# Switching regulator output voltage dropping when load connected?

I need a 2 V source capable of providing up to 5 A to supply current to a heating element; I am trying to use a Texas Instruments PTR08060WVD (datasheet) to achieve this, and with an open circuit I do indeed get 2 V, however as soon as I connect the load, the output voltage drops to around 1.3 V (and is a little unstable).

Any ideas what's causing this & how to fix? Circuit diagram is below:

• Have you checked with a 'scope how it behaves? Is it really outputting ~1.3V under load or is it oscillating? Commented Aug 7 at 13:41
• @Unimportant Checked and there was not oscillation under load but it was noisier
– Dip
Commented Aug 7 at 14:31

If the resistance you have for the heating element is the hot resistance you're probably pulling too much current from the device.

For 0.3 Ω + 10 mΩ at 2 V you'd have 6.645 A, more than 10% over the device's rating. To keep it below 6 A the MOSFET would have to be dropping around 200 mV. That's with a fixed 0.3 Ω resistance, a heating element is not a fixed resistance, they are generally a much lower resistance when cold than when hot. So what may be happening is the element is trying to draw far more than 6 A and the regulator can't supply the full voltage.

• The cold resistance is 0.24ohm but is usually around 0.3, my LTSpice sym says the pulse current through the heating element should be 5A
– Dip
Commented Aug 7 at 14:32

I suspect that you are trying to draw well over 5A, which is why the output voltage falls. The maximum voltage across the 0.3Ω load is:

$$V = IR = 5A \times 0.3\Omega = 1.5V$$

This is suspiciously close to the 1.3V you have witnessed, if the MOSFET were switched fully on.

You must apply exactly $$\V=IR=5A \times 10m\Omega = 50mV\$$ across the 10mΩ sense resistance, which will be hard to get right, especially if the op-amp has millivolts of input offset. Any potential over 50mV at the op-amp's non-inverting input will cause this over-current condition. In fact, depending on the op-amp, this might not even be possible. A TL081, for instance, won't work with inputs closer than 1V to its negative supply.

On top of that, after the heater has eaten 1.5V, most of the the remaining 0.5V is across the transistor, which will therefore be dissipating $$\P=IV=5A \times 0.5V=2.5W\$$, which seems a waste. It's 33% of the heater's power.

It makes no sense to use both PWM and the voltage controlled current sink that you have implemented here. Use either one or the other. I presume you are trying to limit current to 5A, given a voltage source of 2V, but this is better achieved with PWM from the correct voltage to start with, like this:

simulate this circuit – Schematic created using CircuitLab

This ensures minimum power dissipation in the MOSFET, will switch faster than a slow-slewing op-amp, and doesn't rely on a precise voltage applied across the 10mΩ current sense resistance.

Don't forget to account for the $$\R_{DS(ON)}\$$ of the MOSFET, which will contribute to the total resistance in that path. You'll find you may need to raise the supply a little to compensate for a non-zero $$\V_{DS}\$$.

If you must use 2V, then consider limiting current with another resistor, to obtain the expected 5A:

simulate this circuit

The same caveat regarding $$\R_{DS(ON)}\$$ still applies, but instead of reducing the supply voltage, you could reduce R2 instead.

Perhaps the PTR08060WVD is going into overcurrent protection, resulting in a hiccup-mode (as described in their datasheet).

The maximum output current is rated at 6A. 2V/(0.33R+0.003R) ~= 6.45A, which already exceeds the maximum current during 'normal operation' of this circuit.

Whilst the datasheet specifies 10A as it's overcurrent protection limit, I imagine the inrush current (charging the output capacitor and cold start heating element) and current ripple are enough to trigger this protection. Which causes it to eternally hiccup.

This means the current should be dropped; either by using a higher output voltage or increasing the load resistance.