This is my circuit in the file
This designed i have using ir2011 to work well.
I try to using ir2110 to work..But
I got one problem , when i remove the feedback resistor R39=47k
The output waveform is 1Vpp sine wave.
But got some DC offset about 500mV.
The waveform after feedback Low-pass filter , It looks like triangle-wave , 40Vpp.
When i reconnect the feedback resistor R39 .
The output of error amp , become 5v output.
Let the output of LT1016 going to saturated.
I connect the -40v to ir2110's Vcc , serise a R33=10ohm.
I saw some designe connect the 10ohm between -40v and Vdd not Vcc.
which way is correct ??
When i remove the feedback R , R33 , the ir2110 burn out easily.
Always see the R33 burned.
I have fix the circuit for 3 days.
Does anyone has any suggestion.???
This designed i have using ir2011 to work well.
I try to using ir2110 to work..But
I got one problem , when i remove the feedback resistor R39=47k
The output waveform is 1Vpp sine wave.
But got some DC offset about 500mV.
The waveform after feedback Low-pass filter , It looks like triangle-wave , 40Vpp.
When i reconnect the feedback resistor R39 .
The output of error amp , become 5v output.
Let the output of LT1016 going to saturated.
I connect the -40v to ir2110's Vcc , serise a R33=10ohm.
I saw some designe connect the 10ohm between -40v and Vdd not Vcc.
which way is correct ??
When i remove the feedback R , R33 , the ir2110 burn out easily.
Always see the R33 burned.
I have fix the circuit for 3 days.
Does anyone has any suggestion.???
Hi,
Try this:
Give +5V to Vdd.
For Vcc let's give -40V with a 10ohm resistor. Let's put a 1uF folcap to the ground from Vcc.
Regards,
Try this:
Give +5V to Vdd.
For Vcc let's give -40V with a 10ohm resistor. Let's put a 1uF folcap to the ground from Vcc.
Regards,
The power supply of your schematic is drawn impossibly wrong and the labels don't match. Please check this carefully and repost your actual circuit if you are serious about asking for help.
Regards -- analogspiceman
Regards -- analogspiceman
Does R33 really have to be there ? The supply voltage of such a driver should be as "stiff" as possible IMO in order to work reliably. Or is there a cap between Vcc and COM in the App note ?
Regards
Charles
Edit: What is your switching frequency BTW ?
Regards
Charles
Edit: What is your switching frequency BTW ?
Thanks for all...
1. I see the datasheet of ir2110 , said , -0.3 < Vdd < Vss+25
If i connect Vdd to +5V , It will higher than Vss+25.
Is that right?
2. I'm sorry , I lost to modify the output voltage label.
I have modify it .
3. could you tell me , what is IMO ??
No , i didn't have cap between Vcc and Com.
Is that effect my whole circuit wrong ??
I will add it and try again ...
The sw freq is 200kHz , but now i search for the LF357.
now , i using tl084 to generate , but the waveform not good.
4. I saw a circuit created by Pierre , he use a IR2113.
He connect the 12vbias direct to Vcc , and serise a 10ohm to connect to Vdd.
which way is right??
1. I see the datasheet of ir2110 , said , -0.3 < Vdd < Vss+25
If i connect Vdd to +5V , It will higher than Vss+25.
Is that right?
2. I'm sorry , I lost to modify the output voltage label.
I have modify it .
3. could you tell me , what is IMO ??
No , i didn't have cap between Vcc and Com.
Is that effect my whole circuit wrong ??
I will add it and try again ...
The sw freq is 200kHz , but now i search for the LF357.
now , i using tl084 to generate , but the waveform not good.
4. I saw a circuit created by Pierre , he use a IR2113.
He connect the 12vbias direct to Vcc , and serise a 10ohm to connect to Vdd.
which way is right??
= In My Opinon
Apart from the things already mentioned - your C28 is a little large with 1n if your switching frequency is really 200 kHz. This will give you a pole in the feedback loop of 160 kHz approx.
Regards
Charles
160KHz too high ???
doesn't the purpose of the feedback low-pass filter is to reduce the high frequency noise ???
so , we have to set the limit freq slightly lower than s.w freq.
am i getting something wrong ???
doesn't the purpose of the feedback low-pass filter is to reduce the high frequency noise ???
so , we have to set the limit freq slightly lower than s.w freq.
am i getting something wrong ???
