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D2004
Joined: 14 Mar 2004
Posts: 21
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 14 Mar 2004 16:08 What is the function of this circuit? |
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Hi,
What is the function of the attached circuit? Anybody can give a clue or reference on that? No transistor ratio or size given. Thanks a lot.
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flatulent
Joined: 19 Jul 2002
Posts: 4875
Helped: 324
Location: Middle Earth
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14 Mar 2004 20:04 Re: What is the function of this circuit? |
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| Which nodes are connected to the outside world? It looks like the two outer bottom transistors have equal drain currents but with different Vgs values.
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D2004
Joined: 14 Mar 2004
Posts: 21
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15 Mar 2004 4:29 Re: What is the function of this circuit? |
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Thanks for your reply.
I got the circuit diagram just like this. However, we may suppose the output node is the M5's gate/source. So, it looks like a current source. And I think the key is the resistor R1...
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15 Mar 2004 4:29 Ads |
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electronrancher
Joined: 24 Mar 2002
Posts: 505
Helped: 34
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| 15 Mar 2004 23:39 |
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it appears to be a crude cmos bandgap of the deltaVgs style. could also be used as a ptat current source with a lot less trouble.
m7 / m8 force equal currents into the two inner branches. m1 m2 act as diodes, with m2 sized to some multiple of m1 so it's vdsat is smaller - the voltage drop is made up by the resistor.
lets start from power-up, where we can assume the gates of m7/m8 are at 0, flooding the inner branches with current.
at first, the gate of m4 is higher than m3's gate so m4 pulls more current out of m5. this is mirrored out of m6 to pull the gates of m7 / m8 high and reduce the current. if the current is too low the opposite happens - m3 pulls down, allowing more current into m7/m8.
all in all, it looks pretty interesting but unless you use W/L on the order of 500/5 the matching will be bad. other problems are:
-bad power supply rejection
-tempco large compared to using regular diodes
-very affected by process variation
but for a homework question it's a good one. hope you get an A!
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D2004
Joined: 14 Mar 2004
Posts: 21
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26 Mar 2004 18:03 Re: What is the function of this circuit? |
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Hi electronrancher,
Thanks for your great analysis. I cannot agree with you more on this circuit.
Could you give a further explanation on the another circuit that I post in the forum Analog IC Design & Layout?
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xiongshoufen
Joined: 27 May 2004
Posts: 67
Location: shanghai in china
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| 23 Nov 2006 2:51 |
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| electronrancher wrote: |
it appears to be a crude cmos bandgap of the deltaVgs style. could also be used as a ptat current source with a lot less trouble.
m7 / m8 force equal currents into the two inner branches. m1 m2 act as diodes, with m2 sized to some multiple of m1 so it's vdsat is smaller - the voltage drop is made up by the resistor.
lets start from power-up, where we can assume the gates of m7/m8 are at 0, flooding the inner branches with current.
at first, the gate of m4 is higher than m3's gate so m4 pulls more current out of m5. this is mirrored out of m6 to pull the gates of m7 / m8 high and reduce the current. if the current is too low the opposite happens - m3 pulls down, allowing more current into m7/m8.
all in all, it looks pretty interesting but unless you use W/L on the order of 500/5 the matching will be bad. other problems are:
-bad power supply rejection
-tempco large compared to using regular diodes
-very affected by process variation
but for a homework question it's a good one. hope you get an A! |
m7 and m3 form positive feedback?
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ramho100
Joined: 16 Nov 2006
Posts: 5
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24 Nov 2006 15:40 What is the function of this circuit? |
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Hi,
could anybody help me by providing a ckt to sense any liquid(may be non conductive) level using change in resistance value(thru a float)
Thanks
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hackjiang
Joined: 14 Nov 2005
Posts: 75
Helped: 1
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| 25 Nov 2006 3:36 |
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| xiongshoufen wrote: |
| electronrancher wrote: |
it appears to be a crude cmos bandgap of the deltaVgs style. could also be used as a ptat current source with a lot less trouble.
m7 / m8 force equal currents into the two inner branches. m1 m2 act as diodes, with m2 sized to some multiple of m1 so it's vdsat is smaller - the voltage drop is made up by the resistor.
lets start from power-up, where we can assume the gates of m7/m8 are at 0, flooding the inner branches with current.
at first, the gate of m4 is higher than m3's gate so m4 pulls more current out of m5. this is mirrored out of m6 to pull the gates of m7 / m8 high and reduce the current. if the current is too low the opposite happens - m3 pulls down, allowing more current into m7/m8.
all in all, it looks pretty interesting but unless you use W/L on the order of 500/5 the matching will be bad. other problems are:
-bad power supply rejection
-tempco large compared to using regular diodes
-very affected by process variation
but for a homework question it's a good one. hope you get an A! |
m7 and m3 form positive feedback? |
Yes, but the neg feedback formed by m4,m6 and m8 is stronger.
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