I'm designing a board based on TI's DRV110 solenoid driver. The datasheet explain that:

"The DRV110 is able to regulate \$V_{IN}\$ voltage from a higher external supply voltage, \$V_S\$, by an internal bypass regulator that replicates the function of an ideal Zener diode. This requires that the supply current is sufficiently limited by an external resistor between \$V_S\$ and the \$V_{IN}\$ pin."

\$V_{IN}\$ is effectively always regulated to \$15\mathrm{V}\$. The DRV110 can sink between \$1\$ and \$3\mathrm{mA}\$, but also requires enough current to drive other connected components. In my design, this \$I_\mathrm{aux} = 1.5\mathrm{mA}\$ just to drive a MOSFET and resistor network.

The datasheet recommends that for a \$V_S\$ (source voltage) of \$24V_\mathrm{DC}\$, we add a series resistor (\$R_s\$) of \$9\mathrm{k}\Omega\$ between \$V_S\$ and \$V_{IN}\$. This makes sense to me. At \$24\mathrm{V}\$ source, the device has to regulate down to \$15\mathrm{V}\$, a drop of \$9\mathrm{V}\$. We now are able to drop the \$9\mathrm{V}\$ over the resistor, and at \$9\mathrm{k}\Omega\$, the current through the internal zener is \$1\mathrm{mA}\$.

  1. This wouldn't deliver enough current if I need \$I_\mathrm{aux}\$ though, correct? We would need \$3.6\mathrm{k}\Omega\$ instead so we would have a total of \$2.5\mathrm{mA}\$.

  2. Am I also limited to some voltage here, higher than \$24\mathrm{V}\$ but not necessarily the full \$48\mathrm{V}\$ that the device can accept? At \$48\mathrm{V}\$ we drop \$33\mathrm{V}\$ down to \$15\mathrm{V}\$, and that \$33\mathrm{V}/9\mathrm{k}\Omega = 3.66\mathrm{mA}\$ which is too much for the DRV110 to sink.

Things get weird, though, because the DRV110 has a wide input range (\$V_S\$ from 6 to \$48\mathrm{V_{DC}}\$) and I believe in the reference design they tried to add an external regulator for a more robust design. Below is the reference design's notes plus the schematic. Note that in the reference design, the \$I_\mathrm{aux}\$ was higher so the required current is more like \$9\mathrm{mA}\$:

"In the reference design, \$R_S = R_1 + R_2\$, and the minimum input voltage = \$19.4\mathrm{V}\$. Therefore, $$R_S = \frac{19.4 – 15}{ 1 + 0.11 + 8} = 480 \Omega.$$

The Zener diode \$D_1\$ clamps the voltage to \$20V\$ using the series resistor \$R_1\$. At a rated input voltage of \$24\mathrm{V_{DC}}\$, the Zener diode \$D_1\$ regulates the voltage to \$20\mathrm{V}\$. Then the drop across \$R_1\$ is \$4\mathrm{V}\$ and across \$R_2\$ is \$5\mathrm{V}\$. This drop ensures that \$R_1\$ takes most of the loads due to an increase in input voltage.

The above design values at the rated input voltage of \$24\mathrm{V_{DC}}\$ gives the ratio \$R_2/R_1 = 5/4\$, leading to \$R_2 = 300\Omega\$ and \$R_1 = 178\Omega\$. These values ensure that with the increase in input voltage, the current sinking of the DRV110 remains constant."


simulate this circuit – Schematic created using CircuitLab

I just don't understand the above logic.

  1. Wouldn't the current to the DRV110 always be $$ \frac{(20-15)\mathrm{V}}{300\Omega} = 16.7\mathrm{mA}, $$ significantly more than the design called for?

Now say that the input voltage was lower (supposedly they designed this for as low as \$19.4\mathrm{V}\$). Let \$V_s = 21\mathrm{V_{DC}}\$. The current through \$R_1\$ is \$(V_s - 20\mathrm{V})/ 178\Omega\$ which would be \$5.6 \mathrm{mA}\$ and this is less than the current through \$R_2\$ which is fixed at \$16.7\mathrm{mA}\$.

  1. Here, would \$D_1\$ stop clamping to \$20\mathrm{V}\$ and the current would be based on the series resistance of \$R_1 + R_2\$? In this scenario, that would be \$(21 - 15)\mathrm{V} / 478\Omega\$ which would be \$12.5\mathrm{mA}\$, again too high.

I think my understanding of the zener function and/or the IC current draw is flawed here. The network is supposed to be current limiting, so perhaps I need to stop thinking about it as determining the current draw? At this point I've been thinking about it too much and I'm not making any headway.

Thanks everyone!

