Op Amp Calculations
in [Design]
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From: Fred Bartoli on 5 Oct 2005 05:35 "Terry Given" <my_name(a)ieee.org> a ?crit dans le message de news:1128463369.256572(a)ftpsrv1... > Fred Bloggs wrote: > > > > > > Terry Given wrote: > > > >> Fred Bloggs wrote: > >> > >>> > >>> > >>> Terry Given wrote: > >>> > >>>> Fred Bloggs wrote: > >>>> > >>>>> > >>>>> > >>>>> Terry Given wrote: > >>>>> > >>>>>> Fred Bloggs wrote: > >>>>>> > >>>>>>> > >>>>>>> > >>>>>>> Terry Given wrote: > >>>>>>> > >>>>>>>> > >>>>>>>> I recently built about 50,000 of this circuit, with a feedback > >>>>>>>> cap too (mathcad rather than mathematica, and a pencil to start > >>>>>>>> with for the analysis), and 15 inputs thru 100k resistors. the > >>>>>>>> effect of the 14 "grounded" resistors shifted the center > >>>>>>>> frequency by about 10% - Aol was about 50. power consumption > >>>>>>>> (and cost) constraints meant I couldnt use a faster opamp, so > >>>>>>>> instead I stopped assuming and started calculating :) > >>>>>>>> > >>>>>>> > >>>>>>> What was the transfer function you were shooting for, and which amp? > >>>>>>> > >>>>>> > >>>>>> a summing band-pass (ish) filter. 40 x TLV274. > >>>>>> > >>>>>> I didnt want to AC-couple the inputs (that would have cost me 240 > >>>>>> capacitors) so I used the bridged-T feedback network with an RC > >>>>>> shunt to give a DC gain of about 1/16 - any DC is basically > >>>>>> common-mode, and the next stage was AC coupled. 3 Rs and 2 Cs was > >>>>>> a whole lot cheaper than an RLC. But 100k/14 = 7k in parallel with > >>>>>> the -ve shunt arm, enough to move Fc 10% or so. > >>>>>> > >>>>>> SPICE clearly showed it, so I went back and re-did my opamp > >>>>>> analysis using Dostals approach (originally I did it using the > >>>>>> Woodgate approximation), and voila - out popped the same answer. > >>>>>> Mr HP3577 also agreed with spice and mathcad. Dostals method also > >>>>>> allowed me to directly calculate the phase margin. Since then, I > >>>>>> have analysed all opamp circuits thusly - but I use the Woodgate > >>>>>> approach with pencil & paper as a bullshit detector :) > >>>>>> > >>>>>> Cheers > >>>>>> Terry > >>>>> > >>>>> > >>>>> > >>>>> > >>>>> > >>>>> > >>>>> You can achieve a wild increase in effective GBW by going to > >>>>> current mode feedback. The peaking and rapid rolloff due to that > >>>>> low impedance -ve shunt is eliminated from any frequency bands > >>>>> usable with the voltage feedback circuit. The output gain of 2x > >>>>> deals with the CMR input range of the TLV274- requires about a volt > >>>>> of headroom to V+ - facilitation odds and ends not shown... > >>>>> View in a fixed-width font such as Courier. > >>>>> > >>>>> . > >>>>> . > >>>>> . >--[Ri]-+-------+------[R1]--+--[R2]------------+ > >>>>> . | | | | > >>>>> . o | | [R3] +--[R]---+-->Vout > >>>>> . | | +5V | | | > >>>>> . | | | C | | |\ | > >>>>> . o | | +-----||-+ +--|-\ | > >>>>> . | | | | | >--+ > >>>>> . | | +-----[Rc]-----+------|+/ > >>>>> . o | | | | | |/ > >>>>> . | | | | | > >>>>> . >--[Ri]-+ | +-----------+ | [R] > >>>>> . | | | | | | > >>>>> . >--[Ri]-+ +------|>|---+ | | | > >>>>> . | | | | | | | > >>>>> . >--[Ri]-+ +--------------------------+ > >>>>> . | | | | | | > >>>>> . | | | | | | > >>>>> . | | |\| c | | > >>>>> . | +-|+\ |/ | | OA TLV274 > >>>>> . | | >-+---| c | > >>>>> . 