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I have One Input Circuit

Input voltage : 6.5V Zener breakdown is 3.6V.

Does the 3.9K and 4.7k ohm resistor acts as a voltage divider? or what purpose does the 4.7k ohm serve?

And can someone help me how to calculate the current through 68k ohm resistor?

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  • \$\begingroup\$ What is this? Are the 6.5V and 3.3V inputs? Or is the 3.3V your output? Where did you find this circuit? What is it supposed to be doing? \$\endgroup\$ – MCG Apr 26 at 12:12
  • \$\begingroup\$ Yes. 6.5V and 3.3V are inputs. I just want to understand how this circuit works and to calculate the current through the resistors. Can you help me? \$\endgroup\$ – Newbie Apr 26 at 12:47
  • \$\begingroup\$ @MCG - Actually the 6.5V is the input. The diode and the 3.3V section is the internal clamp diode present in my microcontroller and that port is connected to the 3.3V supply. Hence, that section. The 6.5V is my input signal and it is fed to the MCU port which has internal clamp diode and that port is 3.3V port. I want to understand the current flow through the circuit and the purpose of the 4.7kohm resistor. Can you please help me understand? \$\endgroup\$ – Newbie Apr 26 at 12:56
  • \$\begingroup\$ The 4.7k resistor is probably there to deal with situations where the 6.5 V source is not connected. Maybe because it's a remote device that can be removed. Maybe because of its behavior in the first ms after power up. Maybe something else. \$\endgroup\$ – The Photon Apr 26 at 14:54
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I gather the circuit is something like this, where the \$6.5\:\text{V}\$ is a power rail and so is the \$3.3\:\text{V}\$ (but from a different power rail.)

schematic

simulate this circuit – Schematic created using CircuitLab

If so, yes -- you can combine \$R_1\$ and \$R_2\$ and the \$6.5\:\text{V}\$ rail to make up a new Thevenin source with \$V_\text{TH}\approx 3.55\:\text{V}\$. Since this is just below the zener voltage you mentioned (which suggests some kind of purposeful intention), the zener should technically be doing nothing much unless the \$6.5\:\text{V}\$ rail rises much -- when it will help hold its cathode at about \$3.6\:\text{V}\$.

Given that, and given that \$D_2\$ probably requires a few hundred millivolts to support a noticeable current (that, and given the limitations also imposed by the large valued \$R_3\$), I think any current likely to occur through \$R_3\$ and \$D_2\$ would be in the very low microamp range (or less.) Even if the \$3.3\:\text{V}\$ supply is powered down, there won't be much current. (Obviously, I'm assuming the grounds are shared.)

If you actually want to compute the current through the diode, you can use the Shockley diode equation and combine it with \$R_3\$. The resulting equation will use the LambertW equation to provide a closed solution. (See diode + resistor, for example.) Or you can use an iterative approach without it. But I'm not sure you care about any of that. (Do you?)

And what's the context for this circuit? That might go a long way in figuring out why this arrangement was created in the first place.

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  • \$\begingroup\$ I have one doubt. From the circuit you simulated above, only consider, Input=5.7V, R1 and R2 same value and Zener with Vbr=3.6V. Dont consider R3, D2 and 3.3V. Resistor R2 is the Load. I found out the current through R1 to be (5.7-3.6)/3900=0.5mA. Now, I find out the current through R2 which is 3.6/4700=0.7mA. How is this possible? My approach of calculating the current is also correct right. What is the mistake in my calculation? \$\endgroup\$ – Newbie Apr 27 at 10:15
  • \$\begingroup\$ @Newbie I didn't simulate the circuit. I only discussed what little I could without knowing the context. I asked you to provide that context. I still don't know it. I certainly didn't guess that \$R_2\$ was the load. The problem with your calculation is that there would not be 3.6 V where you say. Instead, there would be about \$5.7\:\text{V}\cdot\frac{4.7\:\text{k}\Omega}{3.9\:\text{k}\Omega+4.7\:\text{k}\Omega}\approx 3.12\:\text{V}\$ there. (The zener wouldn't fire.) That's why your current calculations for \$R_1\$ and \$R_2\$ don't pan out. You used the wrong voltage for the calculations. \$\endgroup\$ – jonk Apr 27 at 12:58
  • \$\begingroup\$ @Newbie No problem. I hope it makes sense to you to use the Thevenin equivalent for the pair of resistors we are discussing (\$R_1\$ and \$R_2\$) and the voltage source under discussion. (Even if \$R_2\$ is the load.) \$\endgroup\$ – jonk Apr 28 at 6:23

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