I have a portable radio and I would like to record the output to a digital voice recorder. The digital voice recorder specification says the input level is "-70 dBv". The output of the portable radio says "245 mV (-10 dB)" and also "Output impedance 10 kilohms or less".

My problem is that the radio simply has a "Rec Out" jack and no way of adjusting the output level. Similarly, the digital recorder has no way or adjusting the mic input level. Given the specs mentioned above, is there a way I can tell if I'll be lucky enough that the recorded audio will be at a useable level, without distortion?

Thanks. Also, I suppose, if the radio has plenty of output (even too much) then I could make up an attenuator cable, if I can work out the right value of resistor.

Just for your info, "-70 dBv" means 70 dB below 1 Vrms. dB = 20 * log10(V / Vref), where here Vref = 1 volt. You can go the other way to find V given dB via V = 10^(dB / 20) Plugging in -70 we get 0.316 mV, roughly 1/1000 of the radio output of 245 mV. So you'd need a 1000:1 attenuator *if* you actually needed to match these levels. But as Phil says, the AGC on the mic input may handle that.

Note that 20 dB represents a voltage ratio of 10,

40 dB is a ratio of 100, 60 dB is a ratio of 1000, and so on. So from your initial specs you could just notice that there was a difference of 60 dB between -70 and -10, so no need for any real math here.

If you do want to make an attenuator, you need a series resistor (R1) from the radio to the recorder input, and a shunt resistor (R2) from the recorder input to the common (shield) connection.

The attenuation (K) is equal to R2 / (R1 + R2), but for

1000:1 or even 100:1, you can approximate this as R2 / R1. Then you can choose an arbitrary R2 (say, 100 ohms) and for K = 1/1000 solve for R1, which will be 1000 * R2 = 100k.

For 100:1 you can use R1= 100*100 = 10k.

Hope this helps!

Bob Masta DAQARTA v6.02 Data AcQuisition And Real-Time Analysis

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