Electronic Load Achieves 0 Ω

1. An electronic load typically employs a transistor across the input terminals so the current flows from drain (collector) to source (emitter).

To achieve 0 Ω, the terminal voltage must be 0 in the presence of a current. Under this condition the pass transistor suffers a loss of operating voltage and cannot conduct a current. To maintain conduction of the pass transistor, an auxiliary power supply is connected (Fig. 2). The required voltage, VB ≥ I_In(max) × R_S + V_DS@I(max), is sufficient to maintain forward conduction with V_In = 0 and 0 < I_In ≤ I_In(max). This condition corresponds to an effective zero input resistance.

2. To maintain conduction of the pass transistor for a zero-load condition, the circuit uses an auxiliary power supply.

It can be shown that R_In = α × k × R_S for 0 ≤ α ≤ 1 and where k multiplies R_S so it can be made small to reduce power loss. If, for example, R_S = 1.0 Ω, k = 100, then 0 ≤ R_In ≤ 100 Ω for 0 ≤ α ≤ 1. If I_In(max) = 1 A and V_DS@I(max) = 2 V, then V_B ≥ 3 V.

In virtually all applications, the pass transistor must be heatsinked as the power dissipated is I_In(max) × (V_In(max) + V_B) – I_In(max)² × R_S. If in this example V_In(max) = 15 V, the pass transistor would have a maximum dissipation of 17 W. This would appear as a 15-W resistor adjustable over 0 to 100 Ω.

You may be wondering why an additional power source was introduced between the source of the MOSFET and the current sense resistor to achieve a “0-Ω load” condition (essentially a short across input voltage) in Figure 2. After all, why would you need an electronic load that can go down to 0 V (i.e., short circuit) condition?

If you wanted to short an electronic load, you could always use a relay directly across the input voltage. Or if you wanted it to be all solid state, you could place another MOSFET that has much higher current capacity (and very low R_DS(on) that approaches a few milliohms) directly across the input and turn it on using a switch. This approach would be much simpler (and cost effective) than adding a second power supply in series with sense resistor. So, why would you use the additional power source?

Here’s the answer. This electronic load can check low-voltage power supplies (i.e., 3.3 V and lower) where a voltage burden (i.e., 1 V) would not permit testing under short-circuit condition. This is useful in testing the short-circuit response of power supplies, current trip level, and effectiveness of overload protection.

Furthermore, even with a low R_DS(on), a MOSFET will not conduct at 0 V (V_DS). At high currents, a MOSFET requires non-zero voltage (see note A) to sustain conduction. This circuit will maintain VDS right into a virtual short circuit.

Finally, a hard switch such as a relay does not permit a smooth transition of resistance. The circuit described here acts more like a rheostat but with the ability to be voltage controlled (see note B) from a remote point. The controller does not have to carry the load current. It would require multiple relays to apply a range of loading. This circuit can provide smooth (stepless) effective resistive loading over a large range including 0 Ω.

The circuit can be scaled up to higher voltages and currents. Figure 3 shows an application.

3. An application circuit like this one achieves a zero-load condition for testing low-voltage power supplies.

Note A: For example, an RFP30N06 MOSFET has a specified RDS(on) of 0.047 Ω. At 30 A this device requires at least 2 V VDS. This would not be an effective short circuit on a 3.3-V, 30-A power supply. On a 1.8-V supply, it would not conduct this current.

Note B: The effective resistance can be controlled with a voltage-controlled amplifier (see “Op Amp And Two JFETs Form A Voltage-Controlled Amplifier”).