SSM2040 Filter Analysis Part 1 - Ryan Williams

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Robert Gibbs
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1 SSM2040 Filter Analysis Part 1 - Ryan Williams The following analysis is use to etermine the maximum bias current neee for the iscrete OTA cells in the ssm2040 filter (René Schmitz version) to achieve a 20KHz cutoff frequency. This information will then be use to set a current limiting resistor in the voltage to current converter. Then an analysis of the voltage to current converter is use to verify that the filter has a 1V/octave response. Some steps are skippe, especially in the OTA cell. The erivation of the equations use to escribe these circuit blocks is escribe elsewhere (check my web site). A few other assumptions are mae about the reaer's knowlege, but I'll point these out later. Kohm : 1000ohm Mohm : 1000Kohm ma :.001A ua :.001mA nf : 10 9 F :.026V na :.001uA The above circuit shows a simplifie version of the OTA block use in the SSM The +5V core is replace by groun to simplify the analysis slightly. The current source (Is1) represents the bias current for the amplifier. The goal here is to etermine the range of current that will be useful for musical applications. The OTA core itself (Q1,Q2,Q3,Q4) converts the ifferential voltage at the bases of Q1 an Q2 to an output current at Q3 an Q2's collectors. The output current is scale by the bias current (Is1). With the aition of the capacitor (C1) an the arlington follower (Q5,Q6) the circuit forms a current controlle integrator with negative feeback taken from Vout. The circuit for the OTA is very similar to that of the CA3080 an LM There are several erivations of these equations on the net so I won't eal with that now but you can check my website for links to some papers that will help. The equation for the output current into C1 is as follows:

2 I o I abc tanh 220ohm V in + V out 10Kohm + 220ohm 2 The Vt variable equals 0.026V at room temperature (again see my site for links talking about this). Since our signal input is fairly small (ue to the attenuating resistors 220ohm an 10K) we can approximate the tanh function. This will simplify the analysis significantly. For small values tanh(x)x so our new equation for Io looks like this: 220 I o I abc ( V in V out ) 10Kohm + 220ohm The constant comes from the 1/(2*Vt). Next the basic equation for a capacitor current/voltage relationship gives this: I C V > I t o 1nF V c t I have labele Vc as the voltage across the capacitor. I also assume that the initial voltage across the capacitor at time t0sec is 0V. The output voltage is approximate as two ioe rops below the capacitor voltage but we will see shortly that this makes no ifference in the output voltage. V out V c 1.2V Plugging the equations for Vout an Io into the OTA equation gives this: 1 nf t V out + 1.2V I abc V in V out 10 Kohm 220ohm + 220ohm The next step assumes you have some knowlege of the laplace transform, an frequency response in general. I have use the laplace transform to solve the ifferential equation above for Vout as a function of Vin an Iabc. This is not the only way to solve such an equation but since we nee our result in the frequency omain (as oppose to the time omain) this is my preffere metho. I'll spell it out in a few steps incase you are not familiar with the laplace transform. If this is completely new to you I suggest reaing a little bit online. try google: "solving ifferential equations with the laplace transform" an probably some hits on "frequency response" laplace1 nf t V out + 1.2V laplace I abc s 1 V in V out nf V out 10 Kohm 220 ohm ohm I abc V in V out 10 Kohm 220 ohm ohm The first laplace transform is of the left han sie of the equation. The secon shows the right han

3 sie. This is one of the most simple laplace transforms you woul ever o. Generally they are much more complicate. The 1.2V on the left han sie is gone because the laplace transform of a constant is 0 (for time t>0sec). The result on the right han sie is exactly what we starte with. The transfer function of the circuit in the frequency omain can now be foun: 220 ohm Kohm ohm s 1nF V out I abc V in V out V o V in 1nF s + I abc I abc shoul be in this form > V o 1 V in s + 1 ω c V o 1 V in 1nF I abc s + 1 The transfer function is a simple first orer lowpass filter. The SSM2040 filter uses 4 stages of this block to create a 4th orer filter, but the cutoff frequency of all 4 of them is the same. To fin the cutoff frequency I have put the transfer function so that the enominator is in the form (omega*s+1) where the cutoff frequency is foun as omega/(2*pi): I abc A f c ( I abc ) : π Hz f c ( 1nA) Hz f c ( 300uA) KHz I have solve fc for some Iabc values to fin the minimum an maximum usable bias currents. The minimum shoul be somewhere aroun 1nA an the maximum shoul be aroun 300uA. In the original circuit the current was limite to a few ma which is far too large for this circuit. Now, just for fun, I have plotte the frequency response of the filter for several ifferent values of Iabc. The following steps are only use to scale the plot appropriately. min : 2π F( s, I abc ) : max : 2π I abc A s + 1 n : 100 i : 0.. n r : r i n s( i) : min e Hz ln max min

4 10 Frequency Response for some Iabc values 1 F( s( i), 100nA) F( s( i), 1uA) F( s( i), 10uA) F( s( i), 100uA) F( s( i), 300uA) s( i) Hz We have now seen that the bias current shoul be limite to 300uA. If the current is not limite, then high CV values can cause pops in the auio output. I have seen this on my filter an this kin of behavior (or estruction) might be expecte from an IC OTA as well. The currents I am talking about are about 2.5mA which is way to high. The filter has some CV feethrough at higher bias currents. This is seen as a DC offset at the filter's output. It is likely cause by the low precision OTA circuit use. The pops I hear on my filter may have been partially ue to the fe through CV rapily moulating the output when the bias current reache some large value. At the high en of the CV potentiometer range, the current changes very quickly with even the smallest ajustment.

