A Solid State Photodiode Gamma Radiation Detector

Photodiodes convert light into current and this current can be coverted into voltage and amplified. Sounds simple right? Well, when trying to detect Gamma photons, the design of a photodiode detector is not so simple. This circuit  is not paticularly complicated, but the design took some effort and many of the component values are critical.

Image of unshielded detector



Video Demo:

This circuit behaves like a classic Geiger counter, but unlike a traditional Geiger Counter you can actually measure the output pulses over time to roughly determine average energy of the gamma photons.  With the addition of a microcontroller, an A/D converter  and LCD – you can display all kinds of interesting data. Later, I plan to add this to my design. If you build this detector – you will need a source of radiation for testing. A cheap smoke detector can provide this (Americium-241) or  import gas lantern mantles which have thorium in them. Check out ebay.

When designing such a detector, the first issue to resolve is the selection of the photodiode. The  important features are: the diode area, leakage current(dark current), and capacitance.You want the capacitance to be low as possible, the diode area large as possible and the reverse bias leakage current as small as possible.  Unfortunately, the larger the diode area, the worse the leakage and capacitance become. A large diode is desirable to provide more area for photons to strike. Gamma photons create a transient current charge on the photodiode, so if the capacitance is to large, the photon strike transient  is absorbed and not detected. By reverse biasing the photodiode, the depletion region is increased and the capacitance is significantly reduced. The reverse biased mode of operation is called photo-conductive mode. However, the reverse bias creates leakage current which generates noise. The gamma photons produce very small charges and the output is very close to the noise floor; so care must be taken to minimise all noise.  Having said this, reducing capacitance appears to be more beneficial than the noise of the dark current is detrimental. This is because you need the frequency response to capture the transient of a gamma photon event. Using the highest bias possible is generally preferable. Not only do the gamma strikes produce higher output, but the stability of the amplifier circuit is improved allowing for more charge gain.

The photodiode I am using is very large. It is 10mm x 10mm and has 80 pF of capacitance under bias and about 2-4 nA  of leakage current. It is the PS100-7-CER-PIN  the performance is excellent but it is very expensive. A much lower cost alternative photodiode is the BPW34.  However, it is less than 1/10 the  area and will require closer proximity to obtain photon strikes. On the bright side, its capacitance is much lower.

The first stage of  amplification for the  photodiode is a transimpedance amplifier. Which is a current to voltage amplifier. The impedance of the photodiode is incredibly high requiring an OP amp with input bias currents of picoamps or less (minimizes loading of the photodiode) and bandwidth greater than 1Mhz as the photon strikes generate pulses from 100KHZ to 50Khz or so. The noise figure of the op amp needs to be very low. Also the OP amp should have low input capacitance so as not to create further high frequency loss. The OP amp I chose was the LMP7721 which has a bandwidth of 17 mHz, an input bias current of 3 femtoamps(amazing) and an input capacitance of about 8pF. Other amplifiers will work. Another good choice is the LM6211 which has even less input capacitance and picoamp bias current. Both of these amps have very good current and voltage noise figure specs.

The feedback resistor R4 is gigantic at 47 Meg and with all of this gain in tandem with the input capacitance of the photodiode, the whole thing will become an oscillator so a compensation capacitor is required across the feedback resistor. This capacitor value can be calculated by means of the formula:

Ccomp = 1+sqrt(1+4Pi*Rf*Cin*unity gain bandwidth)/(2Pi*Rf*unity gain bandwidth).

Ignoring the two “+1’s” :

Ccomp = sqrt(4Pi*47M*80Pf*17MHz)/(2Pi*47M*17MHz)

In my case, the Cin is 80pF, unity gain freg is 17Mhz and Rf 47Meg. When you do the calculation you will end up with .2 pF which is so tiny, it is not practical. To solve this, I use the resistor network formed by R5 and R6 to allow for the use of larger value capacitors. As long as the resistors R5 and R6 are much smaller than Rf, the effective capacitor value is reduced by the resistor ratio. The value of the compensation capacitor is critical. If it is to large gain is reduced and of course if it is to small, the amplifier will be unstable.

The second amplifier stage is the first stage of a LM358. It has a low frequency roll off at about 20 Khz to eliminate 60Hz hum and other low frequency noise. This is  acomplished by the first order filter formed by C9 and R13. The second stage of the LM358 is a comparator with  capacitor C12 added to beef up the pulses for better flashes and or clicks from a small speaker or piezo transducer.

As shown in the schematic, the photodiode and first stage of amplification, need to be shielded with foil to eliminate light and electrical noise. Because of this, it can only detect gamma rays  or cosmic radiation , which is super charged alpha particles.

For fun, lets discuss another way to think of this detector:

In this case Cf is  .5pF (.22pF + stray) and Rf is R4(49meg). The R feedback value is large so that it does not load down the photodiode but still provides feedback such that voltage at the input node stays at zero and the charge is forced onto the capacitor Cf by the feedback amplifier. It is not large to set gain in the traditional sense.  If you look  at this circuit as just a simple inverting amplifier, my point is easier to understand. For voltage gain, the (feedback resistor/ source impedance) would be the voltage gain of this inverting OP Amp topology.  Here, the photodiode source impedance can be 100 megaohms up to  gigaohms. So a feedRf of 10 meg or 47 meg doesn’t make much difference.  The transformation of charge to voltage is often refered to as sensitivity rather than gain. The units would be mV/MeV (millivolts per mega electronvolts). This is a unsual amplifier in that we are trying to convert a discrete number of collected electrons into a voltage pulse.

