Need help understanding differential input stage

Bitrex · 14th June 2009, 12:13 AM

I have a question regarding the design of differential input stages - the schematic I'm looking at is here from the "solid state wiki":
http://sound.westhost.com/project3a.htm

The question I have from the diagram is that if I'm understanding the circuit correctly, the base of Q4 is sitting about a diode drop below Vcc or 35 volts. Now when the input to amplifier is positive going, the output is negative going and the voltage at the base of Q4 turns on the transistor more, causing the output at Q4's collector to rise.

However, when the input to the amplifier is negative going, it looks to me like the voltage at the base of Q4 will rise from Vcc minus a diode drop and cut off the transistor! So how does this input stage (which seems a fairly common circuit) generate the "other half" of the output?

sawreyrw · 14th June 2009, 12:50 AM

It doesn't matter what "polarity" the input voltage is, because the closed loop circuit causes the voltage at the base of Q2 to be whatever it needs to be to maintain the output at the correct level. In other words, Q1 and Q4 will still conduct as needed to achieve the correct output voltage.

Mr Evil · 14th June 2009, 01:20 AM

What you're missing is that Q4 has a very high voltage gain, so it only takes a tiny change in its base voltage to get to the desired output voltage, nowhere near enough to turn it off.

Bitrex · 14th June 2009, 01:42 AM

I think I see now. There's feedback at DC into the inverting input of the differential pair Q2, so that DC voltage is coupled into the input transistor through the common connection. So when the input voltage rises from the source signal the output voltage rise is applied as a "bias" voltage to Q1, and the bias voltage follows the input voltage along keeping the transistors in conduction. I think I have that part down now.

What ensures that the output stays biased at 0 volts though, and not some other value? If there's a good reference online that someone can throw me that describes the operation of this topology in detail, I'd very much appreciate it.

Rafael L · 14th June 2009, 02:26 AM

Look

http://ifile.it/dk3ytpv

Bitrex · 14th June 2009, 03:51 AM

Thanks Rafael, that's just the kind of reference I've been searching for.

14th June 2009, 12:13 AM	#1
Bitrex diyAudio Member Join Date: Jan 2009	Need help understanding differential input stage I have a question regarding the design of differential input stages - the schematic I'm looking at is here from the "solid state wiki": http://sound.westhost.com/project3a.htm The question I have from the diagram is that if I'm understanding the circuit correctly, the base of Q4 is sitting about a diode drop below Vcc or 35 volts. Now when the input to amplifier is positive going, the output is negative going and the voltage at the base of Q4 turns on the transistor more, causing the output at Q4's collector to rise. However, when the input to the amplifier is negative going, it looks to me like the voltage at the base of Q4 will rise from Vcc minus a diode drop and cut off the transistor! So how does this input stage (which seems a fairly common circuit) generate the "other half" of the output?

14th June 2009, 01:20 AM	#3
Mr Evil diyAudio Member Join Date: Aug 2004 Location: Behind you	What you're missing is that Q4 has a very high voltage gain, so it only takes a tiny change in its base voltage to get to the desired output voltage, nowhere near enough to turn it off. __________________ https://mrevil.asvachin.eu/

Thread Tools	Search this Thread
Show Printable Version Email this Page	Search this Thread: Advanced Search

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Differential I/V stage? And RX input	guzzler	Digital Source	29	19th December 2006 05:32 PM
Adding a third input to a differential input stage?	maudio	Solid State	15	11th October 2006 02:16 AM
Differential input stage	Dynsdale	Tubes / Valves	8	3rd September 2005 07:29 PM
differential input stage	xuesong	Solid State	3	27th December 2004 02:08 PM

New To Site?	Need Help?
Register to Participate Search Privacy Statement Contact Us	Frequently Asked Questions Did you forget your password? Mark Forums Read


	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

14th June 2009, 12:50 AM	#2
sawreyrw diyAudio Member Join Date: Apr 2006 Location: Minnesota	It doesn't matter what "polarity" the input voltage is, because the closed loop circuit causes the voltage at the base of Q2 to be whatever it needs to be to maintain the output at the correct level. In other words, Q1 and Q4 will still conduct as needed to achieve the correct output voltage.

14th June 2009, 01:42 AM	#4
Bitrex diyAudio Member Join Date: Jan 2009	I think I see now. There's feedback at DC into the inverting input of the differential pair Q2, so that DC voltage is coupled into the input transistor through the common connection. So when the input voltage rises from the source signal the output voltage rise is applied as a "bias" voltage to Q1, and the bias voltage follows the input voltage along keeping the transistors in conduction. I think I have that part down now. What ensures that the output stays biased at 0 volts though, and not some other value? If there's a good reference online that someone can throw me that describes the operation of this topology in detail, I'd very much appreciate it.

14th June 2009, 02:26 AM	#5
Rafael L diyAudio Member Join Date: Nov 2008 Location: Brazil	Look http://ifile.it/dk3ytpv

14th June 2009, 03:51 AM	#6
Bitrex diyAudio Member Join Date: Jan 2009	Thanks Rafael, that's just the kind of reference I've been searching for.

Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)