No - toooooooooo low !
Theoretically the feedback signal should be as accurate as possible. i.e. not filtered at all with this carrier-based topology. I.e. it's bandwidth shouldn't be restricted.
But there will always be overshoot etc from the switching stage that the feedback loop can't cope with and these should be filtered out. So one uses a first-order lowpass as a compromise. But this one shouldn't have a too low cutoff frequency.
Regards
Charles
Titan,
Upon first glance I see a missing capacitor, you need a ~1uF cap between Vcc and Com, and the bootstrap capacitor should be between 0.1 ~ 0.33uF, yours is too large. No need for a 4.7-ohm resistor in series with the diode either, just omit that. You mosfet gate resistors are very different, the low side being 10x the value of the high side, this will not work well.
I suggest you study the IR2110 and IR2011 datasheets a bit closer and take a look at the IRDAMP reference design, that should give you some pointers.
Best regards,
Sander Sassen
http://www.hardwareanalysis.com
Upon first glance I see a missing capacitor, you need a ~1uF cap between Vcc and Com, and the bootstrap capacitor should be between 0.1 ~ 0.33uF, yours is too large. No need for a 4.7-ohm resistor in series with the diode either, just omit that. You mosfet gate resistors are very different, the low side being 10x the value of the high side, this will not work well.
I suggest you study the IR2110 and IR2011 datasheets a bit closer and take a look at the IRDAMP reference design, that should give you some pointers.
Best regards,
Sander Sassen
http://www.hardwareanalysis.com
1. My s.w freq= 200KHz , the cutoff-freq = 160KHz
If it too low , how many is suitable for me ??
2. 1uF bootstartp cap is too large ???
so , i only need a 100n for it ???
If i remove 4.7ohm , did i have to remove the 10ohm either ??
3. yes , i have read the IRDAMP reference yet .
and i also read many diyaudio member's designe.
It have many different bootstrape cap using.
So , i 'm not really sure what value is correct .
4. I'm sorry , my circuit's gate resistor get some wrong.
I forget to correct the circuit gate value.
If it too low , how many is suitable for me ??
2. 1uF bootstartp cap is too large ???
so , i only need a 100n for it ???
If i remove 4.7ohm , did i have to remove the 10ohm either ??
3. yes , i have read the IRDAMP reference yet .
and i also read many diyaudio member's designe.
It have many different bootstrape cap using.
So , i 'm not really sure what value is correct .
4. I'm sorry , my circuit's gate resistor get some wrong.
I forget to correct the circuit gate value.
I would increase the filter-cutoff by a factor of at least five. It is better to have a cleanly switching output stage from the beginning anyway so you'd have to filter less.
Regards
Charles
titanchen68 said:
The number of careless mistakes in your schematic doesn't inspire one to believe that you care very much about your design or that you will build the hardware any more carefully than you drew your schematic. Back to work, please.
The labels were only a cosmetic problem. Take a close look at the diode bridge - the ac and dc pins are swapped. Also, do you realize that you have shorted a section of the transformer's output windings (grounded in two places)?
You cannot really expect people to waste their time advising you on an idea so poorly presented.
Just to let you know that 100 ohms at gate is too much. If you need dead time implement it at the input of the IR2110 with resistor diode and cap. reduce to 22 ohms each mosfet and adjust your dead time at the input to have minimal current flow in the PSU at idle...Next, move your feedback at the output of the coil and remove low pass filter in the feedback. A small capacitor will be needed to get stability of you feedback, like in analogue amplifier. Add snubber at the speaker output to avoid instability. Why did you use 4 mosfets? 2 should be enought for +/- 55V....That's low power! Try too to drive the triangle wave directly with no buffer, just resistor pad. Dont forget that lower will be the triangle wave, better will be the open loop gain and the performance in closed loop!
Simple design, but work! Look at my web site!
Bye
Fredos
www.d-amp.com
Simple design, but work! Look at my web site!
Bye
Fredos
www.d-amp.com
analogspiceman said:
Thanks for your replying
1.The circuit i have build by using IR2011 long time ago.
And the resistor lable i got mistake for it before.
I modify two of them , and forget others.
I'm care about my design.
2. I connect the one of the transformer output to the -50v.
That because i want to get a 20vDC dependent on -50v.
so , i connect the ac20v's gnd to -50vdc.