5 The above circuit is the current sink that generates Iabc for each of the OTAs. The 4 NPN collectors (Q2,3,4,5) are Iabc1, 2, 3, an 4. The resistor R7 (not given a value yet) is use to limit the maximum current of the current sinks (more on this later). First we fin the output voltage of the opamp (U1) as a function of the Vcv control voltage: V opamp V cv : 15 V 51 Kohm 200 Kohm 51 Kohm V cv 100 Kohm The voltage at the base of the Q1 is: V base V cv : V opamp ( V cv ) 1Kohm 27Kohm + 1Kohm To fin the Iabc current I'll look at Q1 with the opamp circuit forme aroun it (U2) plus one of the Iabc NPN transistors (Q2 for example). The other transistors have the same base-emitter voltage an collector current. The trick to this circuit, is that the collector current through Q1 is hel constant by the opamp (U2). Q1 is in the negative feeback path of U2, Because the opamp will keep the voltage at it's inputs equal, the noninverting input will always be hel at 0V (unless the opamp output excees it's range). The current trough Q1 can then be foun as: I c1 : 15V 1Mohm The equation for an NPN transistor's collector current is:

6 V be I c I s e where Is is the saturation current, Vbe is the base emitter voltage an Vt is the voltage at room temperature. This equation generates an exponential current as a function of it's base emitter voltage. This is not satisfactory for our purposes because both Is an Vt change with temperature. Is is is probably not precisely given in a ata sheet but might be somewhere aroun 10^-15. This value changes quite a bit with temperature. The changes in Vt are fairly small an will be ignore for now. They can be correcte with a PTC ("tempco") resistor or electronically with a more complicate circuit. To solve the Iabc current we nee to o a little trick which cancels out Is from the equation. Diviing I2 by I1 will achieve this. note: this step coul be a subtraction but ue to log rules, ivision is equivalent: V b1 V e1 I c2 I c1 I s e V b2 V e2 I s e The Is variables in each transistor shoul be the same if we use two matche transistors on the same chip (the CA3046 or HFA3046 in our case). If the transistors are han matche an thermally connecte (glue together) then we can usually make the same assumption. Because the emitter voltages of both transistors are equal, an the base of Q2 is groune, we can simplifiy that equation to: V b1 I c2 I c1 e now set IabcIc2. Ic1, an Vt are plugge in as constants an Iabc is foun to be: V base ( V cv ) : I c1 e I abc V cv I abc ( 2V) na I abc ( 1V) na The two currents shown above Iabc at 2V CV an 2*(Iabc at 1V) are use to check whether the bias current will allow 1V/octave. We can see that the current at 2V CV is almost right at twice the current at 1V an that is very close to 1V/octave. If resistor tolerances are off then the value coul be slightly ifferent. I woul reccomen using 1% resistors an possibly ecreasing R4 (27K) just a small amount to allow greater than 1V/octave. Then the CV attenuator pots coul set 1V/octave. A trimmer coul be use as well if very goo scaling is important. This may be the case if you are planning to tune the oscillating frequency of the filter to a VCO.

7 I abc ( 1.793V) na I abc ( V) ua f c ( I abc ( 1.793V) ) Hz f c ( I abc ( V) ) 20 KHz The above values show that 1.793o V CVs cover just about the whole auio range. The panel control of the current source (not shown) can be ajuste to output between -15V an +15V. The extremes of this range are not usable. fc for these two values are shown below: f c I abc ( 15V) Hz f c ( I abc ( 15V) ) KHz well below the auio range. well above the auio range. The panel control coul be moifie to limit the minimum an maximum Vcv values but I'll leave this alone because it will be nice to a or subtract an offset from the panel if neee. the pot is usuable over most of it's range (-7 or -8o +12V). Instea the current limiting resistor R7 shoul be ajuste. To fin a reasonable value we take the sum of all the Iabc values an Ic1, then ivie the biggest possible voltage rop across R7 by the sum of these currents. I have ae 200uA to each value of Iabc (an arbitrary number I chose to account for any error). We can work own from there to fin a resistor value that exists. As long as the bias currents stay within about 600uA then I on't think there will be many problems. I c uA R Kohm 0.7V ( 13.5V) R 0.7V V R : I c uA The 13.5V shown is for a TL072. That is about the lowest voltage that a TL072 can output before clipping (assuming a +-15V supply). -0.7V is one ioe rop below groun at the emitter of each NPN transistor. I have chosen a resistor of 6.8Kohm an the current through each transistor will be limite to: R : 6.8Kohm ( ) V 4 R ua max bias current a plot of the bias current vs Vcv is shown below. if ( I abc ( V cv ) < uA, I abc ( V cv ), uA ) I bias V cv :

8 Ibias vs. Vcv ua I bias ( V cv ) V cv Volts Yes, I i cheat an manually clip the plot at 470uA. One other point to note is that a normal 0..+5V control voltage cannot sweep the filter across it's entire auio range. If this is wante, then a smaller input resistor on the U1 opamp woul achieve this. Perhaps for the secon CV input. The first woul be labele 1V/oct an the secon coul be labele CV1.

Triangle to Sine Conversion with OTAs Ryan Williams :

Triangle to Sine Conversion with OTAs Ryan Williams : http://www.sdiy.org/destrukto This note describes the basic operation of a triangle to sine converter using OTAs. I have made no attempts to show a