2.5V>-+-----|-/ | |\ |/ | > >>>>> . |/| | e+ c > >>>>> . | | |\ |/ > >>>>> . | [Rb] e+ > >>>>> . | | |\ > >>>>> . | | e > >>>>> . | | | > >>>>> . GND---------+--+-----------+------------ > >>>> > >>>> > >>>> > >>>> > >>>> > >>>> I'll study that a bit later. unfortunately it also achieves a wild > >>>> increase in parts count and cost - this circuit is replicated many, > >>>> many times :) > >>> > >>> > >>> > >>> > >>> I can't comment on your application since only you know what that is. > >>> You may find this strange, but the idea is to overcome the > >>> limitations of low inverting input shunt impedance and not to improve > >>> your product. > >>> > >> > >> What about the typo? the 2nd transistor shorts out the +5V supply.... > >> > >> Cheers > >> Terry > > > > > > It is a common collector pre-drive for the third transistor which is CE. > > If extreme gain accuracy is not needed it can be pulled, the CE should > > be a high beta type at low Ic. > > > > shouldnt there be something to limit current though? +5V...Vce...Vbe > with nary a resistor in sight. ultimately the base current could be as > high as the opamp output current (assuming negligible contribution from > the summing junction).... > > nice artwork BTW > That, (which can be corrected by moving Q2's collector) plus Q1/Q2 operate at vanishingly low IC so that the input impedance has nothing to do with that of a CFB, even at moderate audio frequencies. Take IC3=1mA, HFE=100 for all the Qs. That makes IC1=25nA and gmQ1=1uA/V and a total transconductance one third of it or gm = 0.33uA/V. The TLV GBW product is 3MHz, so at 10kHz the input impedance is 3M/300=10K and inductive (a 0.16H inductance). With a 3pF input it'll resonate at a low 230kHz and give a 2nd order roll-off above this frequency. The 3rd stage miller capacitance will also roll-off in that frequency range. Hardly a usable CFB amp :-) -- Thanks, Fred.
From: Fred Bloggs on 5 Oct 2005 09:03 Fred Bartoli wrote: > "Terry Given" <my_name(a)ieee.org> a ?crit dans le message de > news:1128463369.256572(a)ftpsrv1... > >>Fred Bloggs wrote: >> >>> >>>Terry Given wrote: >>> >>> >>>>Fred Bloggs wrote: >>>> >>>> >>>>> >>>>>Terry Given wrote: >>>>> >>>>> >>>>>>Fred Bloggs wrote: >>>>>> >>>>>> >>>>>>> >>>>>>>Terry Given wrote: >>>>>>> >>>>>>> >>>>>>>>Fred Bloggs wrote: >>>>>>>> >>>>>>>> >>>>>>>>> >>>>>>>>>Terry Given wrote: >>>>>>>>> >>>>>>>>> >>>>>>>>>>I recently built about 50,000 of this circuit, with a feedback >>>>>>>>>>cap too (mathcad rather than mathematica, and a pencil to start >>>>>>>>>>with for the analysis), and 15 inputs thru 100k resistors. the >>>>>>>>>>effect of the 14 "grounded" resistors shifted the center >>>>>>>>>>frequency by about 10% - Aol was about 50. power consumption >>>>>>>>>>(and cost) constraints meant I couldnt use a faster opamp, so >>>>>>>>>>instead I stopped assuming and started calculating :) >>>>>>>>>> >>>>>>>>> >>>>>>>>>What was the transfer function you were shooting for, and which >>>>>>>> > amp? > >>>>>>>>a summing band-pass (ish) filter. 40 x TLV274. >>>>>>>> >>>>>>>>I didnt want to AC-couple the inputs (that would have cost me 240 >>>>>>>>capacitors) so I used the bridged-T feedback network with an RC >>>>>>>>shunt to give a DC gain of about 1/16 - any DC is basically >>>>>>>>common-mode, and the next stage was AC coupled. 3 Rs and 2 Cs was >>>>>>>>a whole lot cheaper than an RLC. But 100k/14 = 7k in parallel with >>>>>>>>the -ve shunt arm, enough to move Fc 10% or so. >>>>>>>> >>>>>>>>SPICE clearly showed it, so I went back and re-did my opamp >>>>>>>>analysis using Dostals approach (originally I did it using the >>>>>>>>Woodgate approximation), and voila - out popped the same answer. >>>>>>>>Mr HP3577 also agreed with spice and mathcad. Dostals method also >>>>>>>>allowed me to directly calculate the phase margin. Since then, I >>>>>>>>have analysed all opamp circuits thusly - but I use the Woodgate >>>>>>>>approach with pencil & paper as a bullshit detector :) >>>>>>>> >>>>>>>>Cheers >>>>>>>>Terry >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>>You can achieve a wild increase in effective GBW by going to >>>>>>>current mode