# TL494/TL594 measure current with grounded load

Multiple examples of current limiting with TL494/TL594 suggest to use current sensing resistor between load(-) and GND. But I want load to be grounded. Can current sense resistor be placed above load(+)? Can I use external Rail-to-rail OpAmp this way to measure load current and then use it for TL494 feedback loop?

• Can current sense resistor be placed above load(+)? Yes, it's called high side current measurement. – Huisman Jun 19 at 19:47
• Since you appear to have the circuit already in LTspice, add small variations in value (like 1%, not all the same so variations but like you would see in real resistors, no two are identical!) to R1, R2, R3 and R4 and see how that affects the accuracy of your circuit. As Huisman suggests: look up "high side current sensing" and see how it is done and think about why your solution will have issues. – Bimpelrekkie Jun 19 at 19:54
• To add to @Bimpelrekkie's comment, LTSpice can perform a "Monte Carlo" simulation wherein the simulator simulates your circuit multiple times, and during each simulation the simulator randomly changes the component values within their specified tolerances. You end up with a family of curves. Here's a quick how-to that describes how this is done: electronicsbeliever.com/… – Jim Fischer Jun 19 at 20:41
• Yes you can but error increases and signal is cut in half. WHat are our specs for resolution, range, accuracy or stackup error tolerance , cost? None of schematics in answers offered so far will not work but yours does. Of course a better high side to rail error amp will work too. – Tony Stewart Sunnyskyguy EE75 Jun 19 at 21:44
• Also the TL494 is obsolete with better 1% 5V accuracy and UVLO on TL594 – Tony Stewart Sunnyskyguy EE75 Jun 19 at 21:49

The configuration of the amplifier as shown in the question is not quite right, as differential amplifiers will yield better results. If the load doesn't mind a small voltage bump from the sense resistor, low side configurations are preferable. Either way, use one of these configurations:

Another option (that requires less resistors) is to use instrumentation amplifiers, check the common mode range.

Find the max voltage through the sense resistor and then calculate the gain of the differential amplifier.

• BTW both you examples have outputs that exceed Vcc and gnd range. – Tony Stewart Sunnyskyguy EE75 Jun 19 at 20:50
• Do they? They are examples and use ideal op amps, there is no vcc or gnd. It's assumed that you know how to set these properly – Voltage Spike Jun 19 at 20:51
• Ok but not ideal examples . Both inputs need to be inverted on both examples – Tony Stewart Sunnyskyguy EE75 Jun 19 at 20:53
• That makes more sense, I'll edit the pic – Voltage Spike Jun 19 at 20:56
• Hmm ... even TI got the high side backwards which requires another inverter and to level shift so as to bring the output from zero to Av*IRs output between 0 and Vcc. Note to O.P. They use pairs of resistor ratios for common mode rejection with differential gain. Use 50mV to 100mV at max current or less. – Tony Stewart Sunnyskyguy EE75 Jun 19 at 20:59

Given the voltage drop in the other answer of the non-inverting differential amplifier high side goes above 10V this is not suitable.

Also looking at how R tolerance degrades CMRR and introduces DM gain error, you would need 0.1% R arrays like 10k hi and 2k low for \$3 it is not a good choice.

Your circuit does not provide this CMRR of a differential balanced gain and also attenuates the signals to 1/3.

Therefore use a high side current sense IC of your choice,with 0 to 5V out with internal gain or whatever range of V to PWM Sensitivity you need.

Note that R8 does nothing.