Hi all,
I would like to dc couple the gain stage and output stage in the schematic below. The textbook that I am using in school does not cover E mosfets very well and I'm having some problem figuring out what Vb is for Q2. Vg for Q1 is 6.94v. In the book that I have it talks about finding "k" for the mosfet. After looking on the spec sheet and using the equation k = Id(on)/(Vgs(on)-Vgs(th) with the variables being Id(on) = 200mA at Vgs = 10v and Vgs(th) = 2.1. I came up with the answer k = .003x 10(-4) mA/V(squared). Please understand that these calculations are a bit beyond where we are in class. Basically I'm looking to bias the output of Q2 at 2/3 Vcc and direct couple it to Q5. Any help would be appreciated. Also I'm looking for opinions on an idea that I have. This is a headphone amp and rather than use a large output capacitor for impedance matching I was thinking of using This transformer:
http://www.sowter.co.uk/specs/8665.htm
I realize that I will have to AC couple the transformer but with a load of 32 ohms using a 12:1 voltage ratio the impedance of the primary comes out to about 4600 which means the coupling cap only need to be: C = 1,000,000/2pi x 10 hertz x 4604 ohms) = 3.45 uF. I think that 5uF will work pretty well. I think that that would beat using a 680uF cap on the output. Opinions?
I would like to dc couple the gain stage and output stage in the schematic below. The textbook that I am using in school does not cover E mosfets very well and I'm having some problem figuring out what Vb is for Q2. Vg for Q1 is 6.94v. In the book that I have it talks about finding "k" for the mosfet. After looking on the spec sheet and using the equation k = Id(on)/(Vgs(on)-Vgs(th) with the variables being Id(on) = 200mA at Vgs = 10v and Vgs(th) = 2.1. I came up with the answer k = .003x 10(-4) mA/V(squared). Please understand that these calculations are a bit beyond where we are in class. Basically I'm looking to bias the output of Q2 at 2/3 Vcc and direct couple it to Q5. Any help would be appreciated. Also I'm looking for opinions on an idea that I have. This is a headphone amp and rather than use a large output capacitor for impedance matching I was thinking of using This transformer:
http://www.sowter.co.uk/specs/8665.htm
I realize that I will have to AC couple the transformer but with a load of 32 ohms using a 12:1 voltage ratio the impedance of the primary comes out to about 4600 which means the coupling cap only need to be: C = 1,000,000/2pi x 10 hertz x 4604 ohms) = 3.45 uF. I think that 5uF will work pretty well. I think that that would beat using a 680uF cap on the output. Opinions?
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Hi!
I have checked the schematics and I believe that you will have approx. 15 V DC at Q2 collector. Try to simple remove C3, R8 and R9 and connect R10 directly to Q2 collector.
Why transformer? You can drive any type of headphone including high impdedance ones. Gain of 11 and 30 volts of power supply voltage = very loud!
I have checked the schematics and I believe that you will have approx. 15 V DC at Q2 collector. Try to simple remove C3, R8 and R9 and connect R10 directly to Q2 collector.
Why transformer? You can drive any type of headphone including high impdedance ones. Gain of 11 and 30 volts of power supply voltage = very loud!
if he removes C3 R8 and R9 , whats to stop a loud input signal turning off TR2 and sending +30V to the gate of the FET ?
- which would likely complain (pop...)
ray
- which would likely complain (pop...)
ray
..ive just checked the spec's for the FET , I know the answer to my previous question.......
just before you jump on me!
I had a figure of 20V for Vdg in my head for some reason.
it's one of them days.....
that'll teach me to post before I check😎
ray
just before you jump on me!
I had a figure of 20V for Vdg in my head for some reason.
it's one of them days.....
that'll teach me to post before I check😎
ray
So if I change R5 to 10K and remove C3(should be C2) and R8 I should get 20v to the gate resistor of Q5. I don't think that I should remove R9. The gate of Q5 should be referenced to signal ground shouldn't it?
G, if you don't believe me, measure the DC voltage at the collector of Q2 and also the gate voltage at Q5. The voltages are between ground and the points mentioned. If the DC volatges are close, you can omit the coupling capacitor without any harm.
Beware of the clipping characteristics, can be nasty, probely not very symmetrical. You can't remove the input and output coupling caps.
Noticed also the AC feedback. The amps needs to roll off in the bass in order to get flat frequency response so it's can be tricky to change component values without calculations.
Beware of the clipping characteristics, can be nasty, probely not very symmetrical. You can't remove the input and output coupling caps.
Noticed also the AC feedback. The amps needs to roll off in the bass in order to get flat frequency response so it's can be tricky to change component values without calculations.
It's not a matter of not believing you Peranders. Not at all. I'm just used to seeing a resistor from signal to ground at the input of a device. I got the 20 volts here:
http://headwize.com/projects/showproj.php?file=szeke1_add_prj.htm
It's recommended in the article to bias the input at 2/3 Vcc. But hey. 15 volts works for me too.
http://headwize.com/projects/showproj.php?file=szeke1_add_prj.htm
It's recommended in the article to bias the input at 2/3 Vcc. But hey. 15 volts works for me too.
The amp you are refering to uses 15 volts so 2/3 of Vcc is OK. The design trick is to set the DC level so you will get symmetrical clipping. MOSFET's has 3-4 V Vgs gives you 7.5+4 = 10.5 V in this example.
When you have AC coupled stages you can choose any DC level you want. When you remove these caps you must do some sort of compromise. High supply voltage is a good thing here and 30 volts makes it a lot easier.
When you have AC coupled stages you can choose any DC level you want. When you remove these caps you must do some sort of compromise. High supply voltage is a good thing here and 30 volts makes it a lot easier.
Hi Peranders. Look I know this sounds dumb to you but what do you mean symetrical clipping? There is only one output device. Do you mean symetrical clipping between Q2 and Q5? I'm not arguing. I just want to understand what you are talking about. Checked out your site by the way. Good stuff. If I ever get past this project I would like to inquire about purchasing some PCB's from you.
Hi G
Symetrical clipping refers to the positive and negitive swings of a sinewave entering clipping at the same level. ( +- 10 volts peak from 0 signal as an example)
later
Bruce
Symetrical clipping refers to the positive and negitive swings of a sinewave entering clipping at the same level. ( +- 10 volts peak from 0 signal as an example)
later
Bruce
Ok. That I understand. Thanks. What are the variables in DC coupling that would cause asymetrical clipping in between stages?
I've looked for info on DC coupling audio stages but the info I found was rather vague.
I've looked for info on DC coupling audio stages but the info I found was rather vague.
If you want symmetrical clipping you must set the DC levels "just in the middle". If the first stage (just an example) swings between 5 and 23 volt you should thrive to get the DC level to ((23-5)/2) +5 = 14
The voltage can swing between +-9 Volts with a center around 14 volts and will be symmetrical, equal voltage around the "zero" level, 9 volts.
Did you get the symmetrical thing?
It can be hard to get all things right DC wise when the design is unsymmetrical. I myself have never designed nothing else than symmetrical amps! I like those!
The voltage can swing between +-9 Volts with a center around 14 volts and will be symmetrical, equal voltage around the "zero" level, 9 volts.
Did you get the symmetrical thing?
It can be hard to get all things right DC wise when the design is unsymmetrical. I myself have never designed nothing else than symmetrical amps! I like those!
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