help with Class AB amplifier - Page 1
Forum section: electronics / projects

author: rex1232
author_rank: Contributor
timestamp: November 22, 2019, 07:10:55 pm
content: Hey guysI am in need of help with understanding the part of the circuit which is marked in the red circel thing (R9 all the way to D10)The first part by that I mean the differential amp stage (Q2,Q3) and the constant current source (Q1, D8 D9) I understand.But Q16 and Q17 looks like another diff amp to me, but is it?I have been told the part which looks like a constant current source (D10,R11,R13,Q6) is actually a current mirror.Can anyone explain what function this part has?ps. (someone told me that the top part (d10 etc) was used as a temperature regulated current source, which would work by D10 changing its voltage drop over it if the temperature changed.) Is there any truth to that?Thank u in advance

author: magic
author_rank: Super Contributor
timestamp: November 22, 2019, 08:31:06 pm
content: Swap D10 and R11 (which makes no difference to the functioning of the circuit) and you will see it's a current mirror, with 180Ω emitter degeneration. R9,Q16,Q17 are indeed a second differential stage, C3 is likely some sort of feedback from later stages.It seems Q7,R12,R14 are Vbe multiplier and the whole circled part is the VAS. Q17 pulls the output down, Q16 pulls it up through the mirror. C3 is the VAS compensation capacitor.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: November 23, 2019, 03:04:15 am
content: Hi,Audio amplifier normally have three stages:1) Input stage Q1, Q2 and Q3This is a long-tailed pair with a current source, Q1, D8, D9 in the emitter.2) Voltage amplifier stage VASThis is Q6, Q16 and Q173) Current amplifier stageQ10, Q12, Q14 and Q15There are two Darlington emitter followers, voltage gain=1Q8 and Q9 are for overcurrent protectionQ7 is vbe multiplier to compensate for vbe in the output transistor stage.In the VAS, D10, R11, R13 Q6 is a current mirror. it would look more like a current mirror if D10 was replaced with a diode connected transistor.Q16 and Q17 is another differential amplifierThe current mirror load gives higher gain than if resistors were used.Regards,Jay_Diddy_B

author: xavier60
author_rank: Super Contributor
timestamp: November 23, 2019, 05:24:26 am
content: The negative feedback path is also important to be aware of. It sets the overall gain to some practical value within the audio pass band plus some extra.As well, it stabilizes the output DC to near zero volts.

author: David Hess
author_rank: Super Contributor
timestamp: November 23, 2019, 03:41:24 pm
content: Quote from: rex1232 on November 22, 2019, 07:10:55 pmYes, Q16 and Q17 form another differential amplifier. Note that the operating point (tail current) is set by the common mode output voltage of the previous differential amplifier.QuoteYes, it is a current mirror. As magic points out, swapping D10 and R11 makes this more apparent.QuoteThat part is the VAS stage. The current mirror converts the differential output current from the differential pair into a single ended output voltage. Douglas Self refer to this configuration as the Hitachi push-pull VAS and apparently it became popular in the 1980s for driving the large gate capacitance of power MOSFET output amplifiers.QuoteD10 roughly compensates for the Vbe and temperature coefficient of Vbe of Q6. The temperature compensation is incidental and not required for proper operation in this application. Often this configuration gets away with leaving out D10. A more precision design would use the base-emitter junction of another BD140 in place of D10 but that would be more expensive for little if any gain.

author: bson
author_rank: Unknown
timestamp: November 23, 2019, 05:06:52 pm
content: Quote from: rex1232 on November 22, 2019, 07:10:55 pmYes, it's an inverting diff amp. A 2-stage differential input is common with BJT inputs due to their low gm(and hence low gain).

author: David Hess
author_rank: Super Contributor
timestamp: November 23, 2019, 06:27:42 pm
content: Quote from: bson on November 23, 2019, 05:06:52 pmFor a given current, bipolar transistors have a greater gmthan FETs. Audio amplifiers with bipolar inputs usually lower their gmfor stability reasons. This is why when a simple bipolar and JFET operational amplifier is built on the same process, the JFET operational amplifier is higher bandwidth.Common 4 stage designs with two differential stages are not particularly better performing than 3 stage designs with one differential stage. In this case the second differential stage operates as a higher output current VAS stage.

author: rex1232
author_rank: Contributor
timestamp: November 23, 2019, 06:38:39 pm
content: Does that mean that C3 is what is called a miller capacitor, in this circuit?

