Constant Current verses Constant Voltage

This text was written to explain my reasoning in producing an electronic constant current drive circuit for 12 volt quartz halogen lamps (used in low voltage down lights, motor vehicles, solar systems and many other applications).
To maintain good lamp efficiency the filament temperature must be correct ! (The halogen/tungsten cycle will only work correctly at the correct temperature).

Diagram #1 is a parallel lamp circuit fed from a constant voltage supply.
Diagram #2 is a series fed lamp circuit with a constant current supply.
A 12 volt 12 watt lamp is the same as a 12 watt 1 amp lamp.
Wire loss is given by I²R in both cases. These formulas apply throughout I=P/V , P=V²/R, P=IV, P=I²R, I=V/R, R=V/I, V=IR.
( I = current in amps, P = power in watts, V = voltage in volts, R = resistance in ohms, Z = impedance )

Constant Voltage parallel circuit.	Constant Current series circuit.
Lamp Voltage must be specified.	Lamp Current must be specified.
Lamp Wattage must be specified.	Lamp Wattage must be specified.
Current varies with load.	Voltage varies with load.
Current surge occurs at turn on.	No current surge occurs.
Current limit required.	Voltage limit required.
I²R loss varies with wattage.	I²R loss unchanged with wattage.
Low Z supply.	Hi Z supply.
More detail	More detail
If lamps are added the supply voltage remains the same and the current increases.	If lamps are added the supply current remains the same and the voltage increases.
Wire (copper) loss ( I²R) goes up as more lamps added.	Wire (copper) loss ( I²R) remains the same as more lamps added.
As extra feed wire is added, the lamp at the end of the wire suffers a voltage and current reduction (lamps loose efficiency). Energy wasted in the wire is not compensated for.	As extra feed wire is added the lamps voltage and current remain the same (full lamp efficiency). Energy wasted in the wire is compensated for with extra energy from the supply.
A constant voltage supply must be current limited. eg. Fuse Without a preset current limit, and with a short circuit on the supply very high currents would flow (undesirable).	A constant current supply must be voltage limited to a maximum potential output voltage.eg. Electronic limited. Without a preset voltage limit and with no load on the supply the voltage would ionise the air between the power supply output terminals, current would flow at the rating of the supply. eg 1 amp (undesirable).

Diagrams 3 & 4 show how the lamps have to be specified for both circuit types.

Due to the fact the current in a lamp varies with filament temperature, the above voltages and currents were measured on a actual test set up.

Lamp voltage	12 volts per lamp	Full brilliance
Lamp current	1.66 amps	Full brilliance
Wire loss I²R	1.66Watts	Low copper loss ( full 40 watts to lamps )
Voltage out of constant current supply	25.66 volts	12 volts per lamp and 1.66 volts for wire loss. ( voltage will vary to maintain 1.66 amps )
Voltage into constant current supply.	12 volts

The above set up (Diag # 5 & 6) was demonstrated at ASTRO South Australia Nov 2000 meeting.

It can be seen when the supply resistance equals the load resistance equal power is dissipated in the supply and load! Maximum power transfer occurs when the source impedance equals the load impedance.
In real life, for the sake of efficiency the load Z is normally higher than the source Z.

With the 12 volt battery above a 45 watt load (3.87 amps) seems to be a reasonable maximum load as the voltage only drops from 12v to 11.61v. The supply only suffers a 1.49 watt heat load.

It can be seen a low output source resistance/impedance is an advantage for constant voltage power supplies. Imagine if the load impedance equalled the impedance of a power station, (maximum power into load) the power station would waste a lot of power!

If we wanted a constant current supply we could simply start with a high voltage supply and simply insert a series resistor to set the supply impedance ( large zeners could be placed across the output to limit the voltage).
The circuit to the left will deliver 1 amp into a short circuit and .91 amp into 100 ohms (load resistor) at 91 volts.
When driving 10 ohms at .99 amp the 1000 ohm resistor dissipates 980 watts and the 10 ohm load we are trying to drive dissipates 9.8 watts ( less then 1% efficient). The current regulation is down to about 10% at 100 volts output. It can be seen we are wasting our time with this method of producing a constant current supply. A higher voltage supply and higher value series resistor will improve the current regulation and reduce the efficiency.

Using modern circuit techniques it it is possible to produce a highly efficient constant current supply, the block diagram above (Diagram #9) is one possible solution. Due to the electronic control the output resistance is very high, a 1 ohm load will produce fundamentally the same current as a 100 ohm load. It should be noted at zero output even with automatic standby the efficiency will be < 90%. Standby current however could be very low, (microamps) having no affect on the battery life. High efficiency can be expected for normal loads.

If is P_ina fixed small amount (say 1mW) as P_out appoaches zero, the efficiency will approach Zero!

Lamp voltage	9.16 volts (very low)	Poor brilliance, low lamp efficiency.
Total lamp current	2.84 amps
Wire loss I²R	8 Watts	High copper loss (only 26 watts to lamps)
Supply voltage	12 volts

R₁	V across R₁	Power into R₁ (Heat into battery)	R₂	V across R₂ Volts into load	Power into R₂ (Power into load)	Current
	V= IR₁	P=I²R₁		V=IR₂	P=I²R₂	I=12/(R₁+R₂)
.1 ohm	.099 volt	.098 watt	12 ohms	11.88	11.76 watts	.99 amp
.1 ohm	.196 volt	.384 watt	6 ohms	11.76	23 watts	1.96 amps
.1 ohm	.387 volt	1.49 watts	3 ohms	11.61	44.9 watts	3.87 amps
.1 ohm	6 volts	360 watts	.1 ohms	6 volts	360 watts	60 amps