Electronic – How to reduce voltage from 40VDC to 5VDC

Am I correct to assume that the overall circuit will only draw 1A from the power source, but 25% of that power will be dissipated as heat by the voltage regulator?

A linear voltage regulator works in the way described in the question. Switching regulators work differently, described further down.

The regulator draws current from the power supply, 1 Ampere in the example plus some marginal operating overhead of the regulator itself, and power dissipated by each load is calculated by P = I² x R, or P = V x I, or P = V² / R, whichever is more convenient to calculate. In this case, as current through elements in series is equal through each element and the combination, and the voltages are known:

• P_load = 6 * 1 = 6 Watts
• P_reg = (8 - 6) * 1 = 2 Watts

The first problem I am seeing with this setup is that my power source is under the rated input voltage

A type of linear regulator known as a Low Drop Out (LDO) regulator is designed to work with lower voltage headroom. One of those should be used instead of the 7806, although in practice, most 7806 regulators will actually function even with 2 Volt headroom - Regulation quality may suffer, i.e. it may "drop out of regulation".

The other option I found is to use something from the RC world called a UBEC.

Standard UBECs are actually switching regulators or buck regulators. The way these work is, a high frequency oscillator "switches" the supply voltage on and off, this oscillating voltage is transformed up or down using either inductors, magnetic or piezoelectric transformers, or possibly in some other way, and then rectified and smoothed out to deliver the desired output voltage, in a method akin to using PWM to regulate "effective" current or voltage.

A switching regulator thus does not waste power proportionately to the voltage difference between input and output. Instead, this technology delivers anywhere from 80 to 93% efficiency, since the voltage that needs to be "reduced" is not being dropped across a resistive load at all.

In other words, a nearly constant 7 to 20% (design-dependent) of the final output power is the overhead that appears as heat and inaudible vibration at the switching regulator.

Yes, an UBEC can be used for the purpose described, or a "DC-DC buck regulator module" can be sourced from sites like eBay, often for much less than an UBEC, and possibly with better performance.

Heat generated at the regulator will be less than for a linear regulator, at the mentioned operating load.

_{(Question subsequently fixed) First off, to get 6 Volts from the alkaline AA cells, they would need to be wired in series, not in parallel. Paralleling batteries will increase current delivery capacity but keep the voltage the same as a single cell.}

While pre-built buck regulator modules are available at the same source (eBay.com) as the boost regulator mentioned in the question, and at a similar price, the following are the trade-offs:

• These buck regulator modules have a measurably higher efficiency in practice than the boost regulator linked in the question - The boost module gets quite warm (I use that specific module for charging my cellphone from a single AA cell while traveling)
• The boost regulator is significantly smaller in size
• The boost regulator works until the batteries are far more depleted, than any boost regulator strategy would allow.

Hence, going by the last two points, the boost regulator module is certainly a preferable approach, if the device running a bit warm is not a concern.