AD831 Mixer help please

[SwingeyP]

Hello.

I am very new to RF design and could do with some help & advice. I am using an AD831 mixer as shown in the attached circuit.

The video shows the circuit on my development board and what I think is the two mixed signals. This is without any voltage connected to the device 9v?
When I connect a 9v battery the chip starts to get hot. I have checked and double checked the circuit. It's basically what is described on the datasheet. Why should the chip be getting hot. Also the voltage drops from 9v to 8.6v when connected.

It appears that the mixer IS doing something. When I adjust the signal generator you can see the waveform change.
Is this the kind of trace I should expect to see?
Why is the chip getting hot?

Scope settings
Sweep .2 us
Volts/Div 10mv

LO = (from DDS) 4Mhz @ 20mv
RF IN (from signal generator) 7Mhz @ 10mv

I hope this makes sense your help is MUCH appreciated.

Cheers - Paul (M0TVU)

 

[SwingeyP]

Just as an after thought. When the battery is connected the single is amplified.

 

[JimB]

Why is the chip getting hot?

The datasheet tells us that the quescent current is 100/110mA, which with a 9v supply is getting on for 1 watt.
So it will get hot.
Also your little 9v battery will be working quite hard, which is why its voltage drops a bit.

Is this the kind of trace I should expect to see?

Probably.
What you are seeing is a mess of the local oscillator, the signal and the IF output.
If you could look at it with a spectrum analyser, I think that you would see all those signals, and maybe a bit more.

If you have a general coverage receiver, have a listen to what should be the IF frequency and see what you an hear.

As your receiver develops, you are going to follow the mixer with a filter, that should remove all the unwanted stuff and just leave the IF.

Jim GM3ZMA

 

[SwingeyP]

Thanks Jim,

This is not my field of electronics at all and i'm taking it step at a time just playing and learning at the moment. (It's fun).
I can hear the LO (currently 4Mhz) and other frequencies on my receiver. I don't have a spectrum analyser (too expensive at the moment).
I plan to change the LO to 3-3.2 Mhz to mix with the 7 Mhz and feed through a crystal ladder filter of 4Mhz. Lots of experimenting with that and op amps before I decide the best path. This will feed into a MC1349P IF amplifier and then some product detector I guess. I have all the blocks in my head just never connected them before so not sure what I should be seeing at each stage.

 

[RadioRon]

Hello.

I am very new to RF design and could do with some help & advice. I am using an AD831 mixer as shown in the attached circuit.

The video shows the circuit on my development board and what I think is the two mixed signals. This is without any voltage connected to the device 9v?
When I connect a 9v battery the chip starts to get hot. I have checked and double checked the circuit. It's basically what is described on the datasheet. Why should the chip be getting hot. Also the voltage drops from 9v to 8.6v when connected.

It appears that the mixer IS doing something. When I adjust the signal generator you can see the waveform change.
Is this the kind of trace I should expect to see?
Why is the chip getting hot?

Scope settings
Sweep .2 us
Volts/Div 10mv

LO = (from DDS) 4Mhz @ 20mv
RF IN (from signal generator) 7Mhz @ 10mv

I hope this makes sense your help is MUCH appreciated.

Cheers - Paul (M0TVU)

This mixer is intentionally designed to use a fair bit of power to do its job. While it takes about 1 watt of DC power to run it, the output RF power is typically very small, so this is a very inefficient IC from a power consumption point of view (for good reason, read on).
I notice that you can make it less hot if you don't mind a minor reduction in dynamic range, simply by tying the BIAS pin 14 to VPOS through a small resistance. This BIAS pin is used to program the current drain, and they say that current consumption (and therefore heat) can be reduced by half if you tie this pin to your positive supply. I think this might be a useful thing to do while you are still playing around. Later on, if you need to increase performance, you can adjust this. In the meantime, it keeps the chip cooler which will make you feel better.

The reason such chips need so much power is because they are designed to deliver a broad bandwidth, and such design usually requires more current. The reason for this depends on what kind of transistors are used inside. Let's assume it uses bipolar transistors. The Bipolar transistor's ability to amplify high frequencies is a function of current passing through the transistor. The best bandwidth through a transistor amplifier usually requires a fair bit of current. The high frequency performance of a transistor is related to its internal junction capacitances.

Why does junction capacitance affect bandwidth? Here is a simple example. Imagine a theoretical circuit made up of a series connection of a voltage source, followed by a resistor and a capacitor (the second terminal on the capacitor is taken back to the voltage source to complete the circuit). Suppose that you are stuck with the value of capacitor and cannot change it, but you need to deliver the highest possible RF voltage across the capacitor. The only way to do that is to reduce the series resistor so that less voltage is dropped in it. How this relates to a transistor depends on what kind of transistor and what amplifier configuration is being used. If you were using a MOSFET, the gate is essentially a capacitor and you have to develop the largest possible voltage at the gate terminal (and assuming a common source amplifier). So you have to make the preceding stage have lower output resistance to achieve this, and this, in turn, usually means lower bias resistances in the circuit. That's just one example.

 

[SwingeyP]

Now THAT's an explanation.

I did check with the AD forums just to make sure I had the circuit correct and they also confirm basically what you have said here.

Many thanks. - BTW in my opinion this is the best electronics forum out there.

Cheers - Paul (M0TVU)