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I'm trying to make (and try to learn something better) about a simple current regulator. I need to make a complete static analisys before starting with the frequency one. But I've some thoughts about how the base current will introduce errors and how a series resistor to the base will make it vary.

The circuit is this (available already made on the web, but with quite silly descriptions..). The values are not quite meaningful, because I'm still at literal calculations.

schematic

simulate this circuit – Schematic created using CircuitLab

Here I've some troubles to extrapolate all the Kirchoff equations to analyze all the parameters (static ones) and so how Rb will influence the circuit.

For example, I find this:

Ib = Vbe/(Rb-Rs) - Rs*Id/(Rb-Rs)

where Id is the current flowing through the MOS M1.

But from here I recognize that Id = Vbe/Rs if we neglect the base current.

How can I give some meaning to the Ib equation, if it is correct? Should I fix the drain current then make the analisys? Of course, if I substitute the Id with Vbe/Rs, Ib will be 0, so the equation seems to be correct. But how to consider Id with the base contribution? Any help from where to start?

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schematic

simulate this circuit – Schematic created using CircuitLab

There really can write many equations for a circuit.

For Q1's base, we can write equations below, it's a bit different from yours:

$$ V_{be} = (I_{D}-I_{B})R_{s}-I_{B}R_{b}\quad(1)\\ I_{D} = \frac{V_{be}+I_{B}(R_{b}+R_{s})}{R_{s}}=\frac{V_{be}}{R_{s}}+I_{B}(\frac{R_{b}}{R_{s}}+1)\quad(2)\\ I_{B} = ((I_{D}-I_{B})R_{s}-V_{be}) / R_{b}\quad(3) $$

For M1 we can write, assume M1 work in saturated region

$$ V_{GS}=V_{2}-I_{B}\beta R-(I_{D}-I_{B})R_{s} > V_{TN}\quad(4)\\ I_{D}=K_{n}(V_{GS}-V_{TN})^{2}\quad(5)\\ V_{1}-V_{D1}-(V_{be}+I_{B}R_{b}) > V_{GS} - V_{TN}\quad(6) $$

Substitute (3) into (4), then differentiate it by \$I_{D}\$, \$R_{b}\$ can reduce the \$V_{GS}\$ change rate caused by \$I_{D}\$ change.

$$ \frac{dV_{GS}}{dI_{D}}=\frac{R_{s}^2-RR_{s}\beta}{R_{b}}-R_{s} $$

Then you can do more analysis based on these equations, it's really tedious work!!!! If you really want to know how one component's change will affect your circuit behavior, you can go for help from simulators, such as sensitivity analysis in PSpice.

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  • \$\begingroup\$ The ultimate intention to have this analysis, is to understand if and how much is useful the base resistor. Someone said that the bjt can be a bit exposed without any base resistor, in case of oscillations. I was trying to be able to assert or confutate that sentence, by making a quantitative analysis on it. Later I will analyse those equations. \$\endgroup\$ – thexeno Nov 25 '14 at 15:42
  • \$\begingroup\$ Rb indeed can reduce the feedback speed. Because it limits the base current, so limit the rate of the Vgs change. So less sensitive to noise. \$\endgroup\$ – diverger Nov 25 '14 at 15:57
  • \$\begingroup\$ Very helpful. How can I analyze this change? Should i derive the feedback loop equation? How can i quantify an amount of R balanced between speed and noise covering? \$\endgroup\$ – thexeno Nov 25 '14 at 16:23
  • \$\begingroup\$ @thexeno: see my edited. \$\endgroup\$ – diverger Nov 25 '14 at 16:26
  • \$\begingroup\$ @thexeno: For time, you need take into the BJT's junction capacitance and the MOSFET's input capacitance, that will be more equations. \$\endgroup\$ – diverger Nov 25 '14 at 16:33
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I'd start by assuming Rb is significantly bigger than Rs - then you can fairly assume that the source voltage is due 9in the main) to source current and Rs.

Next I'd suggest that the BJT will start conducting when the base voltage (source voltage approximation) is at about 0.6 volts. When does this start happening - when Is*Rs is 0.6 volts.

Then you have negative feedback - as soon as the BJT starts to significantly conduct it's reducing the gate voltage and therefore reducing the source current which in turn means the BJT is going to stop conducting.

You can apply as many formulas as you like but the exact point that the BJT and MOSFET settle down to peaceful equilibrium is not that easily predictable unless you go through plenty of math and I ask - is it worth it?

If you want a current source that is predictable build one around an op-amp.

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  • \$\begingroup\$ What you say is what I already know, but thanks. I don't want to use all the maths that is possible to apply, but I'd like to know how every component will change the system. Is it worth? No. But, I want to do this anyway. The opamp analysis is a bit simple, even the frequecny analisys, I suppose. And I made an implementation based in it with frequncy analysis also. Now, I'm on this one and I feel the troubles. :) \$\endgroup\$ – thexeno Nov 25 '14 at 12:00

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