feedback. The peaking and rapid rolloff due to that >>>>>>>low impedance -ve shunt is eliminated from any frequency bands >>>>>>>usable with the voltage feedback circuit. The output gain of 2x >>>>>>>deals with the CMR input range of the TLV274- requires about a volt >>>>>>>of headroom to V+ - facilitation odds and ends not shown... >>>>>>> View in a fixed-width font such as Courier. >>>>>>> >>>>>>>. >>>>>>>. >>>>>>>. >--[Ri]-+-------+------[R1]--+--[R2]------------+ >>>>>>>. | | | | >>>>>>>. o | | [R3] +--[R]---+-->Vout >>>>>>>. | | +5V | | | >>>>>>>. | | | C | | |\ | >>>>>>>. o | | +-----||-+ +--|-\ | >>>>>>>. | | | | | >--+ >>>>>>>. | | +-----[Rc]-----+------|+/ >>>>>>>. o | | | | | |/ >>>>>>>. | | | | | >>>>>>>. >--[Ri]-+ | +-----------+ | [R] >>>>>>>. | | | | | | >>>>>>>. >--[Ri]-+ +------|>|---+ | | | >>>>>>>. | | | | | | | >>>>>>>. >--[Ri]-+ +--------------------------+ >>>>>>>. | | | | | | >>>>>>>. | | | | | | >>>>>>>. | | |\| c | | >>>>>>>. | +-|+\ |/ | | OA TLV274 >>>>>>>. | | >-+---| c | >>>>>>>. 2.5V>-+-----|-/ | |\ |/ | >>>>>>>. |/| | e+ c >>>>>>>. | | |\ |/ >>>>>>>. | [Rb] e+ >>>>>>>. | | |\ >>>>>>>. | | e >>>>>>>. | | | >>>>>>>. GND---------+--+-----------+------------ >>>>>> >>>>>> >>>>>> >>>>>> >>>>>> >>>>>>I'll study that a bit later. unfortunately it also achieves a wild >>>>>>increase in parts count and cost - this circuit is replicated many, >>>>>>many times :) >>>>> >>>>> >>>>> >>>>> >>>>>I can't comment on your application since only you know what that is. >>>>>You may find this strange, but the idea is to overcome the >>>>>limitations of low inverting input shunt impedance and not to improve >>>>>your product. >>>>> >>>> >>>>What about the typo? the 2nd transistor shorts out the +5V supply.... >>>> >>>>Cheers >>>>Terry >>> >>> >>>It is a common collector pre-drive for the third transistor which is CE. >>>If extreme gain accuracy is not needed it can be pulled, the CE should >>>be a high beta type at low Ic. >>> >> >>shouldnt there be something to limit current though? +5V...Vce...Vbe >>with nary a resistor in sight. ultimately the base current could be as >>high as the opamp output current (assuming negligible contribution from >>the summing junction).... >> >>nice artwork BTW >> > > > That, (which can be corrected by moving Q2's collector) plus Q1/Q2 operate > at vanishingly low IC so that the input impedance has nothing to do with > that of a CFB, even at moderate audio frequencies. > > Take IC3=1mA, HFE=100 for all the Qs. That makes IC1=25nA and gmQ1=1uA/V and > a total transconductance one third of it or gm = 0.33uA/V. Not quite, essentially all the incremental base drive voltage is developed across hie of the first transistor. Your reasoning is wrong and your estimate is high. > > The TLV GBW product is 3MHz, so at 10kHz the input impedance is 3M/300=10K > and inductive (a 0.16H inductance). > With a 3pF input it'll resonate at a low 230kHz and give a 2nd order > roll-off above this frequency. Yeah- right, i,in= A(jw)*gm*vin or Zin= re,Q1/A(jw), for the TLV274 with A(jw)=6.28E6/(jw), this becomes Zin= re,Q1 * jw /6.28E6 so that at 1KHz you have re,Q1E-3. Taking re,Q1=beta^2*0.026/IcQ3,Q this makes Zin about 2.6K at 1KHz. But you have a fairly huge gain of vout=A(jw)*gm*beta^2*2*Rc*vin, and since gm=1/(re,Q3*beta^2), Av=A(jw)*2*Rc/re,Q3, so that Yf becomes Av/Rf, and the current divide ratio is Av/(Av+A(jw)*gm*Rf) or 1/(1+Rf/(2*Rc*beta^2))- this is looking more like a CMF all the time- a broadband split neglecting a few poles here and there. > > The 3rd stage miller capacitance will also roll-off in that frequency range. That is very easily taken care of- but it's not my job to educate you...all you need to know is that the CE rolloff is not an issue. > > Hardly a usable CFB amp :-) Nah- "hardly usable" is what your boss says about you. SPICE does not agree with your conclusions- what does your IQ non-enhancer, Mathematica, tell you now? The one drawback to the triple Darlington is the same peaking effect due to error amplifier impedance growth with frequency, it is possible to replace Q1 with a PNP emitter follower feedback to OA input, interchange OA (-) and (+), but then this creates another headache by introducing a dominant pole in the loop.
