to expect a reduction or an increase
in the overall sensitivity:
O.S.' = O.S. + 20 log(V_in/V_tx_ref)
6. if the distance from S and E is different from the stated 50cm,
we have to expect another alteration of the O.S.:
O.S.' = O.S. + 20 log(50cm/d)
where d = effective distance from tx-rx (in cm).
My biggest perplexity is: at which V_in is the value referred to? I read
carefully the page but I can't figure out... perhaps the max Input
voltage? I don't think so.
Perhaps the V_ref isn't referred to the receiver side, but to the
transmitter one? So how can we calculate the V_out?
Perhaps all the reasoning is wrong...
is there anybody experienced with these devices who can help me?
Thanks in advance, and sorry for the length of the message!
Pierpaolo