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I have this circuit:

enter image description here

The effective resistance at the reference node of D4801 is given as

Rb = R4802||(R4805+(R4807||(R4806+R4808+44KOhm))) = 9.129KOhm

I am not able to understand how this formula is arrived.

Shouldn't it be, Rb = R4802||(R4805+((R4806+R4807)||R4808+44KOhm)))

Please tell me how the first Rb is correct and the above is incorrect?

How should we factor or consider the R4806 resistor into this formula? I find it confusing on how to include the R4806 and R4807 into the effective resistance formula.

Any tips on how to find a easy way in these type of circuits?

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  • \$\begingroup\$ Neither are accurate - the 44k should realistically be more like 22k because of the forward conduction of the base emitter. If you want better accuracy in your calculations when dealing with diode (base emitter) impedances, use a sim. \$\endgroup\$
    – Andy aka
    Commented Oct 18, 2020 at 12:44
  • \$\begingroup\$ When you say, "Neither are accurate - the 44k should realistically be more like 22k", the 22k refers to Ra or Rb ? \$\endgroup\$
    – user220456
    Commented Oct 18, 2020 at 12:46
  • \$\begingroup\$ Leave out the '480's and add some air into your equations: \$ R_b = R_2 || (R_5 + (R_7 || (R_6 + R_8 + 44k))) \$. \$\endgroup\$
    – Transistor
    Commented Oct 18, 2020 at 12:46
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    \$\begingroup\$ @Newbie I'm referring to Ra because Rb is shunted by B-E. \$\endgroup\$
    – Andy aka
    Commented Oct 18, 2020 at 12:47
  • \$\begingroup\$ @Transistor, could you please tell me how it is done? \$\endgroup\$
    – user220456
    Commented Oct 18, 2020 at 12:48

2 Answers 2

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The Rb is the resistance used for calculating the Voltage division of the 5V between R8403 and "Rb". It is derived by looking down from R8403 looking at the impedances. You first see: R4802 || X. When looking at X you see the SCR (which is infinite when switched off so you dont include it) and you see R4805 in series with a certain Y circuit. So you get: R4802 || (R4805 + Y). When following Y you see a split path. Which means parallel R4807 with Z. So you get: R4802 || (R4805 + ( R4807 || Z). When following Z, you see a series path to ground: R4806 + R4808 + RA and RB. The capacitor has infinite resistance so again this parallel branch is left out. The transistor shorts the RB. So Z is R4806 + R4808 + RA. Which In total leads to: Rb = R4802 || (R4805 + ( R4807 || (R4806 + R4808 + RA)))

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  • \$\begingroup\$ The upper transistor Q4802 has also infinite impedance. Addition: If the goal is to calculate the Voltage at the gate of the SCR, you should keep in mind that there is a Voltage drop of around 0.7 V (look for exact value in datasheet) between base and emitter of the lower Q4802 (same name is not very convenient) \$\endgroup\$ Commented Oct 18, 2020 at 13:18
  • \$\begingroup\$ Thank you for the simple, clear and detailed explanation. Accepted the answer! \$\endgroup\$
    – user220456
    Commented Oct 18, 2020 at 13:33
  • \$\begingroup\$ not so queeksdraw Newbie \$\endgroup\$
    – D.A.S.
    Commented Oct 18, 2020 at 13:55
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Neither are accurate. yes it must include R4802, but the Zener control has high gain and the PNP has hFE so the negative feedback reduces your node effective input impedance. ( such as if you were to inject noise from some Rs)

But since the Zener control current is small, it has negl. Loading effects. (Verify)

Ticky Tacky Technical Details

enter image description here

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  • \$\begingroup\$ Could you please expand and explain how to understand the formula ? Could you also provide your explanation on how you would do this? \$\endgroup\$
    – user220456
    Commented Oct 18, 2020 at 12:50
  • \$\begingroup\$ Do you know how an Op Amp inverting input has 0 ohm differential impedance ~ yet dc current in uA? Similar effect. Except input is Zener differential control to 0V \$\endgroup\$
    – D.A.S.
    Commented Oct 18, 2020 at 12:53
  • \$\begingroup\$ I am not able to understand your answer. I want to just understand how the effective parallel resistance at the reference of the D4801 is to be arrived? If not, how the formula actually tries to ? \$\endgroup\$
    – user220456
    Commented Oct 18, 2020 at 12:58
  • \$\begingroup\$ Prerequisite answer to comment above makes it easier \$\endgroup\$
    – D.A.S.
    Commented Oct 18, 2020 at 13:01
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    \$\begingroup\$ Can you arrive an Req in your head yet? I used Irfanview to edit image in 10 minutes \$\endgroup\$
    – D.A.S.
    Commented Oct 18, 2020 at 15:19

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