Hello there, when I have doubt, I always come back to diyaudio and get great support.
I am building a new 100 W power amp, with matched transistors at input stage, driver stage and output stage, and match resistors, capacitor value also.
My simple question at the finish stage is this,
How do I arrange the power transistors's emitter resistors?
If in each channel, the amp use two pair of output power transistors, what will be a better arrangement use two 3W/5W emitter resistors?
Case 1: Put 0.22 ohm, 0.22 ohm on the positive voltage side, and 0.24 ohm, 0.24 ohm on the negative side , or Case 2: Put 0.22 ohm, 0.24 ohm on either side? Assuming all the NPN power transistors are matched in HFE, the PNP outputs are matched, and the difference between NPN and PNP is well matched in 2 % HFE difference (Vce 30V, Ic 1A)
Actually on the board, it designs to use three pairs bi-polar power transistors. The rule should be the same. Present arrangement 0.2234, 0.2239, 0.2237 in NPN, 0.2240,0.2240,0.2240 ohm in PNP group for right channel. And in left channel I am arranging 0.2241,0.2242, 0.2242 in N side, 0.2243, 0.2243, 0.2244 ohms in PNP side , they are KOA PPR58 J power resistors, not common cement resistors.Surprisingly close matched resistor value, better than 1%.
I am building a new 100 W power amp, with matched transistors at input stage, driver stage and output stage, and match resistors, capacitor value also.
My simple question at the finish stage is this,
How do I arrange the power transistors's emitter resistors?
If in each channel, the amp use two pair of output power transistors, what will be a better arrangement use two 3W/5W emitter resistors?
Case 1: Put 0.22 ohm, 0.22 ohm on the positive voltage side, and 0.24 ohm, 0.24 ohm on the negative side , or Case 2: Put 0.22 ohm, 0.24 ohm on either side? Assuming all the NPN power transistors are matched in HFE, the PNP outputs are matched, and the difference between NPN and PNP is well matched in 2 % HFE difference (Vce 30V, Ic 1A)
Actually on the board, it designs to use three pairs bi-polar power transistors. The rule should be the same. Present arrangement 0.2234, 0.2239, 0.2237 in NPN, 0.2240,0.2240,0.2240 ohm in PNP group for right channel. And in left channel I am arranging 0.2241,0.2242, 0.2242 in N side, 0.2243, 0.2243, 0.2244 ohms in PNP side , they are KOA PPR58 J power resistors, not common cement resistors.Surprisingly close matched resistor value, better than 1%.
When you have multiple pair of output devices it's best to have matching emitter resistors between the devices that are sharing the load. This helps to prevent current hogging by one device. Instead of trying to measure the actual resistance of each resistor, connect them all in series and pass some current through them (~1A), then measure the voltage drop. This will likely give you a more accurate match.
Thanks Jwilhelm, it's best to have matching emitter resistors between the devices that are sharing the load, Is it better in case 1 or case 2? Which case shares the load?
To make more clear, We have PowerTransister 1 with Re1, PT2 with Re2
for the positive volts, PT 3 with Re3 and PT4 with Re4 for the negative volts.
Should we arrange Re1 0.22ohm, Re2 0.22ohm, and set Re3 0.24ohm, Re4 0.24ohm (case 1)
or we should average out and arrange Re1 0.22 ohm, Re2 0.24 ohm, and arrange Re3 0.22ohm, Re4 0.24ohm (case 2)??? That confuse me a lot, assume PT is matched well.
To make more clear, We have PowerTransister 1 with Re1, PT2 with Re2
for the positive volts, PT 3 with Re3 and PT4 with Re4 for the negative volts.
Should we arrange Re1 0.22ohm, Re2 0.22ohm, and set Re3 0.24ohm, Re4 0.24ohm (case 1)
or we should average out and arrange Re1 0.22 ohm, Re2 0.24 ohm, and arrange Re3 0.22ohm, Re4 0.24ohm (case 2)??? That confuse me a lot, assume PT is matched well.
Use equal emitter resistor for every output devices. If You want better sharing, use higher value (0,33 of 0,47ohms), but use equal value for every transistor.
Sajti
Sajti
Case 1 would be better than case 2. The goal is to have all the output transistors fed form the positive rail sharing equally and all of the output transistors fed from the negative rail sharing equally. It's not ideal that the positive and negative emitter resistors aren't perfectly matched but the negative feedback circuit will remove any offset caused by this imbalance, assuming this is a conventional design amplifier.
Case 2 will be better than Case 1. This is because matching positive and negative phase is more important than current sharing within one phase.
Matching positive and negative phase is irrelevant if an output device burns up from current hogging.
Hi Fjc,
Where does an idea to use 0.22R on the positive side and 0.24R on the negative side come from?
Normally, you would want to use equal values - e.g. 0.22R - for all the output devices.
Cheers,
Valery
Where does an idea to use 0.22R on the positive side and 0.24R on the negative side come from?
Normally, you would want to use equal values - e.g. 0.22R - for all the output devices.
Cheers,
Valery
I assumed he took actual measurements of parts he has on hand, but this may not be the case.
I did some simulation on my V4 amplifier.
All 0R22: THD is 0.000571%
Case 1: THD 0.000571%
Case2: THD 0.000517%
Sajti
All 0R22: THD is 0.000571%
Case 1: THD 0.000571%
Case2: THD 0.000517%
Sajti
Out of curiosity, what does the simulation say if you have only one output pair connected?
Well, in any case, option 2 would be better. Even though the idle current distribution (as well as the power dissipation distribution with the signal on) between the pairs will alter a bit - the difference is around 4% - nothing to worry about. "Immaterial".
I thought this would be lower due to less devices adding to the noise. I guess this would depend on output power levels, ect.
This would depend on the heat sink. In my testing on non-ideal heat sinks the temperature really deviates between devices when you load them down. With our normal DIY oversized heat sinks it's much less of an issue though.
Less than 10% difference in resistor value means "hogging" is a somewhat excessive description. However, 10% difference could mean 5% 2nd order distortion so more work for the feedback loop.jwilhelm said:
ballast resistor optimization.
Hi - The exact pairing of the resistors versus transistor beta is not critical. Keep in mind that the voltage follower power stage has 100% voltage feedback
that will tend to equalize current sharing. A close match of the driver input resistors and maintaining concurrent temperatures in the output transistors should help.
As others have suggested, equal values in the upper and lower drivers is current practice and increasing the resistance values upward to .33 or .47 ohms will provide better tracking.
The idle current for CFP (Sziklai) is much lower than EF or QEF pairs. 🙂
Hi - The exact pairing of the resistors versus transistor beta is not critical. Keep in mind that the voltage follower power stage has 100% voltage feedback
that will tend to equalize current sharing. A close match of the driver input resistors and maintaining concurrent temperatures in the output transistors should help.
As others have suggested, equal values in the upper and lower drivers is current practice and increasing the resistance values upward to .33 or .47 ohms will provide better tracking.
The idle current for CFP (Sziklai) is much lower than EF or QEF pairs. 🙂
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