# Voltage across tape recorders

Here there's a question that I'd like to double check with you lot.

According to an article on impedance, the voltage across each of the tape recorders in the image is divided by two. Is that so?

• It could be a good idea to summarize the relevant information from the article in your question, rather than forcing people to read the link to understand what you're asking. Dec 2, 2019 at 3:21
• When impedances are matched, yes two loads cuts the voltage in half. But often drivers are much lower impedance. Dec 2, 2019 at 3:48

The author of that article seems to be confused. The caption says:-

Figure 2. In a matched-impedance system working to the 600 ohm standard, connecting two tape machine inputs to the same console output would cause a level drop of 6dB, because each of the two parallel 600 ohm loads only receives half the signal power.

6dB drop is incorrect, but so is 3dB (half power). It is actually about 3.5dB, ie. slightly less than half power.

We can prove this by examining the equivalent circuits:-

simulate this circuit – Schematic created using CircuitLab

R1 and R3 represent the output impedance of the source, and the other resistors are the loads.

In the first case the signal voltage splits equally between source and load resistors, 1/2 across each causing a 6dB drop. However the source's internal voltage has been set 6dB higher to produce a reading of 0dBVU (at 0.5V in this example) on the load's VU meter.

In the second case with two loads in parallel the total load impedance is 300Ω. Voltage still splits proportionally to the resistances, now 2/3 across the source and 1/3 across the loads (= 2/3 of 1/2). As the total resistance seen by the source is now 900Ω instead of 1200Ω, the current it has to deliver increases by 1/3. This current splits equally between the two loads, so each gets 2/3 of the current of the single termination.

Power = voltage x current, so with 2/3 the voltage and 2/3 the current the power is 4/9 times as much. Decibels are 10xlog10(power ratio), so 4/9 = ~-3.5dB.

The result is a reduction in the signal level at each tape recorder input, as the same source signal current now has to be shared between the two destinations, therefore developing half the voltage across each input impedance.

The author is assuming that the same current flows, but in reality putting two paths for the current to flow means that the total circuit impedance is decreased, and thus more total current flows. If the source beyond that 600 ohm impedance is an opamp capable of supplying the extra current, then you should get 2/3 of the original voltage. If it isn't, you'll get less and possibly a lot of distortion.