# Hard Circuit

Discussion in 'Homework Help' started by ihaveaquestion, Sep 7, 2010.

1. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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Uhhh... I don't even know where to start for b) and c):

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Presumably the Shockley diode equation must be applied in the solution.

Last edited: Sep 7, 2010
3. ### JDT Well-Known Member

Feb 12, 2009
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I think all you really have to consider is the current in the 10k load resistor. This current adds or subtracts from the standing current in the diodes

Start off by considering the circuit when Vin is zero. The current through each diode is 0.5 I. There is no current in the voltage source or the load.

When Vin is non-zero, a current is forced through the load resistor. This un-balances the current in the diodes.

Don't forget that the current sources have infinite impedance. This means that, in your circuit, the earth point has no relevance. It could be drawn as a series connection of the voltage source and resistor across the diode bridge. The current through the load resistor will always equal the current in the voltage source.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The current source(s) appear to bias the diode network. The estimation of current flow across the diode network from the small signal source to the 10k resistor would presumably take into account the dynamic resistance of the diodes at their bias condition. Conduction across the diode network would not be possible in the absence of the biasing - given the diode orientations.

I'm not sure what this problem is all about - i.e. what is the context? I've "guessed" it's got something to do with dynamic resistance. Maybe signal modulation / multiplication?

The absence of any information about what assumptions can be made is also a little disconcerting.

Perhaps the OP might enlighten us.

A bit of simulation might help .....

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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This might add some ideas to the debate ....

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6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The scan wasn't that great for the png format.

I've attached a pdf format file.

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7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Did a simulation using 1n4148 diodes.

For part (c) Vout=~2V peak-to-peak

Part (c) makes some sense since the peak current in the resistor was ~0.1mA (10% of the bias current). The (more accurate) source voltage required was ~2.019V peak-to-peak. This makes the drop across the diode network about 19mV peak-to-peak.

It was difficult to set up a solution for part (b) but I have a value of I=70nA. Vout was about 140uV peak-to-peak.