From: The Phantom on 5 Oct 2005 10:19 On 4 Oct 2005 23:49:02 -0500, The Phantom <phantom(a)aol.com> wrote: >On Tue, 4 Oct 2005 11:44:53 +0200, "Fred Bartoli" ><fred._canxxxel_this_bartoli(a)RemoveThatAlso_free.fr_AndThisToo> wrote: > >> >>"The Phantom" <phantom(a)aol.com> a ?crit dans le message de >>news:r7f3k19amd4737csli9d1c7c7c1d5engqn(a)4ax.com... >>> On Mon, 3 Oct 2005 21:13:16 +0200, "Fred Bartoli" >>> <fred._canxxxel_this_bartoli(a)RemoveThatAlso_free.fr_AndThisToo> wrote: >>> >>> >> >Well, you've forgotten the GBW in all that :-) >>> >> >>> >> No, I didn't forget at all; it's implicit in A. >>> >> >>> > >>> >Well, yes, I somewhat figured that. >>> >But not mentionning A(s) when you explicitly use s somewhere else might >>lead >>> >the not so careful to some error, i.e. forgetting its phase which is -90d >>> >over almost the bandwidth. >>> >>> I think that anyone sufficiently well versed in complex arithmetic >>> as used nowadays to write transfer functions will know that if they >>> want the DC gain, they can just use the TF I gave with A constant, and >>> if they want AC results, then of course they will know that A must be >>> a function of frequency.. >>> >> >>OK, fair enough. >> >>> > >>> > >>> >> >If we set WT=2.pi.GBW then we have >>> >> > >>> >> > -A - B p >>> >> >----------------- with >>> >> > 1 + C p + D p^2 >>> >> > >>> >> > A0(R4+R5) >>> >> >A = ---------------- >>> >> > R3(1+A0)+R4+R5 >>> >> > >>> >> > A0.C1(R4 R5 + (R4 + R5) R6) >>> >> >B = ---------------------------- >>> >> > R3(1+A0)+R4+R5 >>> >> > >>> >> > (A0(R3 + R4 + R5 ) + C1.WT((R3 + R4)R5 + R6(R3 + A0.R3 + R4 + >>R5)) >>> >> >C >>= -------------------------------------------------------------------- >>> >> > (R3 (1 + A0) + R4 + R5 ) WT >>> >> > >>> >> > >>> >> > A0.C1 ((R3 + R4) R5 + (R3 + R4 + R5) R6) >>> >> >D = ------------------------------------------ >>> >> > (R3 + A0 R3 + R4 + R5) WT >>> >> > >>> >> > >>> >> >Less than 2 min to work this out from scratch, incl. sign error >>> >correction. >>> >> >>> >> What sign error is that? >>> > >>> >That was in the starting equations (wrong opamp gain sign). >>> >>> Did you notice that three different people contributed to the >>> posting that you originally replied to? John Woodgate did the first >>> ASCII schematic, John Fields did the second and provided an equation, >>> viz: >>> >>> R4 + R5 Vcc R2 >>> Vout = -Vin --------- + --------- >>> R3 R1 + R2 >>> >>> All I (Phantom) posted was a couple of transfer functions. I didn't >>> use John's equations, and I don't think there was a sign error in what >>> I posted. Most of your reply was directed to me, so I thought you >>> were suggesting I had made a sign error. >>> >>> It would make it easier for others to comment without ambiguity and >>> misunderstanding on your posting if you would cut and paste in the >>> equation you think is in error, perhaps even indicating what you think >>> the correct equation should be. >>> >> >>Oh, I see. >>Sorry for the misunderstanding, I wasn't implying somebody made a sign >>error. >>It was just *me* that made the sign error when first writing the problem and >>got obviously wrong results. >>Error corrected in a snap. > > I understand now. >> >>By saying: >>> >> >Less than 2 min to work this out from scratch, incl. sign