Is this how I would expect a voltage regulator to work?

Question

I have a question just to check that I understand voltage regulators from a conceptual standpoint.

Say I have a power source, which delivers 8V. I need to step it down to 6V for a continuous load of 1A. 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?

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Now, I do plan on doing this, but with an L7806 voltage regulator (see page 11 for L7806 electrical characteristics), an 8V power source, and 2 motors which draw 1A each. The first problem I am seeing with this setup is that my power source is under the rated input voltage (8.6 - 19V, my source is 8V). Next, it is rated for a current draw of up to 1A (but I need to draw 2A).

So I am not sure what to do. From what I understand, if I use 2 voltage regulators to independently provide the power to each motor (each wired in parallel to the power source), then the max draw per regulator would be 1A. I still have the problem of my power source being under the voltage source though, but I may be able to fix that.

The other option I found is to use something from the RC world called a UBEC. For example, this one is rated to output 6V at a maximum of 3A, with better heat dissipation.

So what I am looking for from an answer is (1) a confirmation that I am understanding how voltage regulators work at a basic level (as well as any corrections or improvements to my misunderstanding), and (2) suggestions as to how I can achieve what I need. I would prefer to have an IC solution (like the 7806, but maybe something more appropriate for my case) because it is cheaper than the UBEC.

Answer 1

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.

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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.

Answer 2

A linear regulator, such as the 7806 or the LD1085 mentioned below, acts like a variable resistor between input and output. The control circuitry alters the effective resistance to "drop" excess voltage and this is dissipated as heat. In your case with Vin=8V and Vout = 6V and I load = 2A, your dissipation in the regulator
= I x V
= 2A x (8-6) V
= 4 Watts.
A heatsink will be needed.

As Anindo says, a UBEC is a switching regulator which "transforms" energy at one voltage level to energy at another voltage with relatively low losses. In this case with only 2V drop you will find the linear regulator simple and probably acceptable loss wise.

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The variable voltage regulator ST LD1085 regulator in stock at Digikey for about $US1.65 in ones will meet your needs.

Pricing page

It allows up to 30V input, up to 3A output, drops only 1.3V minimum at 3A and can be configured to supply any voltage in the range 1.5V - 30V.

Use fig3 page 6, right hand diagram to set the output voltage to suit.