|
|
|||||||
| Home | Forums | Rules | Articles | Store | Gallery | Blogs | Register | Donations | FAQ | Calendar | Search | Today's Posts | Mark Forums Read | Search |
| Solid State Talk all about solid state amplification. |
|
Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving |
|
|
|
Thread Tools | Search this Thread |
|
|
#1 |
|
diyAudio Member
Join Date: Feb 2010
|
Hi All,
Reading this forum and great DSelf book I decided to match the input stage transistors in my car amp. It has pretty standard schematics: double (symmetrical) diff. amp, resistive tail current; no current mirrors, just a resistor in one leg. Loaded by symmetrical VAS. Each half contains only one transistor and degeneration resistor in emitter. As I can see the common agreement here is to match transistors "in circuit", but matching 4 transistors at the same time may be difficult? While we can easily monitor/compare Vbe, the Hfe might be "masked out" by the fact that current flew directly between the bases. Ideally I would like to see voltage offset between bases 2mV or less. It will allow me to get rid of NFB capacitor (I decrease amp. gain to 10). Also, matching legs currents would be good addition. But I'm not sure if this possible for this schematics. All impedance related DC offset "eliminated" by matching emitter degeneration resistors and impedances seen by bases. Got 25/25 of 2SA970/2SC2240 waiting to be installed... Thanks for any ideas! |
|
|
|
|
#2 |
|
diyAudio Member
Join Date: Dec 2005
Location: the north
|
Is it 12 Volt supply?
Good you have 25 pieces of transistors. You should find a good matched pair of SC2240. And a good matched pair of SA970. Then use them. To try and find 4 matched of two NPN and two PNP will not happen. NPN and PNP have different HFE and different VBE. This means that there will be more base current for one pair and less for the other. This creates what is called one Input Offset Current and this current creates a voltage drop across input resistance and this gives Offset Voltage which will be multplied with gain of amp. It might be that you therefor must keep the NFB capacitor. To make the Input Offset Current very small it is not unusual to run input transistors at low current, like 0.150 mA. This of course means only a few uA needs to bias via bases. JFET input stages does not have this problem. Because they need almost no bias current at all.
__________________
lineup |
|
|
|
|
#3 | |||
|
diyAudio Member
Join Date: Feb 2010
|
No, +-30V.
Quote:
Quote:
Quote:
|
|||
|
|
| Currently Active Users Viewing This Thread: 1 (0 members and 1 guests) | |
| Thread Tools | Search this Thread |
|
|
|
|
||||
| Thread | Thread Starter | Forum | Replies | Last Post |
| Double transistors in the VAS Why? | GEirin | Solid State | 7 | 25th February 2010 09:08 PM |
| Counterpoint - A Fully Symmetrical I/V stage | Russ White | Digital Source | 22 | 26th October 2008 01:36 PM |
| Symmetrical input stage | hjelm | Solid State | 19 | 22nd April 2004 12:37 PM |
| Matching Transistors | Vivek | Solid State | 4 | 3rd February 2004 04:38 AM |
| jFET input ,full symmetrical class A circuit | promitheus | Solid State | 28 | 20th November 2003 06:12 AM |
| New To Site? | Need Help? |
| Page generated in 0.08787 seconds (76.49% PHP - 23.51% MySQL) with 10 queries |