# Voltage Regulator Properties and Behavior

1. Jan 30, 2013

### Bergenheimer

Hello, simple question here!

When a voltage is regulated down or up in a circuit by, lets say a DC/DC converter, does the corresponding current increase or decrease respectively so that the system remains consistent with the laws of conservation of energy?

In other words, I want to know if i lose power when a circuit component regulates a voltage. For example, if i have 60 volts of power going into a 48V charge controller, is there any power lost here when the charge controller regulates the voltage to the batteries on the other end?

Thanks,

∫ Bergenheimer

2. Jan 30, 2013

### Staff: Mentor

The output power is the regulated output voltage multiplied by the load current.

The input power is the sum of the output power plus the power lost in the regulator (due to inefficiency). For most DC-DC converters, the efficiency of conversion is in the 90%+ range.

3. Jan 30, 2013

### Studiot

Just to be crystal clear.

Any power supply can set or determine either the voltage or the current, but not both.

The second quantity is set by the load and its interaction with the power supply (psu).

So a psu provides

1) A specific voltage, whatever current the load draws.

or

2) A specific current and adjusts the voltage to force this through the specific load.

Battery chargers usually fall into category 2.

The input power is the input current times the input voltage and the output power is the output current times the output voltage.

They are independent, except that obviously there must be enough input to finance the output and the inefficiency losses.

4. Jan 30, 2013

### Bergenheimer

Ok, I have a 48 Volt charger in which i set the float, bulk, and equalizing voltages. It is rated up to 150 amps. The input to the system is a solar panel. The solar panel is rated at 60 volts and outputs about 8 amps. So if 480 watts is put into the charge controller, can I expect 480 watts to be provided to the battery (assuming the battery requires charge)? Or does it only get 48 volts and 8 amps (384 watts).

So i guess what im asking here is this a norton/thevinin equivalent situation (from a circuit analysis point of view).

-Berg

5. Jan 30, 2013

### Studiot

What makes you think the charger will always draw 8 amps from the photovoltaic cells?

That would violate case 1 in my earlier post.

The photovoltaic panel offers a combination of cells that develop 60 volts at the terminals under stated lighting conditions.

The current will depend on the applied load, up to a maximum of 8 amps.

If your charger is set to deliver 8 amps to the batteries at 48 volts it will do this and draw 8amps plus its own requirements, say .2 amps from the photovoltaic panel.

So the input power is 8.2*60 = 482 watts

It will deliver

output power 8*48 = 384 watts

It will waste 98 watts

So its efficiency will be 80%

edit: If your battery charger were only delivering 6 amps to the batteries, the calculations above would use 6, instead of 8 and 6.2 instead of 8.2 in the formulae.

Last edited: Jan 30, 2013
6. Jan 31, 2013

### Bergenheimer

Ok, thanks. I was hoping that the charge controller was able to convert that lost 98 watts (which is voltage potential) into available current to the load. Why is this not possible?

7. Jan 31, 2013

### Staff: Mentor

Please see my post #2. DC-DC converters are basically constant power converters (minus the efficiency loss). The load draws whatever current is defined by its internal resistance I=V/R. This determines the output power, which determines the input power.

You can draw less than the max rated current, but you cannot draw more. If the output power demand exceeds the input power available, then the output voltage will droop. If the output power demand is less than the maximum available input power from the solar panel, then less input current will be drawn from the solar panel.