MrHeckles Posted February 13, 2006 Report Share Posted February 13, 2006 Hi guys,This is part of a motor controller circuit. It is designed to function as an inhibitor from switching directions if the motor is still moving. It works MOSTLY as intended, ie if the motor is spinning forwards (even with no speed demand) and then is flicked into reverse, the circuit will sense the current in the motor, and inhibit the reverse PWM signal. Hence braking the motor, until such time as the voltage into "V in 1" or "V in 2" is less than that set by the 5k pot. It then allows the motor to change directions.Here is the problem. The negative voltage rail is sagging below the designed -9v (by sagging below, I actually mean it is higher). AND there is a difference in this drop between the forwards and reverse or the motor. It results in the voltage preset (from the 5k pot) being different for forwards and reverse. Now, I figure that this drop is due to there being too much load for the ICL7660 to handle, thus reducing it's negative voltage output. My question is this......How can I keep the negative voltage rail at a steady -9v and stop there being variations between the forwards and reverse.Note: Audioguru is probably very familiar with this cursed circuit by now :-) Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 13, 2006 Report Share Posted February 13, 2006 Hi Mr. Heckles,Your top two 4558 ICs are straining with a high current into their low-value negative feedback resistors. The output resistance of the ICL7660 is about 45 ohms, so 20mA into the 4558s creates a 0.9V drop.An application note for the ICL7660 shows a voltage regulator with one. I would use an LM10 which has a voltage reference and its opamp is almost rail-to-rail. If the LM10 is fed with 12V, thenit can pre-regulate the ICL7660. Quote Link to comment Share on other sites More sharing options...
MrHeckles Posted February 13, 2006 Author Report Share Posted February 13, 2006 The LM10 is a little hard to get over here. Is there something else I can use?What part does the voltage reference play in that circuit? Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 13, 2006 Report Share Posted February 13, 2006 Hi Mr. Freckles, err, Mr. Heckles, ;DI don't know which ICs you can buy. I can get nearly anything.The voltage reference is the opamp's reference that it compares the output voltage to. When the output voltage begins to drop then the opamp increases the input voltage to the ICL7660 so its negative output voltage is regulated. Quote Link to comment Share on other sites More sharing options...
MrHeckles Posted February 14, 2006 Author Report Share Posted February 14, 2006 So it doesnt matter what value the voltage reference has, as long as the op-amp knows what it is?The LM10 is an extended range item at RS down here, and is quite expensive. Can I get away with a cheap voltage reference chip (like a TL431C) and a low power op amp?Are there any special properties that the op amp needs?The cheapest low power op amp I found is the TLC271ACP. Would this do? Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 14, 2006 Report Share Posted February 14, 2006 So it doesnt matter what value the voltage reference has, as long as the op-amp knows what it is?Correct. Then you scale the -9V to match it.The LM10 is an extended range item at RS down here, and is quite expensive. Can I get away with a cheap voltage reference chip (like a TL431C) and a low power op amp?An LM10CN costs $2.70US, about a cup of coffee in a restaurant. It does everything you need in an 8-pin DIL package.Are there any special properties that the op amp needs?The cheapest low power op amp I found is the TLC271ACP. Would this do?No. The Cmos opamp's output current is way too low. With a 20mA load, its loss is 5V.The opamp's output powers the ICL7660. Your 4558 opamps are driving a fairly high current into their very low value negative feedback resistors so have a high supply current, maybe totalling 20mA.The LM10 has a voltage reference inside, a low operating current and an output voltage loss of only 0.4V with a 20mA load. Quote Link to comment Share on other sites More sharing options...
MrHeckles Posted February 14, 2006 Author Report Share Posted February 14, 2006 An LM10CN may cost you a cup of coffee but down here RS only sells the LM10CWM (and I cant solder SMT yet), and Farnell sells the LM10CN for $27.33 plus shipping, which is about equal to 10 cups of coffee.I would love to have a nice small all contained chip, but ya gotta work with what youve got.What sort of op-amp am I going to have to look for if I cant get an LM10CN? Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 14, 2006 Report Share Posted February 14, 2006 How can they rip you off so much and why do people there let them?If you have a friend in the US who could mail one to you, the postage wouldn't cost very much and delivery wouldn't be long.If you reduce the current of the 4558 opamps by increasing the values of their feedback resistors and using low power opamps, maybe the current will be low enough for any opamp to drive the ICL7660. Quote Link to comment Share on other sites More sharing options...