3. maybe the circuit in power area , i didn't present very clear.
and i 'm sorry about it.
fredos said:
1. 100ohm too high?? I ever read a thread by a diyaudio member.
He said 33ohm too low , and recommand 100ohm is suitable.
And he said he always simulate by 100ohm gate resistor in LT spice.
But thans , i will try 22ohm also.
2. If i connect the feedback from output of coil.
the feedback network need a phase compensate network, right??
But i don't know yet how to design it....
3. is it your webside " www.d-amp.com "??
but Inside it does not have any circuit ...
now , i remove the R35=4.7ohm , and change the c25 to 0.47u.
And add a 1u/16 cap between Vcc and Com.
It get work. But I don't know why???
Is that 4.7ohm to slow down the charge current ??
In class-d circuit , always have a error-amp.
I get a question , the error-amp sum of the input and feedback square-wave.
Why do this way can reduce the noise and dc-offset ???
Please forgive my poor electronic concept.
In class-AB , negtive feedback will help for stability.
But the input and output has one constand gain to calculate.
But in class-d , i know it is a nagtive feedback
But the output and input seems didn't look like the same , if we feedback in front of the LC filter.
And add a 1u/16 cap between Vcc and Com.
It get work. But I don't know why???
Is that 4.7ohm to slow down the charge current ??
In class-d circuit , always have a error-amp.
I get a question , the error-amp sum of the input and feedback square-wave.
Why do this way can reduce the noise and dc-offset ???
Please forgive my poor electronic concept.
In class-AB , negtive feedback will help for stability.
But the input and output has one constand gain to calculate.
But in class-d , i know it is a nagtive feedback
But the output and input seems didn't look like the same , if we feedback in front of the LC filter.
I faced similar issues when building my Class-D amp.
Have a look at an interesting thread here at DIYAudio "help with feedback" or something like that.
Basically, Charles, whose help is great, helped me getting the feedback network right.
Yes, you need a C in parallel with the feedback resistor to add a zero for compensation.
However, my experience is that with some loads there is ringing at the output, so it is not too estable at this moment.
Charles, when you talk about a PD in the integrator, you mean having a high-valued resistor from - input to output of the opamp, in parallel with a R + C in series, is that what you mean?
A question more, that's for Fredos: if your triangle is small, your loop gain is higher, that's clear. The advantage of that is decreased distortion, theoretically, but that is worst for estability, isn't it?
Let's see if we all together can design a very estable global feedback loop.
Have a look at an interesting thread here at DIYAudio "help with feedback" or something like that.
Basically, Charles, whose help is great, helped me getting the feedback network right.
Yes, you need a C in parallel with the feedback resistor to add a zero for compensation.
However, my experience is that with some loads there is ringing at the output, so it is not too estable at this moment.
Charles, when you talk about a PD in the integrator, you mean having a high-valued resistor from - input to output of the opamp, in parallel with a R + C in series, is that what you mean?
A question more, that's for Fredos: if your triangle is small, your loop gain is higher, that's clear. The advantage of that is decreased distortion, theoretically, but that is worst for estability, isn't it?
Let's see if we all together can design a very estable global feedback loop.
Ohhh ! I just saw that I wrote BS. It should read PI. Which means Proportional-Integral.
This is the circuit with the RC series circuit in the feedback path of the OP-AMP. The large parallel resistor is there to 1.) prevent DC drift and 2.) reduce loop gain at low frequencies. The latter can be used to avoid the rising THD with increasing frequency at the cost of higher overall THD.
Regards
Charles
I did some testings precisely about this last monday, I found that the ringing was much smaller as I increased the switching frequency from 260K to around 340KHz.
My feedback values are: Feedback res: 100K with 33pF in parallel.
Error opamp: 180K in parallel with (6.8K in series with 330pF).
Input resistor to error opamp: 2k2
So DC gain is around 100K/2K2=45V/V
Why that apparent estability improvement by increasing the sw. frequency? Is something wrong in the values?
My feedback values are: Feedback res: 100K with 33pF in parallel.
Error opamp: 180K in parallel with (6.8K in series with 330pF).
Input resistor to error opamp: 2k2
So DC gain is around 100K/2K2=45V/V
Why that apparent estability improvement by increasing the sw. frequency? Is something wrong in the values?
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