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Typical Characteristics

TPS25924 Assuming I(vout) is current out of the OUT pin of the IC.

The test parameters say the V(IN) is 12V for the given characteristics.

According to the abs max ratings specified in the datasheet enter image description here

V(OUT) can have a max value of V(IN)+0.3V = 12.3V (max) V(OUT) can have a min value of -0.3V Safe to assume that it'd be -0.3V <= V(OUT) <= 12.3V

Computing V(IN - OUT) for when V(IN) = 12V. This value would be close to 12V cause there's a NFET(edited) in the IC (small voltage drop in mV). So is it safe to say V(IN - OUT) is positive for all these plots? Assuming the max value would be a case of rev current flowing in the circuit.

Q. What is the behavior with changing R(ILIM)? Decreased R(ILIM) leads to decreased I(OUT)for the same voltage levels.

Q. How can you explain the current shoot-up for the smallest value of V(IN - OUT)? The voltage is positive, which means the voltage at input > output.

Q I'm trying to understand the below table as well enter image description here For the same voltage, resistance is increasing how can the current limits increase as well? Relating it to V = IR, keeping V fixed, R is increasing. How can the current limits increase as well?

I think this can be explained via Figure 21 (Please let me know if this is right)

enter image description here

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  • \$\begingroup\$ Yes, figure 21 shows the relation between current limit and resistor value. At small values of V(IN - OUT) the current is defined by the resistance of the internal NFET, not by the configuration resistor. \$\endgroup\$
    – Jens
    Apr 14, 2023 at 14:30
  • \$\begingroup\$ Thanks! R(on) of MOSFET is the reason for shoot up for I(load) & then once it exceeds the limiting current threshold set by externally connect R(ILIM) it saturates. \$\endgroup\$
    – Harkirat
    Apr 15, 2023 at 3:40

1 Answer 1

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  1. Do not use the absolute max ratings for calculating the vout range. When the device is subject to absolute max conditions, the device will get destroyed very quickly.

See the block diagram on page 13. The IC does not have a PFET, it has an NFET.

Yes, when Iout is going out from Vout pin, Vin-Vout is positive &
Iout = (Vin-Vout)/R Where R is the resistance between Vin and Vout in the IC. So, as Vin-Vout increases, Iout shoots up for low values of Vin-Vout but saturates once the current hits the current limit value that is set using the RILIM.

  1. To know how the current limit is implemented, see the block diagram. There is a fixed current of 10uA pumped into (the external) RILIM to generate a voltage. A scaled version of this voltage is then compared with the voltage drop across the current sense to detect if the current limit has been exceeded. So, clearly higher RILIM means higher voltage and hence allows more current to flow to the output before it triggers the ILIM comparator.

  2. Also, note here that this current limiter action is not dependent on VIN & VOUT provided Vin-out >>0.15V and is purely dependent on the value of RILIM.

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  • \$\begingroup\$ Thanks answered most of my queries. Just another one, according to the recommended spec V(ILIM) is 0 to 3V (Pin voltage). This voltage is independent of the current source voltage, which generates 10ua current, right? I'm guessing this voltage drop from the current source(higher) to the voltage across R(ILIM) (lower) is what allows for a voltage to be generated, which is then compared with the drop across the current sense to determine whether I(load) > I(ILIM)? Also for point 3 how did you come up with 15 mv (is it the forward drop of NMOS that you're considering)? \$\endgroup\$
    – Harkirat
    Apr 15, 2023 at 3:39
  • \$\begingroup\$ The 15mV was a typo. It is actually 150mV where the current stops rising and becomes flat. This is where the limiter is in action \$\endgroup\$
    – sai
    Apr 15, 2023 at 4:29
  • \$\begingroup\$ I do not fully understand what you mean but a current source has no voltage as such. The voltage across a current source varies depending on the load connected. The voltage is fixed by the resistor that you connect outside in this case 10u*RILIM. Rest is correct. This voltage is then compared with I(load)*Rsense to determine if Iload>Ilim \$\endgroup\$
    – sai
    Apr 15, 2023 at 4:34
  • \$\begingroup\$ Uhm, i didn't mean any sort of inherent voltage of the current source but the 6V fixed voltage which is meant to be always greater than V(ILIM) i.e. the pin, to ensure that the current flow is towards R(ILIM). Which then decides the voltage at the non-inverting pin of the comparator. The purpose was to ensure that the 6V is always above V(ILIM) which I think is enforced by the fact that as per rec. conditions V(ILIM) can be 0 to 3V. \$\endgroup\$
    – Harkirat
    Apr 15, 2023 at 5:12
  • \$\begingroup\$ Yes, agreed, the current source will work because we ensure 0-3V at V(ILIM) \$\endgroup\$
    – sai
    Apr 15, 2023 at 5:25

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