author: David Hess
author_rank: Super Contributor
timestamp: November 23, 2019, 06:48:51 pm
content: Quote from: rex1232 on November 23, 2019, 06:38:39 pmYes.

author: rex1232
author_rank: Contributor
timestamp: November 23, 2019, 07:33:47 pm
content: Is another function of Q7 (R12, R14, R28) not biasing?Thank you for the response

author: magic
author_rank: Super Contributor
timestamp: November 23, 2019, 07:41:03 pm
content: I don't think there is any "another function"Yes, it's the bias spreader which determines, and hopefully thermally compensates, the output stage standing current.

author: David Hess
author_rank: Super Contributor
timestamp: November 24, 2019, 02:44:56 am
content: Quote from: rex1232 on November 23, 2019, 07:33:47 pmThat is Q7's only function; Q7 is configured as a Vbe multiplier.The resistor ratio multiplies Q7's Vbe, including the roughly -2mV/C temperature coefficient of Q7's Vbe, to produce a higher voltage which in this case is about 4 x Vbe + -8mV/C from the driver and output transistors. Q7 should be mounted on the same heatsink as the output transistors so that they track in temperature.

author: bson
author_rank: Unknown
timestamp: November 24, 2019, 06:52:36 pm
content: Quote from: David Hess on November 23, 2019, 06:27:42 pmWhat current is that, the DC bias current? Why would that matter - you'd bias for linearity and range, the bias current then is what it is - within reason of course. In other words, BJTs and FETs will obviously have different bias currents.

author: rex1232
author_rank: Contributor
timestamp: November 24, 2019, 09:48:08 pm
content: Hey, the current mirrors I have seen always have 2 transistors where they are connected by the base and also to their collector.How does it work when there only is when transitor? (Q6)Does it have something to do with D10's temperature compensation and Q16, Q17 pulling the Current mirror up and down?Thank you in advance

author: magic
author_rank: Super Contributor
timestamp: November 24, 2019, 10:23:31 pm
content: The diode is an approximation of that missing transistorI'm kinda too lazy to write a whole post explaining how mirrors work, you should be able to find info on the net.Recently W2AEW advertised his video on this topic so what the heck, I will link it, his videos tend to be informative and easy to digest.I pray that he explained both the transistor and the diode versionQuote from: bson on November 24, 2019, 06:52:36 pmIn my (very limited and somewhat biased towards fancy high-spec designs) experience it's not that unusual to see BJT inputs running at a milliamp or more of current per leg, which gives each transistor a gm in excess of 30mS. That's more than most JFETs can do at full IDSS. Many BJT designs degenerate the input pair to keep gm down.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: November 24, 2019, 10:38:48 pm
content: Hi,This little sketch should help show you get from a standard current mirror to the one used in this amplifier:The circuit on the left is the standard implementation. This works best if the two transistors are at the same junction temperature. The transistor on the left is actually wired as a diode.In the centre circuit the transistor has been replaced by a diode.This still has most of the properties of a current mirror, but the mirror ratio will not be 1:1 especially over temperature.In the third circuit the positions of the diode and resistor have swapped. This doesn't change the function in any way.We are now at the circuit in the AB amplifier.Regards,Jay_Diddy_B

author: David Hess
author_rank: Super Contributor
timestamp: November 24, 2019, 10:40:07 pm
content: Quote from: rex1232 on November 24, 2019, 09:48:08 pmIn a current mirror, one of the transistors is wired as a diode with its base and collector shorted. So if close matching is not required, one of the transistors can be replaced with a diode.

author: rex1232
author_rank: Contributor
timestamp: November 25, 2019, 10:41:11 am
content: Hey guys thank you for all the helpI made this block diagram is the signal ways/connection of block to block correct?Thank you in advance

author: rex1232
author_rank: Contributor
timestamp: November 25, 2019, 12:31:17 pm
content: So basically its a cheap version of the "normal" type of current mirrorThank you for the response

author: David Hess
author_rank: Super Contributor
timestamp: November 26, 2019, 12:11:27 am
content: Quote from: rex1232 on November 25, 2019, 12:31:17 pmI suppose. If the transistors were 2N4401s then the savings from replacing one with a diode would be insignificant but with a more expensive BD140, it is a common optimization. In this case the lowered precision is not important.