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If you connect two transistors in parallel then each needs an emitter resistor so that they share the current fairly equally.
The opamp will probably not have enough output current to drive output transistors without adding a driver transistor.

I sketched a circuit with a PNP driver transistor and a PNP output transistor. They have enough current gain for the opamp to be able to drive.
BUT since they are both emitter-followers with a Vbe of about 0.7V and 1V then the output voltage will not go any lower than about +1.5V to 2V.

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Your amplifier will oscillate because its transistors provide voltage gain. Then at a frequency where the opamp's phase shift would cause oscillation, its gain is less than 1 so it does not oscillate, but the total gain with the transistors is higher than 1 so the entire amplifier will oscillate when negative feedback is applied (the phase shift causes the negative feedback to become positive feedback at high frequencies).

Also, common emitter transistors have more phase shift than emitter-followers due to the Miller effect (the capacitance from collector to base is amplified).

Also, your PNP transistors are missing resistors that will turn them off quickly. So yours turn off slowly which adds additional phase shift.

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So, which circuit you finnaly suggest for me with two power transistor if it's need?

Your circuits with an opamp plus common-emitter transistors have too much voltage gain so they will probably oscillate.
I showed two circuits with emitter-follower driver and output transistors that will not oscillate.
But if you use PNP transistors the output voltage will not be able to go near 0V unless the circuit uses an additional negative supply.

I showed a circuit with NPN emitter-followers that can go down to 0V if the inputs and output of the opamp can, without an additional negative supply. Add a second output transistor and emitter resistors for both output transistors and it will work perfectly.
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Is this ok?

It will work well if the driver transistor has a fairly wide bandwidth like a BD139.
In this forum there is a 0V to 30V at a few mA to 3A power supply. The improved version has two 2N3055 power transistors in parallel with emitter resistors and a driver transistor. When a slow TIP31 driver transistor was used the transient response showed bad overshoots and ringing. The BD139 fixed it. The original version used a very old maybe obsolete driver transistor that was fast.

I corrected and made notes on your circuit:

post-1706-14279144906364_thumb.png

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TNX for suggestions, i use 24V Vcc supply voltage in my current version so i must calculate R2 for proper voltage drop on it, i think so.
Should i use MJ15003 NPN output power transistors in combination with some kind of BD139 transistor?
MJ15003 transistor has a small thermal resistance factor so it can decrease heat dissipation to heatsink.

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i must calculate R2 for proper voltage drop on it

You cannot change the voltage drop of R2 by changing its value. R2 quickly discharges the base-emitter capacitance of the output transistor and discharges its leakage current. 1k ohms is fine.

Should i use MJ15003 NPN output power transistors in combination with some kind of BD139 transistor?
MJ15003 transistor has a small thermal resistance factor so it can decrease heat dissipation to heatsink.

A small thermal resistance allows a power transistor to increase its transfer of the heat to the heatsink. Then the transistor will be cooler and the heatsink will be warmer.
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If i use just one power transistor for max. 10A discharched current i should use emmiter resistor also for voltage drop on it, i think that 0.22 or 0.33E is ok.
What about noise from negative feedback of OP, i think that its better to use capacitor between negative pole of OP and ground?

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If i use just one power transistor for max. 10A discharge current i should use emitter resistor also for voltage drop on it, i think that 0.22 or 0.33E is ok.

No. An emitter resistor is not needed for only one output transistor.
Are you discharging a 12V battery? My circuit with an NPN driver transistor and NPN output transistor is a battery charger, not a discharger.
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Instead of power supply we use battery for supply circuit and it stars discharged over load resistor and npn transistor.
Because of adjustable pwm voltage we regulate fixed discharged current, I think that this circuit is ok with npn follower circuit and transistors mentioned above.

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I want to use circuit above but instead of supply on power transistor there is battery which discharges, for charge it we use power supply voltage, so different orientation of circuit.
Is there any calculation for additional RC pole, in my current circuit remains some ripple with one single rc pole (470E + 1u capacitor).
I have to see in my program for frequency I forget it.

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No. An emitter resistor is not needed for only one output transistor.
Are you discharging a 12V battery? My circuit with an NPN driver transistor and NPN output transistor is a battery charger, not a discharger.


sorry but I will very grateful if you give me advice about discharge circuit, I know that my version will wok but probably you have some critical comment, you have more experience!
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  • 6 months later...

No. An emitter resistor is not needed for only one output transistor.


Are you discharging a 12V battery? My circuit with an NPN driver transistor and NPN output transistor is a battery

charger

, not a discharger.

So, what is difference between charger and discharger, my idea was in switching relays which reconnect lines and provide both orientation??

What do you think about that?

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This thread has been for such a long time that most schematics are gone. I do not know what you want to do.

A charger puts current into a battery from a voltage higher than the battery voltage. The current is limited or regulated.

A battery is discharged by a load that draws current out of the battery. The load current is limited.

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This thread has been for such a long time that most schematics are gone. I do not know what you want to do.

A charger puts current into a battery from a voltage higher than the battery voltage. The current is limited or regulated.

A battery is discharged by a load that draws current out of the battery. The load current is limited.

Yes, thanks for all your help, I just was afraid that discharging doesn't work in that kind of circuit orientation but such design work correctly for now, I mean you proposal NPN follower.

 

 

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  • 1 month later...

This is circuit for my battery discharge but there is few problems which appear during test, at first when relay is off (two switches on battery) current flows via bd139 because the power transistors is forward biased there is maybe other solution for this for example another switch or SLT.

I don't know why OP folower doesn't work properly any more, if i put on OP + input pin zero voltage from MCU, is should be the same value on - input pin of OP but it isn't, there is always some value, in last case was 0,6V so 1,2A current was flow through BD139 and power transistors.

If you have any solution please let me know and one more time thanks for all your help!

schemeit-project (3).png

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If you have 0V on the (+) input pin of the LM324 opamp and its (-) input is at a positive voltage of about 0.6V then its output should be as low as it can go which is about 0.01V so thall the transistors should be turned off.

 

If the voltage at the 0.5 ohm resistor is 0.6V then the bases of the output transistors should be about 0.86V and the base of the BD139 should be about 1.6V. Then the input should be 0.6V. When the input voltage is less then the voltage at the 0.5 ohm resistor should be the same. 

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If you have 0V on the (+) input pin of the LM324 opamp and its (-) input is at a positive voltage of about 0.6V then its output should be as low as it can go which is about 0.01V so thall the transistors should be turned off.

 

If the voltage at the 0.5 ohm resistor is 0.6V then the bases of the output transistors should be about 0.86V and the base of the BD139 should be about 1.6V. Then the input should be 0.6V. When the input voltage is less then the voltage at the 0.5 ohm resistor should be the same.

Yes, but if I put 0V on + input pin of OP, on - pin should be 0V or not?

So on power resistor 0,5E should be 0V and no current should flow through battery, why in my case flow around 1A through BD139 when I connect power supply on circuit so 24V??

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