Looking for a BJT with low collector leakage current

In summary, the problem is that the transistor is turning on too early and causing an output offset. The best way to solve the problem is to find a transistor with the highest Vbe.
  • #1
yungman
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I am looking for small signal BJT that has very low leakage current when the base is biased at about 0.2V. I have issue of transistors drawing collector current when the base is at about 0.25V that is way less than the normal turn on of 0.6V to 0.7V. I am currently using BC546 or 556, and KSA992 or KSC1831 transistors, they leak too much. I need transistor with as low collect capacitance as possible ( low Ccb and Cce).

BTW, how do I look for this on the data sheet? Looking at leakage at Vceo?

Thanks
 
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  • #2
Many bipolar transistors leakage are limited by perimeter-dependent leakage. Properly passivating perimeter is expensive (because cannot be done on wafer level) yet low-benefit procedure for general applications, therefore rare. Mostly perimeter passivation is used for high-temperature transistors (rated for 200C or such) - you should search among them.
2N2484 have leakage in nA range.
 
  • #3
How much current are we talking? The base-emitter junction does not suddenly turn on at 0.6-0.7 volts. The current increases exponentially starting at Vbe=0, and increases 10X for every 60 mV increase in Vbe. See the attached curve, which is called a "Gummel curve". Ib is in red and Ic is in blue. The collector current is then beta times the base current. If you look at the attached curve, you will see it is flowing about 100 pA of Vce current at Vbe=0.25V. But this depends on a lot of things, including the size of the transistor, the temperature, and Vce. Bottom line, you should expect it to leak in this configuration, because you have begun to turn it on.

Gummel.png
 

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  • #4
If that 0.2V on the base is not required, then an alternative drive approach for a Common Emitter stage would be a a B-E resistor, then feed the Base signal in thru a diode, or a Common Base stage, or an Emitter Follower.

And I will repeat the question from @phyzguy, how much leakage current are you currently getting?... and other than Zero, what leakage are you after?

Schematic details around the stage would likely generate better answers than does asking for help searching datasheets.

Cheers,
Tom
 
  • #5
Switching circuit turning on too early ?

Have you tried a Darlington .?

old jim
 
  • #6
Thanks guys. The application is for current limiting as shown in the schematic circled in red. When the current is too high, the clamping transistor will turn on and limit the current. But I notice these transistors cause some offset at the output. In my design, I already use a voltage divider to lower the voltage across the base emitter and limit it to about 0.25V, it is still slightly turning on.

Today, I experimented on a proto board by putting a 1M resistor on the collector and put 0.25V across the base emitter. I did not measure any voltage across the 1M resistor. ( I first measure from the collector to ground and measure 1V drop across the 1M resistor by subtracting the collector voltage from the Vcc. But then I realize the input impedance of the DVM is finite and form a voltage divider, so I measured across the 1M resistor instead). So it doesn't seems like the transistor has enough leaking to cause the problem. But it does seems to affect the output offset.
 

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  • #7
Try a 1N34 Germanium diode in series with R8, along with a 47K resistor on Q5 between B & E. Duplicate at R15, Q12. Watch the diode polarities.
 
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  • #8
Tom.G said:
Try a 1N34 Germanium diode in series with R8, along with a 47K resistor on Q5 between B & E. Duplicate at R15, Q12. Watch the diode polarities.
Thanks for the reply. In my design, I already have voltage divider as shown. I can just as easy to lower the resistor values of R74 and R122 to lower the voltage. The problem is then the current limiting is set to a much higher value unless I increase the value of R74 and R122 to compensate. But then I change the open loop gain of the amp. It's a lot of things in play not to mention I already have the pcb. I think the best way is to find a transistor with the highest Vbe. What do you think.

BTW, this is not homework. I am designing and build hifi amps. It is a working circuit, I just looking into the offset issue.
 

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  • #9
yungman said:
I think the best way is to find a transistor with the highest Vbe

upload_2019-2-15_7-28-22.png


well,, now it's ONSEMI... sorry
 

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  • #10
Tom.G said:
Try a 1N34 Germanium diode in series with R8, along with a 47K resistor on Q5 between B & E. Duplicate at R15, Q12. Watch the diode polarities.
Actually I was playing with the values of R74 and R78 in my own schematic in post #8, it will work out. In the existing circuit, the idle current through Q44 is about 17mA, when it goes up to about 40mA, Q41 will start to turn on and limit the current. Say if I increase R74 to 20 ohm and R78 reduce to 30 ohm, I can keep the open loop gain the same without disturbing the poles and zeros. At 17mA idle current, voltage drop across R74 is 0.34V. I would use a schottky diode instead of 1N34. The voltage across the base emitter of Q41 is very close to zero. But if the current of Q44 increase to 30mA, Q41 will start to turn on and limit the current.

It's too late for this run of pcb. I am using SMD components, I looked at the layout, it's hard to modify it and keep it sturdy. There is no room for through hole components. For now, either I find a better transistor, or literally remove the Q41 ( same on the other side) and call it a day. Most amps don't even have this protection anyway. Only time it happens is when something in the following stage burn and clamp to ground or to the other voltage rail. Put it in this way, if the output stage burn, burning one more transistor is not the end of the world!

