# Current controlled VCA and differential amplifier question

1. Aug 22, 2009

### bitrex

Hi everyone. I'm working on a project where I'm using a current controlled differential amplifier as a VCA followed by an op amp in a differential amplifier configuration with gain. I'm using the op amp to make up gain, because I'm dividing down the input signal going into the VCA stage by 1000 to 1 to stay within the transistor's linear operating region. My problem is to get that gain from the op-amp stage I have to use small values of resistors for the input (R1 in the Vout = R2/R1*vin equation) and it's messing up the operating point of one of the VCA diff-amp stages because of the virtual ground at the inverting input of the op-amp. What would be the best way to solve this problem? I'm thinking of just using another op-amp section as a non-inverting buffer, but that of course uses up another op-amp section for each stage and seems wasteful. I'm wondering if there's a better way to go about doing this?

Edit: In my design, I had the VCA set for unity gain at maximum input control current, which necessitated using small collector resistors. Perhaps I could set the VCA to have more gain, and the op amp differential amplifier to have less, and then I could make the input resistors to the op amp larger so they have less effect on the circuit?

Last edited: Aug 22, 2009
2. Aug 23, 2009

### Mike_In_Plano

Bitrex,

Building a VGA from discrete components is a course in frustration because of the demanding matching requirements. You may want to check out an LM13700. I've had great success with this part and it's fairly cheap in quantity.

I'd suggest leaving the internal buffer stage out of your design (just ground the input of the buffer). Feed the output of the LM13700 through a 470 ohm (note 1) into the inverting input of an op amp. Use an 8.2K for feedback, and you should be in business. That will give you about 2Vp-p for 60mV p-p in and 300uA gain set.

Note 1 - The 470 ohm resistor provides protection from capacitive loading on the inverting input.

3. Aug 23, 2009

### bitrex

Thanks for the reply, Mike. Yeah, an LM13700 was my first idea, but the fellow whose project I'm attempting to modify apparently lives in a country where the LM13700 is very expensive or requires an overseas order. I thought I'd see what I could do just using transistors and ordinary op-amps. I've attached a picture of what I came up with - the circuit has approximately unity gain for an input control voltage of 10 volts down at the bottom left. I suppose for best performance, all of those transistors would have to be matched...:yuck:

Edit: oops, I believe I made an error in the schematic. The collector load resistors need to be higher for unity gain.

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4. Aug 24, 2009

### Mike_In_Plano

Bitrix,

I can't help but say that you're creeping me out a bit when you say your customer can't get LM13700's. Most of the free world (North America, Asia, All of Europe) can easily obtain these parts. That makes me worry about the folks that are on the "do not sell" list...

5. Aug 24, 2009

### bitrex

Not a customer of mine, just an Internet audio enthusiast whom I'm in contact with. I think he's from South America, so I believe the problem is less an issue of being on some kind of "do not sell" dual-use list and more one purely of logistics. Probably they truly aren't available locally, and the marginal cost including shipping on a few ICs ordered from the US to the southern hemisphere is prohibitive. Also, shipping to some areas of South America is unreliable, to put it mildly. I used to work for an electronics distributor, and there were always customs problems and long delays shipping equipment to areas of South America, particularly Argentina and Brazil. One order to Brazil I remember was the first and only time I recall DHL returning a package to us not because the address was invalid, but for whatever reason they simply could not physically get the package to its destination! After some discussion with the customer, we sent the package again via the regular postal service and it eventually (miraculously) arrived. It only took two months in transit and customs.

Ironically, shipping packages via DHL to places like Sarajevo and the West Bank was always error-free, with the packages arriving to the customer right on schedule. Of course each of those orders took more time to process due to the "dual-use" type customs forms that had to be filled out...

6. Aug 25, 2009

### Mike_In_Plano

Fair enough. I remember one of our techs from South America complaining that there were roads that didn't have "recognized" names.
Anyway, I tossed together something that may help. It's made of very common parts and you can probably use a 4558 derivative or better yet, a TL082 op amp. It only requires one pair of transistors be matched.

Good Luck!

- Mike

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7. Aug 25, 2009

### bitrex

Nice circuit! I see now one of the major flaws in my design - the input transistors are referenced to ground so the output buffer needs to have its non inverting input referenced to 1/2 Vcc to keep the output linear. I was wondering why I was getting asymmetric gain in my simulation...:rofl: I believe I understand your circuit pretty well, but I have two questions. I think the purpose of Q10 is to correct for the base current/collector current difference in the current source formed by Q9 and U2, but I'm having trouble understanding how it works, could you explain? Also, how do you go about calculating the gain of the circuit for a certain input voltage/control current with the Wilson current mirror as the load of the differential pair? Thanks for all your help!

Edit: Since the impedance looking into the Wilson current mirror is very high, I think that perhaps the gain of the differential amp is just determined by the transconductance of the pair at a certain operating point (1/2*control current/26mV) times R5?

Last edited: Aug 25, 2009
8. Aug 26, 2009

### Mike_In_Plano

Hiya Bitrex,

U2 and Q9 convert your 0-5 volt control voltage into a current source. Q10 functions as a diode should your control voltage go negative.

Q6 and Q7 form a simple current mirror to reflect Q9's current from the emitters of Q1 and Q2.

Q3, Q4, and Q5 form a Wilson current mirror - it has much better accuracy than the simple current mirror. It reflects the current from Q1 and sums it with the opposing current of Q2. With both transistors pulling equal current, the net current available at the collector of Q2 should be about zero.

The collector of Q2 is held at 1/2Vcc by op amp, U1. The difference current between Q1 and Q2 acts on R4 to generate the output voltage.

Since the inputs of U1 operate at 1/2Vcc, R6 offsets the output of U1 to a nominal 0 volts.

As for the gain calculation, I guess it should be R4 / (Re1 + Re2), where Re = 26 / Ie. In actuality, I get lazy these days, and part of that was just seeing what would happen in Spice. So, I never bothered to work through the calculation. I just plugged...

Anyway, I hope this works out for your friend. He'll definitely want to match Q1 and Q2 closely and perhaps glue them together.

- Mike