MrHeckles Posted February 14, 2006 Author Report Share Posted February 14, 2006 We are a small country miles from anywhere. They ship in what sells. What is not common is generally expensive.I do my part in refusing to buy things that are too expensive.The only person I know in the states is my brothers ex-fiance. He called the wedding off a month or so before it was going to happen and broke up with her. I guess I cant really ask her to send me electronic stuff ;DThe upper 4558s are absolute value amps, and they need to keep the same ratio between the resistors. If I do this........Will that reduce the current draw of them?The other option I though of was using a power op amp. Would this work for driving the 7660? The TCA0372DP1 is nice a cheap and has a very low voltage drop vs load current? Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 14, 2006 Report Share Posted February 14, 2006 Hi Mr. Heckles,The absolute value amps will operate the same but now their supply current will be lower.The TCA0372DP1 is a nice one and will work fine with a voltage reference which could even be a zener diode if you want. Quote Link to comment Share on other sites More sharing options...
MrHeckles Posted February 16, 2006 Author Report Share Posted February 16, 2006 Please could you explain to me how this setup works. I'm not sure I understand how you get it to hold -5v. By that I dont understand how the feedback works in this case.Do the + and - inputs both have 1.2v (in the case of having the indicated reference)?How does changing the 910k resistor change the output voltage of the 7660? Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 16, 2006 Report Share Posted February 16, 2006 Hi Mr. Freckles. hee, hee. ;DI checked and double checked and think they made a mistake. The 910k resistor's value must be much higher for a -5V output and much higher again for -9V out. The 100k resistor connected to it could have its value reduced instead.Calculate the resistors using Ohm's Law since opamps hardly have any input current. Watch out for having a voltage at the inputs of the opamp that exceeds its input common-mode voltage rating, The input common-mode rating for most opamps do not allow the voltage to be too close to its positive or negative supply voltage. Quote Link to comment Share on other sites More sharing options...
MrHeckles Posted February 17, 2006 Author Report Share Posted February 17, 2006 How is the value of the 56k ohm resistor selected? Is it that value for a reason or arbitrary?Crunch some numbers (possibly in a completely incorrect way) I figure that if I use a 3.3v zener diode for the reference, then I would need to replace the 100k ohm resistor with a 50k ohm, and the 910k ohm with a 695k ohm, in order to give an output regulation of about -9v.Does this sound correct? Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 17, 2006 Report Share Posted February 17, 2006 How is the value of the 56k ohm resistor selected?It provides current for the shunt voltage reference. Is it that value for a reason or arbitrary?The 56k resistor was calculated from refering to the datasheet of the voltage reference.I figure that if I use a 3.3v zener diode for the reference, then I would need to replace the 100k ohm resistor with a 50k ohm, and the 910k ohm with a 695k ohm, in order to give an output regulation of about -9v.Perfect. However a 3.3V zener diode is a lousy voltage reference. Its voltage changes with its input current and load current. Its voltage changes with temperature change.Therfore I recommend using a 5.6V zener diode. A BZX79C5V6 operates well at only 5mA. Then the divider ratio must be re-calculated and a 1.2k resistor from 12V will supply 5.3mA to the zener diode.Does this sound correct? Quote Link to comment Share on other sites More sharing options...
MrHeckles Posted March 2, 2006 Author Report Share Posted March 2, 2006 I have built it, and it works fine. Seems the datasheet did have an error regarding that value of the feedback resistor. I ended up using a 5.6v zener diode and 400k resistor with a 100k variable resistor for trimming purposes (as the feedback resistor).The caps are 100uF on the 7660 chip, and 10uF on the op-amp.The regulator works well. It reacts fairly slowly to load changes, but on the whole gives a nice steady -9v.Thanks for all your help audiogooru ;D Quote Link to comment Share on other sites More sharing options...
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