There goes to show just because the circuit is used in other production amps does not mean it is correct. There are a few amps that have this circuit like the schematic of Acurus I posted in post #6. That one is even worst, at least when I copy the circuit, I put R74 and R78 divider to lower the voltage across the base emitter of Q41. Those amps must either have worst leakage problem or they run at much lower current and thereby have much lower frequency response than my amp. ( this affect the slew rate of the amp). Acurus is not exactly a cheap amp, I paid over $1000 at the time already.

Thanks
 
Last edited:
  • #11
jim hardy said:
View attachment 238789

well,, now it's ONSEMI... sorry
I need to do some calculation first. Thanks

Alan
 
  • #12
I think so far, the best way is by Tom.G using a schottky diode. As shown in Fig.1 attached, if the idle current is 17mA, voltage drop across R74 is 0.3V, with the 2K pull up resistor at the base of Q41, there should be no voltage across the base emitter of Q41. When the max current of 50mA, there will be 0.6V across the base emitter of Q41 and Q41 will start conducting and limit the current at about 50mA.

Sadly the darlington idea will not work, as shown in Fig.2, collector of Q1 will cause leaking just like the original circuit.
Diode current limiting.jpg


I will keep this in mind when I design and layout the next amplifier.

I am surprised nobody think about this as shown in quite a few schematic of existing amps. Not only leakage is a problem, more seriously, this form a local feedback and might change the characteristic of the stage. My amp is 0.0025% THD @20KHz at large signal, this likely does not help.

Thanks
 

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  • #14
jim hardy said:
hmmm google returns different ONSEMI datasheets for BC517 - seems to depend on time of day.
Some don't have a figure 10.

That one is https://www.onsemi.com/pub/Collateral/BC517-D.PDF

no fig 10 at https://www.onsemi.com/pub/Collateral/BC517-D74Z-D.PDF
Yes, without the emitter resistor is definitely better, but the problem is still there. With darlington, there will be 2Vbe drop, meaning I have to adjust the resistor divider to bias at higher voltage. This will bring us right back to the original problem. For example, if I need to put 0.4V across the base emitter junction, that will be 0.2V across each base emitter of the two transistor inside the darlington, that will cause leakage.

Now, the idea will work if it is a discrete darlington ( using two separate transistors) where the collector of the first transistor can just go to the ground instead of connecting to the current limiting circuit, so the leakage of the first transistor will not affect the leakage. That will work as good as using a diode.
 
  • #15
Tom.G said:
Try a 1N34 Germanium diode in series with R8, along with a 47K resistor on Q5 between B & E. Duplicate at R15, Q12. Watch the diode polarities.

I decided to do it, I change R74 to 20ohm, that lower the idle current to about 15.3mA, keep it cooler. With 20ohm, current limit should be about 44mA. It is not too hard to modify using SMD components. At idle, voltage drop across R74 is going to be 20 X 0.0153= 0.306V, I don't think the schottky diode is going to turn on and the Vbe of Q41 is going to be 0V.

Good idea.

Thanks
 
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  • #16
I think the problem is more fundamental.

The LHS of the circuit shows two constant current generators, a source and a sink, that provide independent bias currents of 2mA to the differential pairs. That current appears across a couple of 1k resistors which set the input voltage to the limited voltage controlled current circuit you have focused on. I do not see collector leakage as a problem with that circuit.

But what happens when output voltage = input voltage? Half the input bias current, 1 mA, flows down through each 1k dropping, 1 volt. Both upper and lower currents are turned on. The current through the duplicated resistors will be about 0.5V/82R = 6 mA each = 12mA total. That 12 mA flows through the quiescent current circuit across the four 1N4148 diodes.

Current leakage through those transistors when they reduce current will be irrelevant.
 

Related to Looking for a BJT with low collector leakage current

1. What is a BJT?

A BJT, or bipolar junction transistor, is a type of semiconductor device commonly used in electronic circuits to amplify or switch signals.

2. Why is low collector leakage current important?

Low collector leakage current is important because it indicates the amount of current that can flow through the collector terminal when the transistor is supposed to be in the off state. This leakage current can cause unwanted power consumption and affect the accuracy of the circuit.

3. How can I find a BJT with low collector leakage current?

You can find a BJT with low collector leakage current by looking at the datasheet of the transistor. The datasheet will list the maximum collector leakage current and you can choose a transistor with a lower value.

4. What factors affect collector leakage current in a BJT?

The collector leakage current in a BJT can be affected by factors such as the temperature, the material used in the transistor, and the design of the transistor. Generally, transistors made with higher quality materials and with better design will have lower collector leakage current.

5. Are there any trade-offs to consider when choosing a BJT with low collector leakage current?

Yes, there are trade-offs to consider when choosing a BJT with low collector leakage current. Transistors with lower collector leakage current may have a slower switching speed or a lower maximum current rating. It is important to consider the specific requirements of your circuit when choosing a transistor.

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