error >>> >correction. >> >>I was just emphasing how easy it is to correct such errors vs the >>paper/pencil method. > > You're absolutely correct. It is really nice, having spent hours on >a really complicated problem, discovering a stupid mistake that would >entail more hours of computation after having discovered the mistake. >to be able to just key in a small change, hit enter, and !bang!, the >final result, a result derived from a large amount of algebraic >drudgery! > >> >> >>> > >>> > >>> >> >>> >> >Isn't that Mathematica lovely? ;-) >>> >> >>> >> Unfortunately, the result doesn't seem to be correct. Put the >>> >> expressions for A, B, C (the numerator of the "C" expression seems to >>> >> be missing a closing parenthesis), and D into the first expression, >>> >> namely: >>> >> >>> >> -A - B p >>> >> ----------------- >>> >> 1 + C p + D p^2 >>> >> >>> >> Then when you have it written out in all it's glory, find the limit >>> >> as A0 --> infinity, and then the limit as C1 --> infinity. You >>> >> *should* get: -((r5*r6 + r4*(r5 + r6))/(r3*r6)) >>> >> >>> >> But, alas, you don't. >>> >> >>> > >>> >But then I do :-) >>> >I've doubled checked the results the brute force way, taking the A,B,C,D >>> >directly from here back into mathematica and all the limits are OK. >>> >>> You set WT=2.pi.GBW, but you haven't indicated what kind of op amp >>> model you're using. Was it the standard one pole model: >>> >>> A(s) = WT/(s + Wa) >>> >> >>Yes it is. > > > Except that you didn't use that particular formulation. You used a >non-standard (in my experience) expression, namely; > > A(s) = WT/((WT/A0 + s)) and this caused your results to contain both >A0 and WT. There's no need for this since WT incudes A0. > > (I'm going to use Wp for the pole frequency in the single pole model >for an op amp in what follows.) > > A standard expression for op amp gain, (which is exactly equivalent >to yours, but not involving *both* A0 and WT explicitly) such as found >on page 19 of the classic Schaumann and Van Valkenburg, is: > > A(s) = WT/(s + Wp) > >another alternative would be: > > A(s) = (A0 Wp)/(s + Wp) > > Solving the circuit with either of these expressions gives results >that explicitly display the effect of Wp, and only one variable that >involves the DC gain of the op amp. > > (I personally like to use the expression: > > A(s) = A0/(1 + sT), where T is the time constant of the pole.) > >> >> >>> or something else? I can't comment further until you tell me your >>> model, but I still think there is something out of kilter with your >>> expression: >>> >>> -A - B p >>> ----------------- >>> 1 + C p + D p^2 >>> >>> after A, B, C and D are substituted. >> >>Well, you can't just make A0 and C1 go to the limit and expect the DC >>response. > > Yes, you can, if you use the second version of the op amp gain I >gave above. If you use the first, standard, expression for A(s), then >you have to make WT go to infinity (but not A0 since it's no longer in >the result expressions) and then C1. > >>You also have to either make WT go to the limit, > > Only if you have used an expression for A(s) which involves *both* >A0 *and* WT, which is not necessary. > >>or make s=0. > > In your Mathematica program, you have these four lines (change mode >temporarily from using s to using p :-) ): > >Limit[sol, A0 -> infinity] >Limit[%, C1 -> infinity] >Limit[%, WT -> infinity] >Limit[%%, p -> infinity] This should be Limit[%%, p -> 0] of course. I proofread all this several times, and still this slipped by. :-( > >The fourth limit operates on the result of the second limit, which >expression already lacks the variable p, I just noticed that this is not true; I think I must have been making small changes to the program (fooling around to explore