High-side current sensing using PNP

Question

I have the following three states:
State 1: When system is active with transducers, current through the load is approximately 350 mA.
State 2: When system is activated without transducers, current through the load is approximately 100 mA.
State 3: When in stand-by mode, current through the load is approximately 70 mA.

I am trying to sense current through the load to differentiate between states using a high-side current sense resistor. Supply voltage: 24 VDC.

I have the schematics as below, which would work if I increased the sense resistor to 3 Ω. This is not ideal, as I would like to keep the voltage drop across R1 under 2 V.

I understand I can use a high-side differential amplifier or current sense amplifier and easily solve this problem, but I would like to learn if there is a way to simplify the solution without an op-amp, since there is no precision required here.

Mainly, I need to verify either in State 1 or in State 2/3. Is there a way to differentiate between all 3 states without an op-amp?

Answer 1

You don't show how the +24V supply or the transducers are connected, so I'm left to guess.

I don't see a way to sense 70mA without amplification, if you insist on a sense resistance under \$\frac{0.7V}{70mA} = 10\Omega\$. The only idea I have is to amplify the sense resistor's voltage, which can be done with two transistors:

^{simulate this circuit – Schematic created using CircuitLab}

This is how \$V_{OUT}\$ varies with load current \$I_1\$:

This has a gain of \$\frac{R_5}{R_4}=10\$, meaning that over a range of 1A, the voltage across sense resistor R1 changes from 0V to 0.47V, and consequently \$V_{OUT}\$ changes by about tens times that.

Even at 1A the sense resistor drops less than half a volt.

There's an offset to \$V_{OUT}\$ which needs to be accounted for, but at least you have clear threshold potentials corresponding to 70mA, 100mA and 350mA.

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This looks like a a current mirror, but it's not being operated in the same way. As with a mirror, Q1 and Q2 emitters have the same potential, and Q1's role is to bias Q2 to achieve that. Unlike a current mirror the potentials at the top of R2 and R4 are not equal, due to the potential difference across sense resistor R1.

This potential difference is what causes Q2 emitter current to vary while Q1 emitter current stays fixed. Q2's collector current varies similarly, resulting in the voltage across R5 varying ten times more than the voltage across R4, and that's where we get a voltage gain of 10. The right-side is really just a common-base class A amplifier.

Unfortunately the current in each path (Q1 and Q2) is non-zero, and even without any load there is current through R1, and a potential difference across it, giving rise to a greater-than-expected permanent offset at the output.

This offset will also vary with supply voltage (+24V here), since quiescent current is highly dependent upon that, so common-mode voltage gain is quite significant.

Answer 2

Here the 350mA load current on 4.7ohm shunt makes 1.645V shunt drop.

1.645V across 330ohm forces 5mA current through 1k ,so it outputs 5V.

Note: The diode compensates the transistor Vbe drop.

If you change 1k to 2k2 the output voltages will be 2.2 times higher (11V, 3.12V, 2.18V) so it will be easier to compare it or work with.

Answer 3

Sticking with only one transistor, your only threshold / reference voltage is its Vbe, so you can detect or trigger on only one current level. For a small-signal part such as a 2N3906, conduction starts at around Vbe = 0.5 V.

The load current mid-point between 100 mA and 350 mA is 225 mA. Using Ohm's Law, the sense resistor should be approx. 2.2 ohms. Without a different shunt arrangement or more complex circuitry, I think that's all you get.

Answer 4

The wildar or wilson current mirrors though more complex are probably a better fit. I have used both. Here is a good presentation. https://slideplayer.com/slide/2315840/