  • \$\begingroup\$ I went over your post a few times and I don't see any immediate flaw in your understanding of zeners. But you are correct in your (2) though. Just because an IC can operate over a wide voltage range doesn't necessarily mean it is possible to use it in a single design that accomodates that entire range. You might have to select a smaller range within that larger range for a single design just due to practical limitations. \$\endgroup\$
    – DKNguyen
    Feb 28, 2019 at 20:52
  • \$\begingroup\$ For example, a buck regulator might be able to accept 6-42V but you are unable make a single design using it that actually accepts 6-42V. You can, however, make a design that accepts 6-24V and another design that accepts 24-48V. \$\endgroup\$
    – DKNguyen
    Feb 28, 2019 at 20:56
  • \$\begingroup\$ Thank you! I understand that the IC cannot simultaneously accept the full range, my question is more about the second half I suppose, about how it works when you are using the external zener regulator first. \$\endgroup\$ Feb 28, 2019 at 20:57
  • \$\begingroup\$ The external zener is to divide power dissipation. In the Current Controlled Driver for 24-V DC Solenoid With Plunger Fault Detection example, Figure 7, zener D5 'dissipates the part from 24V to 20V' and the DRV110 together with the other aux IC's 'dissipate from 20V 15V' Badly phrased, but hope clear enough to understand the idea \$\endgroup\$
    – Huisman
    Feb 28, 2019 at 20:57
  • 1
    \$\begingroup\$ If I read both datasheets, you can perfectly use the DRV110 for the entire range. The design example pdf just shows a 'trick' how to use the zener of DRV110 to stabilize the voltage for the DRV110 as well as other IC's tied to it (the other IC's draw Iaux). Just don't tie other IC's to the VIN pin of the DRV110 if you see problems in power dissipation. \$\endgroup\$
    – Huisman
    Feb 28, 2019 at 21:09

1 Answer 1


1) Correct. $$R_S = \frac{V_{in}-15V} {1mA + I_{aux}} = \frac{24V-15V} {1mA + 1.5mA} = 3.6 k\Omega $$

2) At 48V, you should pick a different resistor. $$R_S = \frac{V_{in}-15V} {1mA + I_{aux}} = \frac{48V-15V} {1mA + 1.5mA} = 13.2 k\Omega $$

3) Yes. It's quite weird. The \$19.4V/24V \approx 4/5\$. But \$4/5*480\Omega = 384\Omega\$. And actually \$470\Omega\$ starts to make more sense. $$\frac{20V-15V}{470\Omega}=10.6mA = 8mA \text{ (for }I_{aux}) + 2.6mA \text{ (for } I_Q)$$ But that requires a big R1 and D1 at higher input voltages.

BTW, I didn't find a hard upper current limit for \$I_Q\$ in the datasheet, only a recommended. Maybe 8mA is allowed and the 300 Ohm works? But maybe it hits the thermal limit when exceeding 3mA?

4) Don't forget to subtract \$I_{aux}\$, but 4.5mA is still more than recommended

EDIT: based on replies
Page 5 shows the recommended values for \$I_Q\$, not the absolute maximum ratings.

An approach to estimate the maximum current is reasoning as follows: Were it a normal zener, the voltage would be clamped to 15V. For the case with the \$300\Omega\$ resistor, up to 8 mA flows through the zener. For a normal zener, this would yield a dissipation of 120mW. 120mW causes using a blunty approach a temperature rise of only \$R_{\theta JA} *120mW = 183.8 °C/W*120mW = 22°C \$.
This applies to a normal zener, but I think their implementation of the ideal zener will not much deviate from this dissipation (If it was worse, why did they not implement a real zener in the DRV110 in the first place??).

So, why the recommendation of max 3mA if it is not for limiting the power dissipation? Took some time, but I think it has to be explained as follows. The footnote says:

The device sinks up to 3 mA with the added supply current.

The device uses up to 3mA at max, not the zener! So, picking a too large \$R_S\$ might drop the supply voltage below the 15V, so the zener is disabled/not functioning. Normal zeners require a current to clamp at the rated voltage (zener datasheets show a test current), which is probably the reason a higher current should be choosen than the sum of \$I_Q+I_{drv}+I_{aux}\$.

But then, it's weird they use 1mA in their calculations, and not the max 3 mA.

  • \$\begingroup\$ Re: Upper limit - On page 5 of the datasheet it mentions \$I_{Q}\$ min is 1mA, max is 3mA, nominal is 1.5mA. Also "The device regulates the supply with an internal zener diode. The device sinks up to 3mA with the added supply current...find appropriate value for Rs." \$\endgroup\$ Feb 28, 2019 at 23:00
  • \$\begingroup\$ Additionally, page 12 it gives equations for the min and max values of Rs according to min/max input voltage, though as we've seen there is a limit to the range a single Rs can cover. The min/max Rs are based on the max/min \$I_{Q}\$ (3 and 1 mA respectively) plus the \$I_{Gate,AVE}\$ which is related to the MOSFET. Mine I calculated as very small (0.06mA) but I also have a resistive pulldown network that will draw current (200ohms in series with the gate and a 10kohm in parallel as a pulldown). \$\endgroup\$ Feb 28, 2019 at 23:04
  • \$\begingroup\$ Ultimately, I just couldn't (and still don't exactly) see the correlation with these values, and those chosen in the separate reference design. Perhaps there is an error in the reference design, or at least in its documentation? \$\endgroup\$ Feb 28, 2019 at 23:05
  • \$\begingroup\$ Re: EDIT - I totally missed that it was "Recommended!" I think that you're correct here, though I believe that when they refer to the device sinking current, they are indeed referring to the internal zener. The recommendation is only when \$V_{IN}\$ is greater than \$V_{ZENER}\$ which is 15V. \$\endgroup\$ Mar 1, 2019 at 17:05
  • \$\begingroup\$ I think it is ultimately trying to say that the device can only regulate the supply down to 15V with at least 1mA to engage the internal zener and if you place too big of an \$R_{SOURCE}\$ you will choke the zener and the supply won't be regulated to exactly 15. Similarly if \$R_{SOURCE}\$ is too big we run into the power dissipation issue you mentioned. It can regulate a 48V supply so its basically the datasheet saying "don't apply 48V directly or you will burn it out, try not to go over 3mA." \$\endgroup\$ Mar 1, 2019 at 17:07

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