some things) and so what I was looking at when I said this was not your original program. > so of course the result of >this limit is the same as the result of the second limit and therefore >accidentally gives the correct DC gain; It gives the correct result because products such as p*C1 have been eliminated by the second limit. > but, in general, just letting >p -> 0 won't give the DC gain if capacitors are involved, I think this is true if any products of p and C1 are present in the expression before letting p -> 0. If you let p -> 0 after the first limit of the four above, then you get the result -(R4 + R5)/R3 (WT drops out, too), which I believe happens as explained below. > as I >explain. > > It is the admittance of the capacitor(s) we want to become infinite >to get DC gain; that is, the product p*C1 must go to infinity. I >notice that if you let p -> 0 first (with C1 still finite), and then >let A0 go to infinity, the result is -(R4 + R5)/R3, because the >product p*C1 has gone to zero. Letting both p -> 0 and C1 -> infinity >would only work if p and C1 vary in such a way that the product p*C1 >-> infinity. Thus, letting p -> 0 won't give the DC gain if C1 >remains finite. But this is not a problem, since letting C1 -> >infinity does the job. > > When I originally took your full expression involving A, B, C, and D >and took the limit as A0 -> infinity and C1 -> infinity, I got the >same result you got after the second limit (above) and I couldn't >understand why WT was still in the result; that's why I initially >thought your result was incorrect. Other than that unnecessary >complication, your result is, of course, correct. > > The basic three equations are easy to write down by inspection as >you did, and Mathematica did just what you told it to, and yes, >Mathematica is indeed lovely. > >> >>Then all goes fine. >> >>But then you can see that it's easy, even for someone "sufficiently well >>versed in complex arithmetic >> as used nowadays" (to take your words), to forget something in the process. >>Hence my preference for writing A(s). It's just one thing less to remember. >> >>From some of your previous posts I believe you have mathematica, so you'll >>find the code herewith.
From: Jim Thompson on 5 Oct 2005 11:03 On Wed, 05 Oct 2005 13:03:55 GMT, Fred Bloggs <nospam(a)nospam.com> wrote: > > >Fred Bartoli wrote: >> "Terry Given" <my_name(a)ieee.org> a ?crit dans le message de >> news:1128463369.256572(a)ftpsrv1... >> >>>Fred Bloggs wrote: >>> >>>> >>>>Terry Given wrote: [snip] >>>>> >>>>>What about the typo? the 2nd transistor shorts out the +5V supply.... >>>>> [snip] >>> >>>shouldnt there be something to limit current though? +5V...Vce...Vbe >>>with nary a resistor in sight. ultimately the base current could be as >>>high as the opamp output current (assuming negligible contribution from >>>the summing junction).... >>> >>>nice artwork BTW >>> >> >> >> That, (which can be corrected by moving Q2's collector) plus Q1/Q2 operate >> at vanishingly low IC so that the input impedance has nothing to do with >> that of a CFB, even at moderate audio frequencies. >> >> Take IC3=1mA, HFE=100 for all the Qs. That makes IC1=25nA and gmQ1=1uA/V and >> a total transconductance one third of it or gm = 0.33uA/V. > >Not quite, essentially all the incremental base drive voltage is >developed across hie of the first transistor. Your reasoning is wrong >and your estimate is high. And it STILL will go up in flames during power-up. Sheeesh! Such DESIGN ;-) [snip] ...Jim Thompson -- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | | http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.
From: Winfield Hill on 5 Oct 2005 12:18
Jim Thompson wrote... > > ... Sheeesh! ... Hey, that's my word! -- Thanks, - Win |