Analysis of diode biasing in the given bridge connection of four diodes

How does the output voltage change when Vs is varied from -10 V to + 10 V in the given circuit?

Given: Vs is input voltage, Voltage at 【A】is the output voltage; all diodes are identical with cutin voltage = 0.7V

PS: TP1 is just an scope which can be ignored in this case.

• why not simulate it? Jan 24, 2017 at 19:00
• Better to assemble it
– user76844
Jan 24, 2017 at 21:27
• @Andy: i did simulate it but could not analyse it. jonk's answer explains it very well. Jan 25, 2017 at 7:27

The voltages across the diodes will vary with the current through them. But they only vary about $100\:\textrm{mV}$ (different diode types may vary a little here) for each factor of 10 change in current. That's not much of a voltage change for quite a lot of change in current. So as a first order approximation for now, you can consider the voltage drops across the diodes as approximately fixed.

When the input is at $0\:\textrm{V}$, there will be about $10\:\textrm{V}$ across both $10\:\textrm{k}\Omega$ resistors to the two power supplies -- or, roughly $1\:\textrm{mA}$. This splits in two directions through the pairs of diodes. So that means about $500\:\mu\textrm{A}$. I suspect that there will be perhaps from $650-700\:\textrm{mV}$ across each diode, then. Let's call it $680\:\textrm{mV}$ just to pick a number.

If you assume that the diodes all have fixed voltages across them no matter what, then the output should follow $V_s$ the entire time. ($V_s$, plus $D_1$'s diode drop, but then minus $D_2$'s diode drop, again.) But there's a problem. As $V_s$ declines towards $-10\:\textrm{V}$, at some point it crosses over a voltage after which $D_3$ simply turns off. Below that point, all that you have is the $10\:\textrm{k}\Omega$ output resistor sourcing current from ground, through $D_4$ and into the negative rail resistor. This sets up a current that will have a magnitude of $10\:\textrm{V}$, less a diode drop, divided by the resistance sum of $20\:\textrm{k}\Omega$. That current creates a voltage drop across the negative rail $10\:\textrm{k}\Omega$ resistor so that the voltage at the cathode of $D_4$ and $D_3$ will be $-10\:\textrm{V}+\frac{10\:\textrm{V}-680\:\textrm{mV}}{10\:\textrm{k}\Omega+10\:\textrm{k}\Omega}\cdot 10\:\textrm{k}\Omega$ or about $-5.34\:\textrm{V}$. This means that $V_s$ will be $-5.34\:\textrm{V}+680\:\textrm{mV}=-4.66\:\textrm{V}$ at this point. So when $V_s \lt-4.66\:\textrm{V}$ nothing more happens at the output. So the output tracks $V_s$ to that point in the negative direction, but no further.

A similar argument also has it that when $V_s \gt +4.66\:\textrm{V}$, the output no longer tracks $V_s$.

\begin{align*} \begin{array}{cc} V_s & V_{out} \\ \\ -10\:\textrm{V} \rightarrow -4.66\:\textrm{V} & -4.66\:\textrm{V} \\ -4.66\:\textrm{V}\rightarrow +4.66\:\textrm{V} & V_s \\ +4.66\:\textrm{V} \rightarrow +10\:\textrm{V} & +4.66\:\textrm{V} \end{array} \end{align*}

You could get a little silly about it and remember that there's a $100\:\textrm{mV}$ change for each factor of 10 current and imagine that as $\vert V_s\vert$ gets closer to that magnitude of $4.66\:\textrm{V}$ that the current gradually shuts off and that while that happens, the voltage drop for the diode also diminishes. But this might add half (remember here we divided by 2, earlier), or another $50\:\textrm{mV}$ of extra range. So I might round things to a magnitude of $4.7\:\textrm{V}$ perhaps. But this is nit-picking and I'd probably should have rounded it, anyway.

