i*******a
发帖数: 13 | 1 由于 Bias Light 会产生很大的电流使大多数 TIA 饱和，哪家公司的TIA具有 >=5A 的
能力？Tx. |
c*u
发帖数: 916 | 2 为什么不自己做一个简单电路。
【在 i*******a 的大作中提到】
: 由于 Bias Light 会产生很大的电流使大多数 TIA 饱和，哪家公司的TIA具有 >=5A 的
: 能力？Tx.
|
f*****0
发帖数: 489 | 3 first of all, why does it have to be a transimpedance amp?
2ndly, what is the problem you want to solve? |
i*******a
发帖数: 13 | 4 I am going to test the quantum efficiency of solar cells using chopped
monochromatic light (a small AC current in the mA range) and bias light (a
large DC current >1 Ampere). Such a large DC current will saturate most TIA.
Any current-to-voltage converter can be used? |
f*****0
发帖数: 489 | 5
light (a
most TIA.
I am still unsure as to why you need a transimpedance amplifier.
I believe we call that a "resistor".
【在 i*******a 的大作中提到】
: I am going to test the quantum efficiency of solar cells using chopped
: monochromatic light (a small AC current in the mA range) and bias light (a
: large DC current >1 Ampere). Such a large DC current will saturate most TIA.
: Any current-to-voltage converter can be used?
|
y*******w
发帖数: 5917 | 6 so basically you want to measure mA AC current amid ~A DC current. The
device you need is a high pass filter. Depends on your chopper frequency, a
simple RC circuit may be enough. mA is huge current, any TIA will do. I
often need to detect ~100pA current at 1MHz, so jealous you have such an
easy project.
Curious, are you EE major? This sounds like a EE 101 problem.
TIA.
【在 i*******a 的大作中提到】
: I am going to test the quantum efficiency of solar cells using chopped
: monochromatic light (a small AC current in the mA range) and bias light (a
: large DC current >1 Ampere). Such a large DC current will saturate most TIA.
: Any current-to-voltage converter can be used?
|
i*******a
发帖数: 13 | |
i*******a
发帖数: 13 | 8 typo: it is micro-A instead of mA. It seems to be an big issue in PV
industry. EE101 may over-simplify the reality such as noise pickup. |
f*****0
发帖数: 489 | 9
ma=10-3 amp;
ua=10-6 amp;
pa=...
but that doesn't matter for this discussion. your original question is
poorly framed. sounds like you want to amplify a weak a/c current signal
burried in a strong DC signal into a voltage output.
to convert a current signal into a voltage output doesn't require a
transimpedance amplifier.
then, you have the issue of separating the a/c signal from the dc
signal. there are many ways to solve that: a capacitor; a transformer; a
dc servo; or even digitizing the inp
【在 i*******a 的大作中提到】
: typo: it is micro-A instead of mA. It seems to be an big issue in PV
: industry. EE101 may over-simplify the reality such as noise pickup.
|
y*******w
发帖数: 5917 | 10 我觉得LZ可能碰到的问题是DC光太强，所以DC光强引起的shot noise可能掩盖了比DC弱
百万倍以上的AC信号，尤其是宽频下。还有就是1/f noise
假设DC是1A，那么其shot noise是6e-10*sqrt(Hz)A,如果带宽1MHz，那么光shot noise
就有0.6uA。所以LZ可以采取以下措施：
1。增加AC量，如果AC只是用作small signal来测量某bias下的效率，那么一般来说
small signal没有必要比bias signal低百万倍。如果只是低千倍，问题显然很容易解
决。
2。加快Chopper，尽量使载频远离DC，然后设计一个比较高级数的高通滤波器，尽量把
DC给除掉。比如一个三级高通，3dB 1kHz，很容易把60Hz（灯泡）以下的信号衰减一百
万倍以上。显然，chopper越快，滤波越容易。
3。高通之后，再在TIA后面使用Lock-in amplifier，一般来说chopper总是和lock-in
amplifier一起使用的，如果你只是测量一个数值，你可以integrate非常长的时间，这
样，甚至pA都可以探测到，不用说区
【在 f*****0 的大作中提到】
:
: ma=10-3 amp;
: ua=10-6 amp;
: pa=...
: but that doesn't matter for this discussion. your original question is
: poorly framed. sounds like you want to amplify a weak a/c current signal
: burried in a strong DC signal into a voltage output.
: to convert a current signal into a voltage output doesn't require a
: transimpedance amplifier.
: then, you have the issue of separating the a/c signal from the dc
|
f*****0
发帖数: 489 | 11
looks like I really f%#&*d you up for good, :). |
n***r
发帖数: 25 | 12 youfellow made very good points here. improve chopper frequency to move
signal away from flicker noise and use lockin amplifier to detect signal.
I guess LZ's major problem is: even TIA is saturated, the signal is still
too weak.
If Idc=1A, Iac=1uA, with R=10ohm Vdc=10V, Vac=10uV.
Vac is easily burried in the noise floor. You may use Lock-in Amplifier to
get some signal out, you wont get great SNR.
So the solution is to seperate Idc from Iac before converting into Voltage.
Of course you can't f
【在 y*******w 的大作中提到】
: 我觉得LZ可能碰到的问题是DC光太强，所以DC光强引起的shot noise可能掩盖了比DC弱
: 百万倍以上的AC信号，尤其是宽频下。还有就是1/f noise
: 假设DC是1A，那么其shot noise是6e-10*sqrt(Hz)A,如果带宽1MHz，那么光shot noise
: 就有0.6uA。所以LZ可以采取以下措施：
: 1。增加AC量，如果AC只是用作small signal来测量某bias下的效率，那么一般来说
: small signal没有必要比bias signal低百万倍。如果只是低千倍，问题显然很容易解
: 决。
: 2。加快Chopper，尽量使载频远离DC，然后设计一个比较高级数的高通滤波器，尽量把
: DC给除掉。比如一个三级高通，3dB 1kHz，很容易把60Hz（灯泡）以下的信号衰减一百
: 万倍以上。显然，chopper越快，滤波越容易。
|
i*******a
发帖数: 13 | 13 Thanks. Yes, that is what I want.I appreciate if you can find some paper for
me. |