A different way to think of this circuit is as a charge amplifier, where a gamma photon strike creates pool of electrons on the photodiode.  From this, we can derive Gamma sensitivity. To do this, we need to know the energy of a gamma photon, the energy required for silicon to break an electron/hole pair (temp sensitive and approximately 3.62 eV at room temp), our feedback capacitance in the amplifier(about .5pF) and the elementary charge of an electron(1.6×10^-19).

To start with: Voltage = Q/C where Q = number of electrons (charge) and C is the Cf in our circuit.

Vout(amplifier) = Qin/Cf(which is effectively integrating all of the charge to voltage at the output).

For example: Americium-241 generates a 59Kev gamma pulse, therefore 59Kev/3.62eV = number of electrons = 16.3K electrons per gamma strike. This assumes all of the possible electrons and holes  are formed , which is an ideal assumption and not realized.

Now to find Q in coulombs, we take 16.3k * 1.6×10^-19  =  2.61 ^-15 coulombs

Output Voltage = Q/Cf(C1) = 2.61 ^-15/.5 pF = 5mV

In my circuit, I amplify this output 100 X so I would expect to see an output of .5v . This is an ideal result and does not take into account stray capacitance, limited amplifier open loop gain, filter losses, and imperfect conversion of electrons in the diode, etc.

When I am lucky I see about 200 mV output from an Americium-241 source. On average, an Americium-241 source will generate a 150mV pulse.  When this circuit is working correctly, there will be about 50mV to 60mV of noise. The practical detection threshold is about 100 mV.


19 thoughts on “A Solid State Photodiode Gamma Radiation Detector

  1. Hello, I have examined your work and I should admit that you did a great job. I am also working on a same kind of circuit and Ihave some problems. Can you help me? I would be very appreciate if you can contact me via e-mail .I really need help may be we can work together. Waiting for your reply. See you

  2. What about using an ADL5304 logamp? I worked on this part and it seems a lot easier to just plug in a photodiode to the front end of the log amp and then buffer the output into your A/D.

    • Hey there

      I appreciate the suggestion. It may work fine. The main issue is maximizing the signal to noise ratio with regard to detecting weak photons….. not so concerned with the benefits of a log response… otherwise the amplifier performance may be very good. The part is very expensive -something like $20 bucks from digikey and even though the implementation is not complicated, this is a 32 pin part with lots of connections so I don’t know that it is an easier implementation. Finally, you need a higher reverse bias than shown using the bias pin the provide. This is required for fast response time to detect the very short pulses. Not saying its a bad idea, it just doesn’t seem better to me (over all). Looks like a great part to use where absolute accuracy is a design requirement.

    • I don’t think it matters – the bias point is AC ground so it is inherently a low Z point in the circuit any way. The fact that the diode is reverse biased requires that the bias only provide leakage current which is very tiny – so yes you can use a 10 meg resistor – but a 1K will work fine also. The reverse dark current is no different either way

  3. Hey, thanks for sharing your work. I have a question if you don’t mind.
    As i understand you don’t use a scintillator, just a pin-diode alone. I’ve read that you can’t detect gamma-quants above 50 keV of energy without a scintillator. This is because compton scattering starts to dominate over a photoeffect at those enrgies. No photoeffect hence no current in the diode. Do you agree with that? Therefore is it not possible to detect 50 keV gamma-rays using your schematic? But i see you claiming that Americium-241 generated a 59Kev gamma pulse. What do you mean by 59 keV? Overall energy or the energy of a quantum?

    • Yes I have read this also, but every gamma source I have tried has worked well, radium, uranium, Americium and Thorium and they generate significantly higher energy photons than 50keV. The 59 KeV is the the energy of the gamma photon generated from Americium decay.

      I am not an expert on semiconductor physics or nuclear physics but I can say experimentally, the circuit seems to work well with a variety of different gamma sources.

      What I am trying to say..is that my experimental observations disagree with this statement.

    • Improved my semiconductors alpha ray and beta ray sensor by using one 2N 3055 cut open for alpha rays, and put this in series with 8 BPW 34’s all soldered in parallel.
      By putting in series, one preamp is used for all 9 sensors providing a wider target for beta and an open target for alpha rays.
      More noise is present after the audio amplifier with the positive result of hearing radioactivity coming closer to a mica window tube’s capability without the high voltage, which could cause an explosion in a explosive environment.

  4. Hi raycharlesring,
    Thank you for sharing. I have a question. what material you use to shield detector?
    I hope you reply.

  5. Yes, scintillators from old X-ray cassettes (thanks Stella!) work.
    I used a dedicated AP diode here but any large area PD with a lens which can be DIY’d using UV cure nail varnish will work fine, good source of these is really old pre-2000 photoelectric smoke detectors.
    Pair of lenses in a module will do the trick, you’ll lose a bit of efficiency if it isn’t focused correctly though.
    Maybe better off using a silicon solar cell shard, most of them with edges insulated will happily take >50V reverse bias.
    Some folks have used large area red LEDs as well, in this case you want to choose one which has a detector threshold as close to the specified scintillator peak as possible (541nm)

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