So let's go with:

\begin{align*} \begin{array}{cc} V_s & V_{out} \\ \\ -10\:\textrm{V} \rightarrow -4.7\:\textrm{V} & -4.7\:\textrm{V} \\ -4.7\:\textrm{V}\rightarrow +4.7\:\textrm{V} & V_s \\ +4.7\:\textrm{V} \rightarrow +10\:\textrm{V} & +4.7\:\textrm{V} \end{array} \end{align*}

So, I post my simulation output here

(Cutin voltage for the diode was 0.4 V in the simulation)

• thank you for the satisfactory analysis, it matches quite well with my simulation result. Thanks again! Jan 25, 2017 at 7:06
– jonk
Jan 25, 2017 at 8:38

There are three conditions with 3 resistors, I'll call R+,R-,Rout all equal here.

1. For input, Vs < +/-0.7V Zin = Rout = 10k
2. For Vs > 0.7V, Zin= R-//Rout = 5k
3. For Vs < -0.7 , Zin = R-//Rout =5k

But since Vs is a voltage source with Zs=0 the output (A) voltage Vout=Vin always but input current, Is doubles when Vs>+/-0.7V.

If there was a source resistance of same value, added to Vs , the voltage would attenuate according to this R divider when it swings outside +/-0.7V and has features of a switched attenuator.

• thank you for your analysis but in this case Vs is varied from -10 V to +10 V. Also, how does Is doubles for Vs > 0.7 V? I don't understand why you mentioned the switched attenuator. Jan 25, 2017 at 7:20
• my misteak ;) I spent more time typing the answer than thinking about it. If I corrected it, it might say, all 4 diodes are conducting from the forward bias of +10,-10V thru R+,R-. So the output (A) always tracks the input (Vs) until Vs reaches diode threshold defined by the R ratio of 50% of 10V. The diode Vf drop of 0.6 is also split by 50%, so the output becomes a didoe limiter for 5V minus 0.3V for all inputs above 4.7V determined by R Ratios and fixed voltages. Is this easier to remember? Jan 25, 2017 at 15:42
• Notice that as the R ratio gets smaller the diode Vf drop ratio becomes smaller and thus the output limiter value approaches only the RL/(RL+Rbias) *V with diminishing error from Vf as in the 1% example ratio above, the output signal is limited to +/-100mV +/- resistor tolerances. but I give credit to @jonk for his excellent analysis Jan 25, 2017 at 16:33
• If the Rout=10 ohms while Rbias stays at 10k, both +/-bias resistors are still conducting thus the output impedance looking in is 5k and the voltage Ratio is 0.001/(.001+5)*10V=2 mV peak as a "passive precision voltage limiter" Jan 25, 2017 at 16:48
• Thank you Tony for a different perspective to look into the analysis. It does make sense now. Sorry for not seeing the reply sooner. Jan 27, 2017 at 9:48

So, I post my simulation output here

(Cutin voltage for the diode was 0.4 V in the simulation)

• Hey! That's tricky! Select my answer, add this, say I gave the answer, and then move the answer to yourself, later! (I don't mind. I just I don't recall seeing this being done before.)
– jonk
Feb 1, 2017 at 17:46
• @jonk haha, sry no intention of tricking lol. Yes, i added the simulation later just to enrich the question. Would be better if the simulation output follows your answer, the answer would look much better. If it can be added to your answer, i will remove mine. Feb 1, 2017 at 19:23
• Hehe. No problem. EE.SE informs me, of course, when "things change" and I was going, "What did I do wrong??" hehe. Then I looked and I thought, "Wow! That's new!" I wasn't bothered at all, but figured I'd just let you know I'd noticed by trying to write in a slightly humorous way so that you wouldn't take it wrong. I really wasn't bothered at all.
– jonk
Feb 1, 2017 at 20:03
• I'll put the image in my answer.
– jonk
Feb 1, 2017 at 20:03
• @jonk I might make a few mistakes in SE as I'm not so often here. Thanks for the headsup :) Feb 1, 2017 at 21:03