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In the below picture, I am struggling with the text, in parenthesis, that says "You could similarly add a bypass capacitor across d1 in figure 1.82"

Can someone explain this to me? What is the capacitor doing if added and why? I realize that at higher frequencies, its impedance, resistance drops, but across d1, how many ohms is a diode that is already forward biased?

The earlier text described using a capacitor to reduce the thevenin equivalent resistance of a voltage divider used to provide a cutoff value for a clamp.

Figure 1.82 uses D1 to put .6 volts at D2 so that it is just on the brink of being forward biased and D1 and D2 respond similarly with changing temperature.

Main question I am stuck on is "You could similarly add a bypass capacitor across d1 in figure 1.82"

What does this do and why? Thanks for any help.

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  • \$\begingroup\$ You find the resistance of a diode as the gradient of the U/I graph at voltages you have it biased at. \$\endgroup\$ – Janka Jan 31 '17 at 20:31
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Paul provided a great answer. I thought I'd take a very slightly different approach. That way, you have several different ways of thinking about the same thing.

At DC, all the current through \$R_1\$ and \$R_3\$ goes through the diode \$D_1\$. This sets up a voltage across the diode. (If you want to see a derivation of a diode's impedance equation, you can look at my post here.) This DC diode voltage also sets the voltage across the added capacitor discussed in the text.

But at high frequency AC, the current through \$R_1\$ and \$R_3\$ goes instead through the added capacitor, bypassing \$D_1\$ almost entirely. It's almost as though you can sink all that current straight to ground! (Lowering the apparent impedance.) This means that the diode voltage stays the same as the current through it also stays the same. So the capacitor, in effect, helps stabilize the voltage across the diode for purposes of higher frequencies. It doesn't do as much for lower frequencies, which still have some ability to affect current through the diode and therefore the voltage across it (though the capacitor does help some.)

Without the added capacitor, high frequency current changes through \$R_1\$ and \$R_3\$ still have to go through \$D_1\$ and therefore they impact the voltage across it, since the voltage does vary a little with changes in its current. It's just that an added capacitor will help if you are mostly caring about higher frequencies. (That is, if you actually want the voltage across \$D_1\$ to be more stable at higher frequencies.)

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  • \$\begingroup\$ I guess I was wondering, what part is similiar? When used this way, it would lower the impediance at high frequencies. It would look like D1 was shorted to ground. When it was previously discussed, its use in the voltage divider was to lower the apparent impediance of the thevenin equivalent circuit. When this was done, and a triangle wave was presented to the clamp, it would flatten the output curve at the clamp voltage in figure 1.88. \$\endgroup\$ – Jeffrey Edward Messikian Jan 31 '17 at 23:11
  • \$\begingroup\$ I guess that in fig 1.88 any input over 5.6 v, at the lower resistance affored by the cap at high frequencies, would have an easier way of going to ground. Hence a flattening at the clamp voltage. In figure 1.82. wouldn't droping d1 at high frequencies fundamentally change the circuit? If high frequencies were presented to figure 1.82, wouldn't d1 and d2 be affected in the same way equally? If d1 was shorted, because of a cap at high frequencies, wouldn't the .6v at r1 d1 disappear? \$\endgroup\$ – Jeffrey Edward Messikian Jan 31 '17 at 23:12
  • \$\begingroup\$ in figure 1.82, i thought the whole point of d1 was to have .6 v at d2 just so that it is almost forward conducting. I was thinking that shorting d1 at high frequencies would lower the impediance, but it would also change the circuit, so why even consider it? Thanks for your help. \$\endgroup\$ – Jeffrey Edward Messikian Jan 31 '17 at 23:12
  • \$\begingroup\$ @JeffreyEdwardMessikian I think the best way for you to think about this is that it is DC behavior sets the biasing point and it stays there. The high speed AC aspect should be thought of as tiny variations which don't have the time needed in order to change the voltage across the capacitor. Since the voltage across the capacitor doesn't change, the voltage across the diode doesn't change and therefore the current through the diode also doesn't change. Don't know if that helps. \$\endgroup\$ – jonk Feb 1 '17 at 3:46
  • \$\begingroup\$ When the author suggested using a capacitor on D1, why wouldn't one on D2 also need to be added? If the behavior of D1 can change, at higher frequencies, and without a cap, why can't d2 do the same? I would think that without a cap on D2, its impediance at high frequencies would change, and on be might need more than .6 v to be on the brink of being forward biased. \$\endgroup\$ – Jeffrey Edward Messikian Feb 1 '17 at 14:31
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Even though D1 is forward biased, it still has a series impedance, and this impedance is nonlinear and voltage (and current) dependent. Adding a bypass capacitor across D1 lowers the circuit impedance, and reduces the high-frequency voltage and current across and through the diode.

Without the capacitor, the high-frequency AC current through D1 will change the average voltage drop of the diode, causing the voltage at the R1 - D1 node to be less positive. This will reduce the forward bias on D2, possibly affecting circuit operation.

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  • \$\begingroup\$ will also the diode impedance be frequency dependent like capacitor and inductor? If yes, how is it related? \$\endgroup\$ – dirac16 Jan 31 '17 at 20:56
  • \$\begingroup\$ A diode has parasitic inductance (not much), and capacitance. In fact, many diodes can be used as a voltage-variable capacitor, where the capacitance is inversely proportional to the reverse voltage on the diode. In this forward-biased case, especially with the big series resistor that's going to dominate any parasitic impedances, the main effect I would be thinking about would be the possible change in bias voltage caused by the high-frequency AC signal across the diode. The bypass capacitor would largely eliminate that. \$\endgroup\$ – Paul Elliott Jan 31 '17 at 21:02
  • \$\begingroup\$ @PaulElliott If there is high frequency in the circuit, wouldn't D1 and D2 see the same frequency? If a capacitor is used on D1, and the .6 is preserved at the R1 D1 node, this same .6 would find its way on D2. But if D2 had no capacitor, maybe d2 didn't need to have the .6 maybe something less. Why not also acap on D2 since both see the same frequency? – \$\endgroup\$ – Jeffrey Edward Messikian Feb 3 '17 at 14:08

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