# Compensating for voltage drop across current-sensing resistor in a TL494-based SMPS

I'm trying to wrap my head around this reference design from this TI document. This is a 5V power supply that is rated at 10A. I see they are using one error amp to limit current and another to stabilize voltage. At 10A voltage drop across R13 would be exactly 1V which is applied to pad 16 and compared with 1/5 of Vref which is 1V. And they stabilize voltage by measuring Vo relative to GND and comparing it to Vref. So Vo relative to GND is kept at constant 5V. But the load is not connected between Vo and GND, it's connected between Vo and R13!

So at 10A the voltage across the load would be 4V, right?

Is there any way to stabilize Vo relative to the shunt resistor R13, not GND, so that output voltage is truly kept constant?

UPDATE: So far there are several ideas, but I don't know if they are any good:

1. Reduce R13 to a much smaller value to the point where its voltage drop is negligible, and then amplify that value (suggested by Andy).

2. Use a current transformer to measure current and neglect the voltage drop in its primary winding.

3. Use an external amplifier circuit to subtract voltage drop across R13 from Vo and then compare this result to some reference.

• Is not the feedback, the pin#1, monitoring the Vout? – analogsystemsrf May 11 '18 at 4:51
• Yes it does, but relative to ground. And load is not connected to the ground directly, but through R13. And voltage drop across R13 can be quite significant. – Beowulfenator May 11 '18 at 6:09

Is there any way to stabilize Vo relative to the shunt resistor R13, not GND, so that output voltage is truly kept constant?

You could use a much smaller value (say 0.01 ohms) and follow this with a x10 amplifier that delivers the same signal level to the TL494 as originally intended. You would need to be careful about choosing the amplifier to ensure that its input offset voltage was much smaller than the 10 mV you would get with a 10 amp current draw.

I guess they implemented it like this as a way to say "look! you can do current limiting with only one extra resistor!" while it doesn't actually work, as you have noticed (ie, at 10A the voltage drop is substantial). Also it is not protected against short between output and ground.

This schematic is very old! No-one would design a switcher like this anymore. Saturated BJTs are very slow switches, which means the frequency will be very low, and this requires a huge inductor, which is bulky, expensive, and lossy.

If you want to design a supply like this, try a modern chip or just buy a readymade switcher module.

• Ultimately I'm trying to modify an old TL494-based ATX power supply to provide variable voltage. So my goal here is to understand how TL494 is controlled. I'm sure there are much better ways to go from 32V to 5V, like MC34063. – Beowulfenator May 11 '18 at 16:26
• if it is an offline supply, with a transformer, then you wouldn't have this problem: current limit would be applied on the transformer primary, and voltage feedback on the secondary side – peufeu May 11 '18 at 16:45

Why not use R1 = 100 ohm , R2 = 4.9k and R13 = 0.01 ohm ?

The typical offset for the amplifier is 2mV , max 10mV , a 10% error for that limiting current can be acceptable.

A different approach would be to change the ground reference for current limiting amplifier since the datasheet states common mode input down to -300mV.

Changed R1 , R2 , R13 for 200mV max on R13 at 10 A

The output voltage will not be affected by voltage drop on the current sense resistor. simulate this circuit – Schematic created using CircuitLab

I believe the solution is simple, but I haven’t fully figured out the implications as it introduces some degree of positive feedback. It is worth at least consideration though.

Just connect the low node of R4 to R13. I would also add a capacitor from the high node of R4 to ground to avoid noise and reduce the positive feedback bandwidth. This impacts the speed of load regulation.

With that connection the reference voltage rises with the output current, just the right amount to keep the voltage across the load constant.

Note that this seems to be the edge of instability, If the rise is just an epsilon faster, the overall feedback would be positive and the output voltage will rise without bound. I.e., the supply output impedance becomes negative.

To avoid that, the positive feedback needs to be reduced to provide a safe margin and a positive output impedance. That can be done by either splitting R13 or R4 into two resistors creating a voltage divider to ground and defining the output impedance. E.g., a 1:10 ratio should set the output impedance to 0.01ohm.