Re: Second Stage of Op-Am (Current to Voltage)

From: Active8 (reply2group_at_ndbbm.net)
Date: 12/28/04 

Date: Tue, 28 Dec 2004 18:35:55 -0500

On Tue, 28 Dec 2004 13:26:01 -0800, Larry Brasfield wrote:

> "Active8" <reply2group@ndbbm.net> wrote in message news:1xevajibb64in.dlg@news.individual.net...
>> On Tue, 28 Dec 2004 12:09:45 -0800, Larry Brasfield wrote:
>>
>>> "Active8" <reply2group@ndbbm.net> wrote in message news:9d5t5ofpc4fj.dlg@news.individual.net...
>>>> On Tue, 28 Dec 2004 05:12:04 -0800, Larry Brasfield wrote:
>>>>
>>>>> "Monty Hall" <chickenkungpao@hotmail.com> wrote in message news:_mbAd.3725$li1.1840@newssvr31.news.prodigy.com...
>>>>>> From the schematic below, can Iout be bidirectional?
>>>>>
<snip>
>>>
>>> It is obvious that either Q2 or Q4 can have the greater
>>> collector current. There's your reversal. Why should
>>> we need Spice to discern this?
>>
>> I didn't. I first explained to Monty that you can't get a reversal
>> of current in this circuit no matter how hard you try - you can't
>> get a voltage at that Iout node that's lower than the voltage of the
>> Miller stage emitter. There's no source.
>

First off, look at the piss poorly posed questions in my other
subthread with the OP. He gave Ic1 - Ic2 = Iout negative as the
condition, and wants to know where the negative Iout is going to
come from - then my answer is that it ain't. For all I know
Intuitive ICs for Telemarketers would've skipped the initial
condidtion of the cap prior to that state, so forget that. I
couldn't assume that on page x where he's still blabbering about a
DC condition like Ic1 - Ic2 = Iout, that he's also talking about a
steady state condition where Cc *might* be obliging and supply some
reverse Iout.

So where's the AC generator anyway? None drawn nor mentioned.

> So, you believe that Q4 cannot conduct at a lower
> Vce than that? How do you think saturated BJT's work?

Q4 can't go below Vee (+ Vce_sat) and that's what the emitter of
Qmiller is connected to. If you *could* go below Vee to *try* to
draw current from Qmiller, Qmiller would be reversed biased.
>

<snip>
>>> Cc is a feedback element in an inner loop here. In order
>>> to find where the excess phase shift of this circuit happens,
>>> it will be necessary to treat Cc as something more than
>>> just "a pole".
>>
>> WTF? Poles and zeros are *exctly* what you want to look at when
>> determining phase shifts.
>
> If the Qnameless/Cc combo provided only a single pole,
> calling "a pole, nothing more" would be fine, I guess. My
> point is simply that it has more effect in applications where
> closed loop stability becomes an issue.

Yup.
>

<snip>
>>>>>> 1. How do you compute Vo'?
>>>>
>>>> Av = 40Ic_Q2(Ro_Q2 || Ro_Q4 || Ri_Qmiller) something like that. 1st
>>>> order quess.
>>>
>>> You seem to have ignored the additional gain that occurs
>>> thru the Q1,Q4 path.
>>
>> No I haven't. See the Q4 impedance term in || with the other loads?
>
> I do and did see it. It is a minor term.

Bull! It's a gain limiting term.
>
<snip>
>>>
>>> What about Q4? Do you think it is in cutoff when Vo
>>> is mid-rail?
>>
>> ROFLMAO. When Vo is mid-rail, currnt flows to the right from the Vo'
>> node. So even if Q4 was in cutoff (and it isn't), it wouldn't
>> matter. The current would come from Q2.
>
> In that state, Q2 and Q4 collector currents are nearling balanced.
> To say the output comes from one only is oversimplified. Once
> at that state, if Vo begins to move positive at a rate faster
> than Ib(Qnameless)/Cc, net current at Vo' will be leaving
> to the left. According to your way of thinking, it would be
> going to Q4.

That's differen't from what the OP was asking (look at the OP) and
as I said, it's an effect of Cc - that may or may not manifest, BTW,
like you said. You mean slewing. That's not the same as the
condition given i.e., Ic1 - Ic2 = Iout negative.

>
>>>>>> +Vcc +Vcc
>>>>>> o o
>>>>>> | |
>>>>>> | |
>>>>>> 2 Ic = Bias Current Io = Bias Current
>>>>>> | |
>>>>>> | |
>>>>>> | |
>>>>>> o------o-----o |
>>>>>> | | |
>>>>>> |< >| o-------o----o Vo
>>>>>> Vin(-) -| Q1 Q2 |- Vin(+) | |
>>>>>> |\ /| Cc --- o
>>>>>> |Ic1 Ic2 | --- |
>>>>>> | | Vo' | |/
>>>>>> | o----------------------o---o-|
>>>>>> | | Iout = Ic2 - Ic1 (+/- 2 Ic)|>
>>>>>> | | o
>>>>>> | | |
>>>>>> | | Ic1 |
>>>>>> | o |
>>>>>> | |/ |
>>>>>> o----o---o-| Q4 |
>>>>>> | | |> |
>>>>>> | | Ic1 o |
>>>>>> | o | |
>>>>>> | |/ | |
>>>>>> |--| Q3 | |
>>>>>> |> | |
>>>>>> | | |
>>>>>> -------o |
>>>>>> | |
>>>>>> | |
>>>>>> -Vee -Vee
>>>>>> (created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de) 

-- 
Best Regards,
Mike
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